https://asone.ai/polymath/api.php?action=feedcontributions&user=Charles+R+Greathouse+IV&feedformat=atomPolymath Wiki - User contributions [en]2021-12-08T16:52:00ZUser contributionsMediaWiki 1.23.5https://asone.ai/polymath/index.php?title=Sylvester%27s_sequenceSylvester's sequence2015-09-30T04:30:11Z<p>Charles R Greathouse IV: fix OEIS link</p>
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<div>'''Sylvester's sequence''' <math>a_1,a_2,a_3,\ldots</math> is defined recursively by setting <math>a_1=2</math> and <math>a_k = a_1 \ldots a_{k-1}+1</math> for all subsequent k, thus the sequence begins<br />
<br />
: 2, 3, 7, 43, 1807, 3263443, 10650056950807, 113423713055421844361000443 (sequence [http://oeis.org/A000058 A000058] in OEIS).<br />
<br />
The elements of this sequence are mutually coprime, so after factoring k of them, one is guaranteed to have at least k prime factors.<br />
<br />
There is a connection to the [[finding primes]] project: It is a result of Odoni that the number of primes less than n that can divide any one of the <math>a_k</math> is <math>O(n / \log n \log\log\log n)</math> rather than <math>O(n / \log n)</math> (the prime number theorem bound). If we then factor the first k elements of this sequence, we must get a prime of size at least <math>k\log k \log \log \log k</math> or so. <br />
<br />
It is also conjectured that this sequence is square-free; if so, <math>a_k, a_k-1</math> form a pair of square-free integers, settling a toy problem in the finding primes project.<br />
<br />
* [[wikipedia:Sylvester's_sequence|The Wikipedia entry for this sequence]]</div>Charles R Greathouse IVhttps://asone.ai/polymath/index.php?title=Moser%27s_cube_problemMoser's cube problem2015-09-30T04:26:55Z<p>Charles R Greathouse IV: fix OEIS link</p>
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<div>Define a ''Moser set'' to be a subset of <math>[3]^n</math> which does not contain any [[geometric line]], and let <math>c'_n</math> denote the size of the largest Moser set in <math>[3]^n</math>. The first few values are (see [http://oeis.org/A003142 OEIS A003142]):<br />
<br />
: <math>c'_0 = 1; c'_1 = 2; c'_2 = 6; c'_3 = 16; c'_4 = 43; c'_5 = 124; c'_6 = 353.</math><br />
<br />
Beyond this point, we only have some upper and lower bounds; see [http://spreadsheets.google.com/ccc?key=p5T0SktZY9DsU-uZ1tK7VEg this spreadsheet] for the latest bounds.<br />
<br />
The best known asymptotic lower bound for <math>c'_n</math> is<br />
<br />
: <math>c'_n \gg 3^n/\sqrt{n}</math>,<br />
<br />
formed by fixing the number of 2s to a single value near n/3. Compare this to the DHJ(2) or Sperner limit of <math>2^n/\sqrt{n}</math>. Is it possible to do any better? Note that we have a [[Upper_and_lower_bounds#Asymptotics|significantly better bound]] for the DHJ(3) <math>c_n</math>:<br />
<br />
: <math>c_n \geq 3^{n-O(\sqrt{\log n})}</math>.<br />
<br />
A more precise lower bound is<br />
<br />
: <math>c'_n \geq \binom{n+1}{q} 2^{n-q}</math><br />
<br />
where q is the nearest integer to <math>n/3</math>, formed by taking all strings with q 2s, together with all strings with q-1 2s and an odd number of 1s. This for instance gives the lower bound <math>c'_5 \geq 120</math>, which compares with the upper bound <math>c'_5 \leq 3 c'_4 = 129</math>.<br />
<br />
Using [[DHJ(3)]], we have the upper bound<br />
<br />
: <math>c'_n = o(3^n)</math>,<br />
<br />
but no effective decay rate is known. It would be good to have a combinatorial proof of this fact (which is weaker than [[DHJ(3)]], but implies [[Roth's theorem]]).<br />
<br />
== Notation ==<br />
<br />
Given a Moser set A in <math>[3]^n</math>, we let a be the number of points in A with no 2s, b be the number of points in A with one 2, c the number of points with two 2s, etc. We call (a,b,c,...) the ''statistics'' of A. Given a slice S of <math>[3]^n</math>, we let a(S), b(S), etc. denote the statistics of that slice (dropping the fixed coordinates). Thus for instance if A = {11, 12, 22, 32}, then (a,b,c) = (1,2,1), and (a(1*),b(1*)) = (1,1).<br />
<br />
We call a statistic (a,b,c,...) ''attainable'' if it is attained by a Moser set. We say that an attainable statistic is ''Pareto-optimal'' if it cannot be pointwise dominated by any other attainable statistic (a',b',c',...) (thus <math>a' \geq a</math>, <math>b' \geq b</math>, etc.) We say that it is ''extremal'' if it is not a convex combination of any other attainable statistic (this is a stronger property than Pareto-optimal). For the purposes of maximising linear scores of attainable statistics, it suffices to check extremal statistics.<br />
<br />
We let <math>(\alpha_0,\alpha_1,\ldots)</math> be the normalized version of <math>(a,b,\ldots)</math>, in which one divides the number of points of a certain type in the set by the total number of points in the set. Thus for instance <math>\alpha_0 = a/2^n, \alpha_1 = b/(n 2^{n-1}), \alpha_3 = c/(\binom{n}{2} 2^{n-2})</math>, etc., and the <math>\alpha_i</math> range between 0 and 1. Averaging arguments show that any linear inequality obeyed by the <math>\alpha_i</math> at one dimension is automatically inherited by higher dimensions, as are shifted versions of this inequality (in which <math>\alpha_i</math> is replaced by <math>\alpha_{i+1}</math>.<br />
<br />
The idea in 'c-statistics' is to identify 1s and 3s but leave the 2s intact. Let’s use x to denote letters that are either 1 or 3, then the 81 points in <math>[3]^4</math> get split up into 16 groups: 2222 (1 point), 222x, 22×2, 2×22, x222 (two points each), 22xx, 2×2x, x22x, 2xx2, x2×2, xx22 (four points each), 2xxx, x2xx, xx2x, xxx2 (eight points each), xxxx (sixteen points). Let c(w) denote the number of points inside a group w, e.g. c(xx22) is the number of points of the form xx22 inside the set, and is thus an integer from 0 to 4.<br />
<br />
== n=0 ==<br />
<br />
We trivially have <math>c'_0=1.</math><br />
<br />
== n=1 ==<br />
<br />
We trivially have <math>c'_1=2.</math> The Pareto-optimal values of the statistics (a,b) are (1,1) and (2,0); these are also the extremals. We thus have the inequality <math>a+b \leq 2</math>, or in normalized notation<br />
<br />
:<math>2\alpha_0 + \alpha_1 \leq 2</math>.<br />
<br />
== n=2 ==<br />
<br />
We have <math>c'_2 = 6</math>; the upper bound follows since <math>c'_2 \leq 3 c'_1</math>, and the lower bound follows by deleting one of the two diagonals from <math>[3]^2</math> (these are the only extremisers).<br />
<br />
The extremiser has statistics (a,b,c) = (2,4,0), which are of course Pareto-optimal. If c=1 then we must have a, b at most 2 (look at the lines through 22). This is attainable (e.g. {11, 12, 22, 23, 31}, and so (2,2,1) is another Pareto-optimal statistic. If a=4, then b and c must be 0, so we get another Pareto-optimal statistic (4,0,0) (attainable by {11, 13, 31, 33} of course). If a=3, then c=0 and b is at most 2, giving another Pareto-optimal statistic (3,2,0); but this is a convex combination of (4,0,0) and (2,4,0) and is thus not extremal. Thus the complete set of extremal statistics are<br />
<br />
(4,0,0), (2,4,0), (2,2,1).<br />
<br />
The sharp linear inequalities obeyed by a,b,c (other than the trivial ones <math>a,b,c \geq 0</math>) are then<br />
<br />
*<math>2a+b+2c \leq 8</math><br />
*<math>b+2c \leq 4</math><br />
*<math>a+2c \leq 4</math><br />
*<math>c \leq 1</math>.<br />
<br />
In normalized notation, we have<br />
*<math>4\alpha_0 + 2\alpha_1 + \alpha_2 \leq 4</math><br />
*<math>2\alpha_1 + \alpha_2 \leq 2</math><br />
*<math>2\alpha_0 + \alpha_2 \leq 2</math><br />
*<math>\alpha_2 \leq 1</math>.<br />
<br />
The Pareto optimizers for c-statistics are (c(22),c(2x),c(x2),c(xx)) = (1112),(0222),(0004), which are covered by these linear inequalities: <br />
*<math>c(22)+c(2x)+c(xx) \le 4</math>, <br />
*<math>c(22)+c(x2)+c(xx) \le 4</math>.<br />
<br />
== n=3 ==<br />
<br />
We have <math>c'_3 = 16</math>. The lower bound can be seen for instance by taking all the strings with one 2, and half the strings with no 2 (e.g. the strings with an odd number of 1s). The upper bound can be deduced from the corresponding [[upper and lower bounds]] for <math>c_3 = 18</math>; the 17-point and 18-point line-free sets each contain a geometric line.<br />
<br />
If a Moser set in <math>[3]^3</math> contains 222, then it can have at most 14 points, since the remaining 26 points in the cube split into 13 antipodal pairs, and at most one of each pair can lie in the set. By exhausting over the <math>2^{13} = 8192</math> possibilities, it can be shown that it is impossible for a 14-point set to exist; any Moser set containing 222 must in fact omit at least one antipodal pair completely and thus have only 13 points. (A human proof of this fact can be [http://www.ma.rhul.ac.uk/~elsholtz/WWW/blog/mosertablogv01.pdf found here].)<br />
<br />
The Pareto-optimal statistics are<br />
<br />
(3,6,3,1),(4,4,3,1),(4,6,2,1),(2,6,6,0),(3,6,5,0),(4,4,5,0),(3,7,4,0),(4,6,4,0), (3,9,3,0),(4,7,3,0),(5,4,3,0),(4,9,2,0),(5,6,2,0),(6,3,2,0),(3,10,1,0),(5,7,1,0), (6,4,1,0),(4,12,0,0),(5,9,0,0),(6,6,0,0),(7,3,0,0),(8,0,0,0).<br />
<br />
These were found from a search of the <math>2^{27}</math> subsets of the cube.<br />
A spreadsheet containing these statistics can be [http://spreadsheets.google.com/ccc?key=p5T0SktZY9DuKZ2DyzO9EOg&hl=en# found here]. Here are more [[3D Moser statistics]]. A [[lookup table]] for 3D Moser sets was constructed.<br />
<br />
The extremal statistics are<br />
<br />
(3,6,3,1),(4,4,3,1),(4,6,2,1),(2,6,6,0),(4,4,5,0),(4,6,4,0),(4,12,0,0),(8,0,0,0)<br />
<br />
The sharp linear bounds are :<br />
<br />
* <math>2a+b+2c+4d \leq 22</math><br />
* <math>3a+2b+3c+6d \leq 36</math><br />
* <math>7a+2b+4c+8d \leq 56</math><br />
* <math>6a+2b+3c+6d \leq 48</math><br />
* <math>b+c+3d \leq 12</math><br />
* <math>a+2c+4d \leq 14</math><br />
* <math>5a+4c+8d \leq 40</math><br />
* <math>a+4d \leq 8</math><br />
* <math>b+6d \leq 12</math><br />
* <math>c+3d \leq 6</math><br />
* <math>d \leq 1</math><br />
<br />
In normalized notation,<br />
* <math>8\alpha_0+ 6\alpha_1 + 6\alpha_2 + 2\alpha_3 \leq 11</math><br />
* <math>4\alpha_0+4\alpha_1+3\alpha_2+\alpha_3 \leq 6</math><br />
* <math>7\alpha_0+3\alpha_1+3\alpha_2+\alpha_3 \leq 7</math><br />
* <math>8\alpha_0+4\alpha_1+3\alpha_2+\alpha_3 \leq 8</math><br />
* <math>4\alpha_1+2\alpha_2+\alpha_3 \leq 4</math><br />
* <math>4\alpha_0+6\alpha_2+2\alpha_3 \leq 7</math><br />
* <math>5\alpha_0+3\alpha_2+\alpha_3 \leq 5</math><br />
* <math>2\alpha_0+\alpha_3 \leq 2</math><br />
* <math>2\alpha_1+\alpha_3 \leq 2</math><br />
* <math>2\alpha_2+\alpha_3 \leq 2</math><br />
* <math>\alpha_3 \leq 1</math><br />
<br />
The c-statistics can also be found from an exhaustive search of Moser sets. The resulting Pareto sets and linear inequalities can be found on Sheet 8 of [http://spreadsheets.google.com/ccc?key=p5T0SktZY9DuKZ2DyzO9EOg&hl=en# this spreadsheet]<br />
<br />
The following are Pareto optimal (3,6,3,1),(4,4,3,1),(4,6,2,1),(2,6,6,0),(3,6,5,0),(4,4,5,0),(3,7,4,0),(4,6,4,0), (3,9,3,0),(4,7,3,0),(5,4,3,0),(4,9,2,0),(5,6,2,0),(6,3,2,0),(3,10,1,0),(5,7,1,0), (6,4,1,0),(4,12,0,0),(5,9,0,0),(6,6,0,0),(7,3,0,0),(8,0,0,0). Here is a [[human proof of the 3D Pareto-optimal Moser statistics]].<br />
<br />
== n=4 ==<br />
<br />
A [http://abel.math.umu.se/~klasm/extremal-moser-n=5-t=3 computer search] has obtained all extremisers to <math>c'_4=43</math>. The 42-point solutions can be found [http://abel.math.umu.se/~klasm/moser-n=4-t=3-p=42.gz here].<br />
<br />
'''Proof that <math>c'_4 \leq 43</math>''': <br />
When e=1 (i.e. the 4D set contains 2222) then we have at most 41 points (in fact at most 39) by counting antipodal points, so assume e=0.<br />
<br />
Define the score of a 3D slice to be a/4+b/3+c/2+d. Observe from double counting that the size of a 4D set is the sum of the scores of its eight side slices.<br />
<br />
But by [http://spreadsheets.google.com/ccc?key=p5T0SktZY9DuKZ2DyzO9EOg&hl=en looking at the extremals] we see that the largest score is 44/8, attained at only one point, namely when (a,b,c,d) = (2,6,6,0). So the only way one can have a 44-point set is if all side slices are (2,6,6,0), or equivalently if the whole set has statistics (a,b,c,d,e) = (4,16,24,0,0). But then we have all the points with two 2s, which means that the four "a" points cannot be separated by Hamming distance 2. We conclude that we must have an antipodal pair among the "a" points with an odd number of 1s, and an antipodal pair among the "a" points with an even number of 1s. By the symmetries of the cube, we may take the a-set to then be 1111, 3333, 1113, 3331. But then the "b" set must exclude both 1112 and 3332, and so can have at most three points in the eight-point set xyz2 (with x,y,z=1,3) rather than four (to get four points one must alternate in a checkerboard pattern). Adding this to the at most four points of the form xy2z, x2yz, 2xyz we see that b is at most 15, a contradiction. <math>\Box</math><br />
<br />
Given a subset of <math>[3]^4</math>, let a be the number of points with no 2s, b be the number of points with 1 2, and so forth. The quintuple (a,b,c,d,e) thus lies between (0,0,0,0,0) and (16,32,24,8,1).<br />
<br />
The 43-point solutions have distributions (a,b,c,d,e) as follows:<br />
<br />
* (5,20,18,0,0) [16 solutions]<br />
* (4,16,23,0,0) [768 solutions]<br />
* (3,16,24,0,0) [512 solutions]<br />
* (4,15,24,0,0) [256 solutions]<br />
<br />
The 42-point solutions are [http://abel.math.umu.se/~klasm/Moser-42-stat.pdf distributed as follows]:<br />
<br />
* (6,24,12,0,0) [[http://abel.math.umu.se/~klasm/moser-n=4-42-12 8 solutions]] <br />
* (5,20,17,0,0) [576 solutions]<br />
* (5,19,18,0,0) [384 solutions]<br />
* (6,16,18,2,0) [[http://abel.math.umu.se/~klasm/moser-n=4-42-e=2 192 solutions]]<br />
* (4,20,18,0,0) [272 solutions]<br />
* (5,17,20,0,0) [192 solutions]<br />
* (5,16,21,0,0) [3584 solutions]<br />
* (4,17,21,0,0) [768 solutions]<br />
* (4,16,22,0,0) [26880 solutions]<br />
* (5,15,22,0,0) [1536 solutions]<br />
* (4,15,23,0,0) [22272 solutions]<br />
* (3,16,23,0,0) [15744 solutions]<br />
* (4,14,24,0,0) [4224 solutions]<br />
* (3,15,24,0,0) [8704 solutions]<br />
* (2,16,24,0,0) [896 solutions]<br />
<br />
Note how c is usually quite large, and d quite low.<br />
<br />
One of the (6,24,12,0,0) solutions is <math>\Gamma_{220}+\Gamma_{202}+\Gamma_{022}+\Gamma_{112}+\Gamma_{211}</math> (i.e. the set of points containing exactly two 1s, and/or exactly two 3s). The other seven are reflections of this set.<br />
<br />
There are 2,765,200 41-point solutions, [http://abel.math.umu.se/~klasm/solutions-4-t=3-41-moser.gz listed here]. The statistics for such points can be [http://abel.math.umu.se/~klasm/Moser-41-stat.pdf found here]. Noteworthy features of the statistics:<br />
<br />
* d is at most 3 (and, except for [http://abel.math.umu.se/~klasm/moser-n=3-t=3-41-d=3.gz 256 exceptional solutions] of the shape (5,15,18,3,0), have d at most 2; here are the [http://abel.math.umu.se/~klasm/moser-n=3-t=3-41-d=2.gz d=2 solutions] and [http://abel.math.umu.se/~klasm/moser-n=3-t=3-41-d=1.gz d=1 solutions])<br />
* c is at least 6 (and, except for [http://abel.math.umu.se/~klasm/moser-n=3-t=3-41-c=6.gz 16 exceptional solutions] of the shape (7,28,6,0,0), have c at least 11).<br />
<br />
Statistics for the 41-point, 42-point, and 43-point solutions can be found [http://spreadsheets.google.com/ccc?key=p5T0SktZY9DuqNcxJ171Bbw&hl=en here]. [http://spreadsheets.google.com/ccc?key=rwXB_Rn3Q1Zf5yaeMQL-RDw&hl=en Here is the full list of Pareto-optimals].<br />
<br />
If a Moser set in <math>[3]^4</math> contains 2222, then by the n=3 theory, any middle slice (i.e. 2***, *2**, **2*, or ***2) is missing at least one antipodal pair. But each antipodal pair belongs to at most three middle slices, thus two of the 40 antipodal pairs must be completely missing. As a consequence, any Moser set containing 2222 can have at most 39 points.<br />
(A more refined analysis can be found at [http://www.ma.rhul.ac.uk/~elsholtz/WWW/blog/mosertablogv01.pdf found here].)<br />
<br />
We have the following inequalities connecting a,b,c,d,e:<br />
<br />
* <math>4a+b \leq 64</math>: There are 32 lines connecting two "a" points with a "b" point; each "a" point belongs to four of these lines, and each "b" point belongs to one. But each such line can have at most two points in the set, and the claim follows.<br />
* This can be refined to <math>4a+b+\frac{2}{3} c \leq 64</math>: There are 24 planes connecting four "a" points, four "b" points, and one "c" point; each "a" point belongs to six of these, each "b" point belongs to three, and each "c" point belongs to one. For each of these planes, we have <math>2a + b + 2c \leq 8</math> from the n=2 theory, and the claim follows.<br />
* <math>6a+2c \leq 96</math>: There are 96 lines connecting two "a" points with a "c" point; each "a" point belongs to six of these lines, and each "c" point belongs to two. But each such line can have at most two points in the set, and the claim follows.<br />
* <math>3b+2c \leq 96</math>: There are 48 lines connecting two "b" points to a "c" point; each "b" point belongs to three of these points, and each "c" point belongs to two. But each such line can have at most two points in the set, and the claim follows.<br />
<br />
The inequalities for n=3 imply inequalities for n=4. Indeed, there are eight side slices of <math>[3]^4</math>; each "a" point belongs to four of these, each "b" point belongs to three, each "c" point belongs to two, and each "d" point belongs to one. Thus, any inequality of the form<br />
:<math> \alpha a + \beta b + \gamma c + \delta d \leq M</math><br />
in three dimensions implies the inequality<br />
:<math> 4 \alpha a + 3 \beta b + 2 \gamma c + \delta d \leq 8M</math><br />
in four dimensions. Thus we have<br />
<br />
* <math>8a+3b+4c+4d \leq 176</math><br />
* <math>2a+b+c+d \leq 48</math><br />
* <math>14a+3b+4c+4d \leq 244</math><br />
* <math>4a+b+c+d \leq 64</math> <br />
* <math>3b+2c+3d \leq 96</math><br />
* <math>a+c+d \leq 28</math><br />
* <math>5a+2c+2d \leq 80</math><br />
* <math>a+d \leq 16</math><br />
* <math>b+2d \leq 32</math><br />
* <math>2c+3d \leq 48</math><br />
<br />
Cubes also sit diagonally in the 4-dimensional cube. They may have coordinates xxyz, where xx runs over (11,22,33) or (13,22,31), while y and z run over (1,2,3). These cubes have a different distribution of 2s than the ordinary slices: (a,b,c,d,e) = (8,8,6,4,1) instead of (8,12,6,1,0) for the side slices and (0,8,12,6,1) for the middle slices. So a different set of inequalities arise. Apply the same procedure as described above for n=3: (Run through the <math>2^{27}</math> subsets of the cube; identify those without combinatorial lines; calculate their statistics in the new xxyz arrangement; retain the Pareto-optimal statistics; retain the extremal statistics; find what inequalities they satisfy.) The inequalities that arise are <br />
* <math>2a+b+2c+2d+4e \le 24</math> and <br />
* <math>4a+b+2c+2d+4e \le 32</math> <br />
within the 3D cube, which when averaged becomes <br />
* <math>4a+b+2c+4d+16e \le 96</math> and <br />
* <math>8a+b+2c+4d+16e \le 128</math> <br />
for the 4D cube. A spreadsheet containing these statistics can be [http://spreadsheets.google.com/ccc?key=p5T0SktZY9DuKZ2DyzO9EOg&hl=en# found here] on sheet 2. Other sheets of this spreadsheet contain results for a 3D cube sitting diagonally in a 5D, 6D or 7D cube. Notice the similarity of the equations that arise for the xxxyz, xxyyz and xxxxyz diagonals. Also notice the (a,b,c,...) statistics for the xxxxyyz diagonals have the same Pareto sets and linear inequalities as the cube's c-statistics.<br />
<br />
The c-statistics for the 3D cube are bounded by a set of 20 inequalities. By considering the fourteen ways a 3D cube sits in the 4D cube, the result is a set of 239 inequalities for the 4D cube's c-statistics. However, all 239 inequalities are satisfied by a 44-point solution given by c(xxxx) = c(2xxx) = c(22xx) = 4, and permutations.<br />
<br />
==== If a Moser set for n=4 has 4 or more points with three 2’s it has less than 41 points. ====<br />
<br />
See [[4D Moser sets with d at least 4 have at most 40 points]].<br />
<br />
==== If a Moser set for n=4 has 1 or more points with three 2’s it has less than 43 points. ====<br />
<br />
See [[4D Moser sets with d at least 2 have at most 42 points]].<br />
<br />
==== If a Moser set for n=4 has 3 or more points with three 2’s it has less than 42 points. ====<br />
<br />
See [[4D Moser sets with d at least 3 has at most 41 points]].<br />
<br />
==== Brute-force calculations for <math>[3]{}^4</math> ====<br />
<br />
An attempt was made to find extremal statistics for the four-dimensional case by brute force. Since a simple routine would check 2^81 possibilities, which is infeasible, shortcuts would be needed.<br />
<br />
The outline of a suggested routine is as follows: There are just over 3.8 million line-free subsets of [3]^3. Since the cube has 3!2^3 = 48 symmetries, they fall into 83158 equivalence classes. To consider all possibilities, it is enough to do the following:<br />
<br />
* Let the 1*** slice be a representative from one of the 83158 classes.<br />
* Let the 3*** slice be one of the 3.8 million subsets.<br />
* Identify which points in 2*** would not complete a vertical or sloping line from the 1*** slice to the 3*** slice.<br />
* Go to a lookup table, and find the Pareto-optimal statistics of all line-free subsets of 2*** with the restriction from the previous step. This lookup table would need 2^27 rows, or 2^27/48 if one made use of the cube's symmetries.<br />
* Deduce a set of possible statistics for the combined **** cube. Update the list of [3]^4 Pareto statistics<br />
* Loop through the 83158 1*** slices and 3.8 million 3*** slices.<br />
<br />
A large part of this calculation is to build the lookup table. Here is a [[Matlab script]] to carry it out.<br />
<br />
Here is a slightly different implementation of the [[4D Moser brute force search]]. It obtained [http://spreadsheets.google.com/ccc?key=rwXB_Rn3Q1Zf5yaeMQL-RDw&hl=en this list of Pareto-optimals] and [http://spreadsheets.google.com/ccc?key=rD2Oc_wyheVOcmCyd-Drdgw&hl=en this list of extremals].<br />
<br />
== Proof that <math>c'_5</math> = 124 ==<br />
<br />
Let (A,B,C,D,E,F) be the statistics of a five-dimensional Moser set, thus (A,B,C,D,E,F) varies between (0,0,0,0,0,0) and (32,80,80,40,10,1).<br />
<br />
There are several Moser sets with the statistics (4,40,80,0,0,0), which thus have 124 points. Indeed, one can take<br />
<br />
* all points with two 2s;<br />
* all points with one 2 and an even number of 1s; and<br />
* 11111, 11333, 33311, 33331. Any two of these four points differ in three places, except for one pair of points that differ in one place.<br />
<br />
For the rest of this section, we assume that the Moser set <math>{\mathcal A}</math> is a 125-point Moser set. We will prove that <math>{\mathcal A}</math> cannot exist.<br />
<br />
:'''Lemma 1''': F=0.<br />
<br />
'''Proof''' If F is non-zero, then the Moser set contains 22222, then each of the 121 antipodal pairs can have at most one point in the set, leading to only 122 points. <math>\Box</math><br />
<br />
:'''Lemma 2''': Every middle slice of <math>{\mathcal A}</math> has at most 41 points.<br />
<br />
'''Proof''' Without loss of generality we may consider the 2**** slice. There are two cases, depending on the value of c(2****).<br />
<br />
Suppose first that c is at least 17; thus there are at least 17 points of the form 222xy, 22x2y, 22xy2, 2x2y2, 2xy22, or 2x22y, where the x, y denote 1 or 3. This gives 34 "xy" wildcards in all in four coordinate slots; by the pigeonhole principle one of the slots sees at least 9 of the wildcards. By symmetry, we may assume that the second coordinate slot sees at least 9 of these wildcards, thus there are at least 9 points of the form 2x22y, 2x2y2, 2xy22. The x=1, x=3 cases can absorb at most six of these, thus each of these cases must absorb at least three points, with at least one absorbing at least five. Let's say that it's the x=3 case that absorbs 5; thus <math>d(*1***) \geq 3</math> and <math>d(*3***) \geq 5</math>. From the n=4 theory this means that the *1*** slice has at most 41 points, and the *3*** slice has at most 40. Meanwhile, the middle slice has at most 43, leading to 41+41+42=124 points in all.<br />
<br />
Now suppose c is less than 17; then by the n=4 theory the middle slice is one of the eight (6,24,12,0,0) sets. Without loss of generality we may take it to be <math>\Gamma_{220}+\Gamma_{202}+\Gamma_{022}+\Gamma_{112}+\Gamma_{211}</math>; in particular, the middle slice contains the points 21122 21212 21221 23322 23232 23223. In particular, the *1*** and *3*** slices have a "d" value of at least three, and so have at most 41 points. If the *2*** slice has at most 42 points, then we are at 41+42+41=124 points as needed, but if we have 43 or more, then we are back in the first case (as <math>c(*2***) \geq 17</math>) after permuting the indices.<br />
<math>\Box</math><br />
<br />
Since 125=41+41+43, we thus have<br />
<br />
:'''Corollary 1''': Every side slice of <math>{\mathcal A}</math> has at least 41 points. If one side slice does have 41 points, then the other has 43.<br />
<br />
Combining this with the n=4 statistics, we conclude<br />
<br />
:'''Corollary 2''': Every side slice of <math>{\mathcal A}</math> has e=0, and <math>d \leq 3</math>. Given two opposite side slices, e.g. 1**** and 3****, we have <math>d(1****)+d(3****) \leq 4</math>.<br />
<br />
:'''Corollary 3''': E=0.<br />
<br />
:'''Corollary 4''': Any middle slice has a "c" value of at most 8.<br />
<br />
'''Proof''' Let's work with the 2**** slice. By Corollary 2, there are at most four contributions to c(2****) of the form 21*** or 23***, and similarly for the other three positions. Double counting then gives the claim. <math>\Box</math><br />
<br />
:'''Lemma 3''': <math>2B+C \leq 160</math>. <br />
<br />
'''Proof''': There are 160 lines connecting one "C" point to two "B" points (e.g. 11112, 11122, 11132); each "C" point lies in two of these, and each "B" point lies on four. A Moser set can have at most two points out of each of these lines. Double counting then gives the claim. <math>\Box</math><br />
<br />
:'''Lemma 4''' Given m A-points with <math>m \geq 5</math>, there exists at least m-4 pairs (a,b) of such A-points with Hamming distance exactly two (i.e. b differs from a in exactly two places).<br />
<br />
'''Proof''' It suffices to check this for m=5, since the case of larger m then follows by locating a pair, removing a point that contributes to that pair, and using the induction hypothesis. Given 5 A-points, we may assume by the pigeonhole principle and symmetry that at least three of them have an odd number of 1s. Suppose 11111 is one of the points, and that no pair has Hamming distance 2. All points with two 3s are excluded, so the only points allowed with an odd number of 1s are those with four 3s. But all those points differ from each other in two positions, so at most one of them is allowed. <math>\Box</math><br />
<br />
:'''Corollary 5''' <math>C \leq 79</math>.<br />
<br />
'''Proof''': Suppose for contradiction that C=80, then D=0 and <math>B \leq 40</math>. From Lemma 4 we also see that A cannot be 5 or more, leading to the contradiction.<br />
<math>\Box</math><br />
<br />
Define the '''score''' of a Moser set in <math>[3]^4</math> to be the quantity <math>a + 5b/4 + 5c/3 + 5d/2 + 5e</math>. Double-counting (and Lemma 2) gives<br />
<br />
:'''Lemma 5''' The total score of all the ten side-slices of <math>{\mathcal A}</math> is equal to <math>5|{\mathcal A}| = 5 \times 125</math>. In particular, there exists a pair of opposite side-slices whose scores add up to at least 125.<br />
<br />
By Lemma 5 and symmetry, we may assume that the 1**** and 3**** slices have score adding up to at least 125. By Lemma 2, the 2**** slice has at most 41 points, which imply that the 1**** and 3**** have 41, 42, or 43 points. From the [http://spreadsheets.google.com/ccc?key=p5T0SktZY9DuqNcxJ171Bbw&hl=en n=4 statistics] we know that all 41-point, 42-point, 43-point slices have score less than 62, with the following exceptions:<br />
<br />
# (2,16,24,0,0) [42 points, score: 62]<br />
# (4,16,23,0,0) [43 points, score: 62 1/3]<br />
# (4,15,24,0,0) [43 points, score: 62 3/4]<br />
# (3,16,24,0,0) [43 points, score: 63]<br />
<br />
Thus the 1**** and 3**** slices must come from the above list. Furthermore, if one of the slices is of type 1, then the other must be of type 4, and if one slice is of type 2, then the other must be of type 3 or 4.<br />
<br />
:'''Lemma 6''' There exists one cut in which the side slices have total score strictly greater than 125 (i.e. they thus involve only Type 2, Type 3, and Type 4 slices, with at least one side slice not equal to Type 2, and the cut here is of the form 43+39+43).<br />
<br />
'''Proof''' If not, then all cuts have side slices exactly equal to 125, which by the above table implies that one is Type 1 and one is Type 4, in particular all side slices have c=24. But this forces C=80, contradicting Corollary 5. <math>\Box</math><br />
<br />
Adding up the corners we conclude<br />
<br />
:'''Corollary 6''' <math>6 \leq A \leq 8</math>.<br />
<br />
:'''Lemma 7''' <math>D=0</math>.<br />
<br />
'''Proof''' Firstly, from Lemma 6 we have a cut in which the two side slices are omitting at most one C-point between them (and have no D-point), which forces the middle slice to have at most one D-point; thus D is at most 1.<br />
<br />
Now suppose instead that D=1 (e.g. if 11222 was in the set); then there would be two choices of coordinates in which one of the side slices would have d=1 (e.g. 1**** and *1****). But the [http://spreadsheets.google.com/ccc?key=p5T0SktZY9DuqNcxJ171Bbw&hl=en n=4 statistics] show that such slices have a score of at most 59 7/12, so that the total score from those two coordinates is at most 63 + 59 7/12 = 122 7/12. On the other hand, the other three slices have a net score of at most 63+63 = 126. This averages out to at most 124.633... < 125, a contradiction. <math>\Box</math><br />
<br />
:'''Corollary 7''' Every middle slice has at most 40 points.<br />
<br />
'''Proof''' From Lemma 7 we see that the middle slice must have c=0, but from the n=4 statistics this is not possible for any slice of size 41 or higher (alternatively, one can use the inequalities <math>4a+b \leq 64</math> and <math>b \leq 32</math>). <math>\Box</math><br />
<br />
Each middle slice now has 39 or 40 points, with c=d=e=0, and so from the inequalities <math>4a+b \leq 64</math>, <math>b \leq 32</math> must have statistics (8,32,0,0,0),(7,32,0,0,0), or (8,31,0,0,0). In particular the "a" index of the middle slices is at most 8. Summing over all middle slices we conclude<br />
<br />
:'''Lemma 9''' B is at most 40.<br />
<br />
:'''Lemma 10''' C is equal to 78 or 79.<br />
<br />
'''Proof''' By Corollary 5, it suffices to show that <math>C \geq 78</math>.<br />
<br />
As 125=43+39+43, we see that every center slice must have at least 39 points. By Lemma 2 and Lemma 7 the center slice has c=d=e=0, thus the center slice has a+b >= 39. On the other hand, from the n=4 theory we have 4a+b <= 64, which forces b >= 31.<br />
<br />
By double counting, we see that 2C is equal to the sum of the b's of all the five center slices. Thus C >= 5*31/2 = 77.5 and the claim follows. <math>\Box</math><br />
<br />
From Lemmas 9 and 10 we see that <math>B+C \leq 119</math>, and thus <math>A \geq 6</math>. Also, if <math>A \geq 7</math>, then by Lemma 4 we have at least three pairs of A points with Hamming distance 2. At most two of these pairs eliminate the same C point, so we would have C=78 in that case. <br />
<br />
Putting all the above facts together, we see that (A,B,C,D,E,F) must be one of the following triples:<br />
<br />
* (6,40,79,0,0,0)<br />
* (7,40,78,0,0,0)<br />
* (8,39,78,0,0,0)<br />
<br />
All three cases can be eliminated, giving <math>c'_5=124</math>.<br />
<br />
=== Elimination of (6,40,79,0,0,0) ===<br />
<br />
Look at the 6 A points. From Lemma 4 we have at least two pairs (a,b), (c,d) of A-points that have Hamming separation 2.<br />
<br />
Now look at the midpoints of these two pairs; these midpoints are C-points cannot lie in the set. But we have exactly one C-point missing from the set, thus the midpoints must be the same. By symmetry, we may thus assume that the two pairs are (11111,11133) and (11113,11131). Thus 11111,11133, 11113, 11131 are in the set, and so every C-point other than 11122 is in the set. On the other hand, the B-points 11121, 11123, 11112, 11132 lie outside the set.<br />
<br />
At most one of 11312, 11332 lie in the set (since 11322 lies in the set). Suppose that 11312 lies outside the set, then we have a pair (xy1z2, xy3z2) with x,y,z = 1,3 that is totally omitted from the set, namely (11112,11312). On the other hand, every other pair of this form can have at most one point in the set, thus there are at most seven points in the set of the form (xyzw2) with x,y,z,w = 1,3. Similarly there are at most 8 points of the form xyz2w, or of xy2zw, x2yzw, 2xyzw, leading to at most 39 B-points in all, contradiction.<br />
<br />
=== Elimination of (7,40,78,0,0,0) ===<br />
<br />
By Lemma 4, we have at least three pairs of A-points of distance two apart that lie in the set.<br />
The midpoints of these pairs are C-points that do not lie in the set; but there are only two such C-points, thus two pairs must have the same midpoint, so we may assume as before that 111xy lies in the set for x,y=1,3, which implies that 1112* and 111*2 lie outside the set.<br />
<br />
Now consider the 160 lines between 2 B points and one C point (cf. Lemma 3). The sum of all the points in each such line (counting multiplicity) is 4B+2C = 316. On the other hand, every one of the 160 lines can have at most two points, and two of these lines (namely 1112*, 111*2) have no points. Thus all the other lines must have exactly two points. <br />
<br />
We know that the C-point 11122 is missing from the set; there is one other missing C-point. Since 1112x, 111x2 lie outside the set, we conclude from the previous paragraph that 1132x, 113x2 and 1312x, 131x2 lie in the set. Taking midpoints we conclude that 11322 and 13122 lie outside the set. But this is now three C-points missing (together with 11122), a contradiction.<br />
<br />
=== Elimination of (8,39,78,0,0,0) ===<br />
<br />
By Lemma 4 we have at least four pairs of A-points of distance two apart that lie in the set. The midpoints of these pairs are C-points that do not lie in the set; but there are only two such C-points, thus two pairs (a,b), (c,d) must have the same midpoint p, and the other two pairs (a',b'), (c',d') must also have the same midpoint p'. (Note that every C-point is the midpoint of at most two such pairs.)<br />
<br />
Now consider the 160 lines between 2 B points and one C point (cf. Lemma 3). The sum of all the points in each such line (counting multiplicity) is 4B+2C = 312. Every one of the 160 lines can have at most two points, and four of these (those in the plane of (a,b,c,d) or of (a',b',c',d') have no points. Thus all other lines must have exactly two points.<br />
<br />
Without loss of generality we have (a,b)=(11111,11133), (c,d) = (11113,11131), thus p = 11122. By permuting the first three indices, we may assume that p' is not of the form x2y2z, x2yz2, xy22z, xy2z2. Then 1112x lies outside the set and 1122x lies in the set, so by the above paragraph 1132x lies in the set; similarly for 113x2, 1312x, 131x2. This implies that 13122, 11322 lie outside the set, but this (together with 11122) shows that at least three C-points are missing, a contradiction.<br />
<br />
== n=6 ==<br />
<br />
We have <math>c'_6 \geq 353</math>. This can be seen for instance by taking the union of the Gamma-cells (6,0,0),(5,0,1),(3,3,0), (3,2,1),(3,1,2),(2,2,2),(1,3,2),(0,3,3),(2,0,4),(0,2,4),(0,1,5), or equivalently<br />
<br />
* The 22 a-points consisting of 111111, 111113, 113333 and permutations;<br />
* The 66 b-points consisting of 111332, 333332 and permutations;<br />
* The 165 c-points consisting of 111322, 113322, 333322 and permutations;<br />
* The 100 d-points consisting of 111222, 133222, 333222 and permutations;<br />
* No e, f, or g points.<br />
<br />
'''Lemma 8''' A 6D Moser set with g=1 has at most 351 points.<br />
<br />
'''Proof''' Define the 22**** score of a 4D set to be 15 a + 5 b + 5 c/2 + 3d/2 + e. Observe that the cardinality of a 6D set is the average of its 30 22**** scores of its 22****-type slices, plus a+b.<br />
<br />
If g=1, then a <= 32 and b <= 96 and so a+b <= 128. Meanwhile, the largest 22**** score for a set with e=1 is 223.5, by the [http://spreadsheets.google.com/ccc?key=rwXB_Rn3Q1Zf5yaeMQL-RDw&hl=en the Pareto optimal list]. The claim follows. (Actually, the [[Maple calculations]] give the better upper bound of 337.)<br />
<br />
'''Lemma 9''' A 6D Moser set with f>0 has at most 353 points.<br />
<br />
'''Proof'''<br />
Currently only via Maple calculations.<br />
<br />
Define the ''11**** score'' of a 4D Moser set to be 4a+6b+10c+20d+60e. The cardinality of a 6D set with f=g=0 is the average of its 60 scores of its "corner slices" - 11***** and permutations and reflections. Define the ''defect'' of a slice to be 356 minus its score. From [http://spreadsheets.google.com/ccc?key=rwXB_Rn3Q1Zf5yaeMQL-RDw&hl=en the Pareto optimal list], the largest scores are<br />
<br />
* 356 (defect 0) - attained by (6,12,18,4,0)<br />
* 352 (defect 4)- attained by (5,12,18,4,0), (5,12,12,4,1), (6,8,12,8,0)<br />
* 350 (defect 6)- attained by (6,11,18,4,0), (5,15,12,3,1), (6,11,12,7,0), (5,15,18,3,0)<br />
* 348 (defect 8)- attained by (6,14,12,6,0), (3,16,24,0,0), (4,12,18,4,0), (4,12,12,4,1), (5, 8, 12, 8, 0)<br />
* all other statistics are at most 346 (defect at least 10).<br />
<br />
This establishes <math>c'_6 \leq 356</math>.<br />
<br />
Call a corner slice "good" if it has score 356 (defect 0), i.e. has statistics (6,12,18,4,0), and "bad" otherwise (so has score at most 352, i.e. defect at least 4). Thus:<br />
<br />
* In a 356-point set, all 60 corner slices must be good.<br />
* In a 355-point set, at most 15 corner slices can be bad. (In fact, the total defect of all the slices is 60.)<br />
* In a 354-point set, at most 30 corner slices can be bad. (In fact, the total defect of all the slices is 120.)<br />
<br />
The (6,12,18,4,0) sets have been [[classification of (6,12,18,4,0) sets|completely classified]]. The basic example is the set consisting of 1111, 1113, 3333, 1332, 1322, 1222, 3322 and permutations (i.e. (4,0,0), (3,0,1), (0,0,4), (1,1,2), (1,2,1), (1,3,0), (0,2,2) in Gamma-notation). This set has d-points 1222, 2122, 2212, 2221. More generally, given any <math>x,y,z,w \in \{1,3\}</math>, there is a unique (6,12,18,4,0) set with d-points x222, 2y22, 22z2, 222w, and these are the only 16 possibilities. Call this set the (6,12,18,4,0) set of type xyzw. It consists of<br />
<br />
* The four "d" points x222, 2y22, 22z2, 222w;<br />
* All 24 "c" points ''except'' for xy22, x2z2, x22w, 2yz2, 2y2w, 22zw;<br />
* The 12 "b" points xYZ2, xY2W, x2ZW, XyZ2, Xy2W, 2yZW, XYz2, X2zW, 2YzW, XY2w, X2Zw, 2YZw, where X=4-x, Y=4-y, Z=4-z, W=4-w;<br />
* The 6 "a" points xyzw, xyzW, xyZw, xYzw, Xyzw, XYZW.<br />
<br />
In particular, the centre 3D slices of a good corner slice have statistics (3,9,3,0), and two opposing side slices have statistics (2,6,6,0) and (4,3,3,1).<br />
<br />
'''Note''': in the 353-point example given above, the 11**** slices are good (and of type 1111), the 13**** slices have statistics (5,12,18,4,0) and consist of the type 3333 slice with the point 133333 removed, and the 33**** slices have statistics (6,8,12,8,0), with the slice consisting of the 6 a-points formed by permuting 1133, the 8 b-points formed by permuting 1112 and 3332, the 12 c-points formed by permuting 1122 and 3322, and all 8 d-points.<br />
<br />
We now show that it is impossible to have a 356-point set, i.e. one in which all corner slices are good. From the above discussion, we see that if the 111*** slice contains its center 111222, then the 113*** slice does not, and vice versa. Thus the d points of the form ***222 consist either of those strings with an even number of 1s, or those with an odd number of 1s.<br />
<br />
Let’s say it’s the former, thus the set contains 111222, 133222, and permutations of the first three coordinates, but omits 113222, 333222 and permutations of the first three coordinates. Meanwhile, we see that the (24,72,180,80,0,0,0) set saturates the inequality 4c+3d \leq 960, which is only possible if every line connecting two “c” points to a “d” point meets exactly two elements in the set. Since the “d” points 113222, 333222 are omitted, we conclude that the “c” points 113122, 113322, 333122, 333322 must lie in the set, and similarly for permutations of the first three and last three coordinates. But this gives at least 15 of the 16 "c" points ending in 22; by symmetry this leads to 225 c-points in all, which is far too many.<br />
<br />
Now we study what happens when two intersecting 4D corner slices of <math>[3]^6</math> are (6,12,18,4,0) sets.<br />
<br />
'''Lemma 10'''. Suppose that the 11**** slice is of type xyzw, and the 1*1*** slice is of type x'y'z'w'. Then x' = x, and y'z'w' is equal to either yzw or YZW. If x'=x=1, then y'z'w'=yzw. <br />
<br />
'''Proof''' Observe that the 11**** slice contains 111222 iff x=1, and the 1*1*** slice similarly contains 111222 iff x'=1. This shows that x=x'.<br />
<br />
Suppose now that x=x'=1. Then the 111*** slice contains the three elements 111y22, 1112z2, 11122w, and excludes 111Y22, 1112Z2, 11122W, and similarly with the primes, which forces yzw=y'z'w' as claimed.<br />
<br />
Now suppose that x=x'=3. Then the 111*** slice contains the two elements 111yzw, 111YZW, but does not contain any of the other six "a" points in 111***, and similarly for the primes. Thus y'z'w' is equal to either yzw or YZW as claimed.<br />
<br />
One can reflect this to get<br />
<br />
'''Corollary 11''' Suppose that the pq**** slice is of type xyzw, and the p*r*** slice is of type x'y'z'w', where p,q,r,x,y,z,w,x',y',z',w' are in {1,3}. Then x'=x iff q=r, and y'z'w' is equal to either yzw or YZW. If x=r (or equivalently if x'=q), then y'z'w'=yzw.<br />
<br />
Now we put some slices together:<br />
<br />
'''Corollary 12''' Suppose that the 11**** slice is of type xyzw, and the 1*1*** and 13**** slices are both good. Then the 13**** slice is of type Xyzw or XYZW. If the 1**1** slice is also good, then the 13**** slice must be of type XYZW.<br />
<br />
'''Proof''' From Lemma 10, the 1*1*** slice is of type xyzw or xYZW. By Corollary 11, the 13**** slice is thus of type Xyzw or XYZW. If the 1**1** slice is also good, then we similarly conclude that the 13**** slice is of type xYzw or XYZW. The claim follows.<br />
<br />
'''Corollary 13''' Suppose the 11**** and 13**** slices are good. Then either they have opposite types (xyzw and XYZW) or else at least six of the eight slices beginning with 1* (i.e. 1*1***, 1*3***, 1**1**, 1**3**, 1***1*, 1***3*, 1****1, 1****3) are bad.<br />
<br />
'''Lemma 14''' If the 11**** and 33**** slices are good and of the same type, then there are at most 350 points.<br />
<br />
'''Proof''' By symmetry we may assume that 11**** and 33**** are of type 1111. This excludes a lot of points from 22****. Indeed, the only points that can still survive in 22**** are 221133, 221333, 221132, 223332, and permutations of the last four indices. Double counting the lines 22133* and permutations we see that there are at most 12 points one can place in the permutations of 221133, 221333, 221132, and so the 22**** slice has at most 16 points. Meanwhile, the two 5D slices 1*****, 3***** have at most <math>c'_5=124</math> points, and the other two 4D slices 21****, 23**** slices have at most <math>c'_4=43</math> points, leading to <math>16+124*2+43*2 = 350</math> points in all. The claim follows.<br />
<br />
'''Corollary 15''' If the 11****, 13****, and 33**** slices of a 355-point set are all (6,12,18,4,0) sets, then either six of the eight slices beginning with 1* are bad, or six of the eight slices beginning with *3 are bad.<br />
<br />
'''Proposition 16''' A 355-point set cannot have all four slices 11****, 13****, 31****, and 33**** slices be good.<br />
<br />
'''Proof''' Suppose not. By Lemma 14, the 11**** and 33**** slices cannot be of the same type, and so they cannot both be of the opposite type to either 13**** or 31****. From Corollary 15 we conclude that at least twelve of the sixteen slices beginning with 1*, 3*, *1, *3 are bad. Since there are at most 15 bad slices, there are only three bad slices beginning with **. By the pigeonhole principle, we may thus assume that **11**, **13**, **31**, **33** are good. But now on applying Lemma 14 and Corollary 15 again we conclude that there are four more bad slices beginning with **, contradiction.<br />
<br />
From this and the score analysis we see that<br />
<br />
'''Corollary 17''' In a 355-point set, exactly three of the four slices 11****, 13****, 31****, and 33**** are good, and the bad slice is a 352-score set. <br />
<br />
In particular, one cannot have two parallel bad corner slices. Combining this with Corollary 13, we see that two adjacent parallel good corner slices must necessarily have opposite types. Combining this with Corollary 17, we obtain an opposing pair of parallel good corner slices of the same type, which is not possible by Lemma 14. Thus a 355 point set cannot exist.<br />
<br />
We now refine the analysis further, with the aim of eliminating 354-point sets.<br />
<br />
'''Lemma 18''' Suppose the 11**** and 13**** slices are good with types xyzw and x'y'z'w' respectively. If x=x', then the 1*x*** slice has score at most 326 (defect at least 30), and the 1*X*** slice has score at most 348 (defect at least 8). Also, the 1**1**, 1**3**, 1***1*, 1***3*, 1****1, 1****3 slices have score at most 350 (defect at least 6). In particular, the total defect of slices beginning with 1* is at least 74.<br />
<br />
'''Proof''' The 1*x*** slice has eight "a" points, while the 1*X*** slice has only four. The other six slices mentioned have six "a" points and at most seven "d" points, and cannot be good by Corollary 12. The claim now follows from [http://spreadsheets.google.com/ccc?key=rwXB_Rn3Q1Zf5yaeMQL-RDw&hl=en the Pareto optimal list].<br />
<br />
'''Corollary 19''' A 354-point set cannot have all four slices 11****, 13****, 31****, and 33**** slices be good.<br />
<br />
'''Proof''' Suppose not. By Lemma 14, the 11**** and 33**** slices cannot be of the same type, and so they cannot both be of the opposite type to either 13**** or 31****. If 13**** is not of the opposite type to 11****, then by (a permutation of) Lemma 18, the total defect of slices beginning with 1* is at least 74; otherwise, if 13**** is not of the opposite type to 33****, then by (a permutation and reflection of) Lemma 18, the total defect of slices beginning with *3 is at least 74. Similarly, the total defect of slices beginning with 3* or *1 is at least 74, leading to a total defect of at least 148. But a 354 point set has a total defect of 120, contradiction.<br />
<br />
One can divide the 60 slices into fifteen families of four depending on their frozen coordinates, for instance one families of four is 11****, 13****, 31****, 33****. We'll call this family the ab**** family, and similarly define the a*b*** family, etc.<br />
<br />
'''Corollary 20''' A 354-point set can have at most one family with a total defect of at least 38.<br />
<br />
'''Proof''' Suppose there are two sets with defect at least 38. The remaining thirteen sets have defect at least 4 by Corollary 19, leading to a net defect of 38*2+13*4=128. But 354-point sets have a net defect of 120, a contradiction.<br />
<br />
'''Proposition 21''' A 354-point set cannot have any family with a total defect of at least 38.<br />
<br />
'''Proof''' Suppose for contradiction that the ab**** family (say) had a total defect of at least 38, then by Corollary 20 all other families have total defect at least 38.<br />
<br />
We claim that the **ab** family can have at most two good slices. Indeed, suppose the **ab** has three good slices, say **11**, **13**, **33**. By Lemma 14, the **11** and **33** slices cannot be of the same type, and so cannot both be of opposite type to **13**. Suppose **11** and **13** are not of opposite type. Then by (a permutation of) Lemma 18, one of the families a*1***, *a1***, **1*a*, **1**a has a net defect of at least 38, contradicting the normalisation. <br />
<br />
Thus each of the six families **ab**, **a*b*, **a**b, ***ab*, ***a*b have at least two bad slices. Meanwhile, the eight families a*b***, a**b**, a***b*, a****b, *ab***, *a*b**, *a**b*, *a***b have at least one bad slice by Corollary 19, leading to twenty bad slices in addition to the defect of at least 38 arising from the ab**** slice. To add up to a total defect of 120, this means that all bad slices outside of the ab**** family have a defect of four, with at most one exception; but then by Lemma 18 this shows that (for instance) the 1*1*** and 1*3*** slices cannot be good unless they are of opposite type. The previous argument then shows that the a*b*** slice cannot have three good slices, which increases the number of bad slices outside of ab**** to at least twenty-one, and now there is no way to add up to 120, a contradiction.<br />
<br />
From Proposition 21 and Lemma 18 we conclude<br />
<br />
'''Corollary 22''' In a 354-point set, the slices 11**** and 13**** can only be good if they have opposite type.<br />
<br />
From this and Lemma 14 we conclude<br />
<br />
'''Corollary 23''' In a 354-point set, every family can have at most two good slices, and thus have a defect of at least eight.<br />
<br />
Since 120 = 15*8, we conclude<br />
<br />
'''Corollary 23''' In a 354-point set, every family consists of ''exactly'' two good slices, plus two bad slices of defect exactly four.<br />
<br />
Now we look at the statistics of center slices of 4D slices (e.g. the 112*** stats of the 11**** slice). Computer search has revealed the following<br />
<br />
* For a (6,12,18,4,0) slice, the central slice must be (3,9,3,0). [The other two slices are (4,3,3,1) and (2,6,6,0).]<br />
* For a (5,12,18,4,0) slice, the central slice must be (3,9,3,0). [The other two slices must add up to (5,9,9,1).]<br />
* For a (5,12,12,4,1) slice, the central slice must be (3,6,3,1). [The other two slices must add up to (5,9,6,1).]<br />
* For a (6,8,12,8,0) slice, the central slice must be (2,6,6,0). [The other two slices must add up to (6,6,6,2).]<br />
<br />
Now we look at a side 3D slice, e.g. 111***. Because at least two of 11****, 13****, 31****, 33**** must be of type (6,12,18,4,0), and such slices have 3D side slices of (4,3,3,1) and (2,6,6,0), we know that the 111*** slice must be adjacent to either a (4,3,3,1) slice or a (2,6,6,0) slice. From the above list, we thus see that 111*** is one of the following possibilities:<br />
<br />
# It is a slice of a (6,12,18,4,0) set, and is therefore either (4,3,3,1) or (2,6,6,0).<br />
# It is a (3,3,3,1) slice of a (5,12,18,4,0) set, with the opposing slice being (2,6,6,0).<br />
# It is a (1,6,6,0) slice of a (5,12,18,4,0) set, with the opposing slice being (4,3,3,1).<br />
# It is a (3,3,0,1) slice of a (5,12,12,4,1) set, with the opposing slice being (2,6,6,0).<br />
# It is a (1,6,3,0) slice of a (5,12,12,4,1) set, with the opposing slice being (4,3,3,1).<br />
# It is a (2,3,3,1) slice of a (6,8,12,8,0) set, with the opposing slice being (4,3,3,1).<br />
<br />
Furthermore, in all six cases, the opposing slice is a subslice of a (6,12,18,4,0) set.<br />
<br />
Possibility 6 has been eliminated by computer, and possibilities 4,5 have been eliminated by hand. As a consequence, the only possible 3D slices are (4,3,3,1), (2,6,6,0), (3,3,3,1), and (1,6,6,0). This eliminates the (5,12,12,4,1) case.<br />
<br />
The (6,8,12,8,0) case can also be eliminated: if (say) 11**** is (6,8,12,8,0), then 111*** and 113*** must now be (3,3,3,1), which forces the 1*1*** and 1*3*** slices to be bad, and hence the 3*1*** and 3*3*** slices are good, hence 311*** and 313*** must be (2,6,6,0) (note that we have no way of matching (3,3,3,1) to (4,3,3,1)), but then there is no consistent choice for the 31**** slice.<br />
<br />
The only possible side slices of a (5,12,18,4,0) set are (4,3,3,1)+(1,6,6,0) and (3,3,3,1)+(2,6,6,0). In particular, it contains exactly one of any opposing "d" points (e.g. 1222 and 3222). Similarly for (6,12,18,4,0) sets. We conclude that the same is true for the 6D set, e.g. exactly one of 111222 and 113222 lie in the set.<br />
<br />
We can define the notion of the "type" of a (5,12,18,4,0) set just as with a (6,12,18,4,0) set. The above observation forces any two adjacent slices (e.g. 11**** and 13****) to have opposing type, and thus any two diagonally opposite slices (e.g. 11**** and 33****) have identical type. By Lemma 14, such slices cannot both be good. Splitting each family into two pairs of diagonally opposing slices, we thus conclude<br />
<br />
'''Proposition 24''' In any two diagonally opposite slices of a 354-point set, one slice is (6,12,18,4,0) and the other is (5,12,18,4,0), and both slices have the same type.<br />
<br />
On averaging, we see that the 354-point set has statistics (22,72,180,80,0,0,0).<br />
<br />
'''Theorem 25''' <math>c'_6 = 353</math>.<br />
<br />
'''Proof''' We repeat the previous argument that there is no 356-point set. Suppose for contradiction that we have a 356-point set, i.e. one in which all corner slices are good. From the above discussion, we see that if the 111*** slice contains its center 111222, then the 113*** slice does not, and vice versa. Thus the d points of the form ***222 consist either of those strings with an even number of 1s, or those with an odd number of 1s.<br />
<br />
Let’s say it’s the former, thus the set contains 111222, 133222, and permutations of the first three coordinates, but omits 113222, 333222 and permutations of the first three coordinates. Meanwhile, we see that the (22,72,180,80,0,0,0) set saturates the inequality 4c+3d \leq 960, which is only possible if every line connecting two “c” points to a “d” point meets exactly two elements in the set. Since the “d” points 113222, 333222 are omitted, we conclude that the “c” points 113122, 113322, 333122, 333322 must lie in the set, and similarly for permutations of the first three and last three coordinates. But this gives at least 15 of the 16 "c" points ending in 22; by symmetry this leads to 225 c-points in all, which is far too many.<br />
<br />
== Connection with Fujimura's Problem ==<br />
Recall the Gamma cells, where <math>\Gamma_{a,b,c}</math> is the set of points whose coordinates include a 1s, b 2s and c 3s. When the Moser problem is specialised to unions of cells, it becomes a Fujimura-type problem but in which one has to forbid isosceles triangles <math> (a+r,b,c+s), (a,b+r+s,c), (a+s,b,c+r)</math> rather than equilateral triangles, and each point <math>(a,b,c)</math> is weighted by the cell cardinality <math>\frac{(a+b+c)!}{a!b!c!}</math>. (In particular, one has to exclude vertical line segments <math>(a+r,b,c+r),(a,b+2r,c)</math>, thus each value of <math>c-a</math> can have at most one cell.<br />
<br />
An integer program was run to find solutions of the Moser problem which were sets of Gamma cells. The resulting lower bounds it found, for various dimensions n, are 2, 6, 16, 43, 122, 353, 1017, 2902, 8622, 24786, 71766, 212423, 614875. More information on these results are stored [http://abel.math.umu.se/~klasm/Data/HJ/ here]. Note that the optimal values are found for n=1 to 4, but the value 122 is below the n=5 optimum value of 124.<br />
<br />
== General n ==<br />
<br />
General solution for <math>c'_N</math>. For any q, the union of the following sets is a Moser set. The size of this Moser set is maximized when q is near N/3, in which case it is <math>O(3^n/\sqrt{n})</math>. Most of the points are in the layers with q 2s and q-1 2s.<br />
<br />
* q 2s, all points from A(N-q,1)<br />
* q-1 2s, points from A(N-q+1,2)<br />
* q-2 2s, points from A(N-q+2,3)<br />
* etc.<br />
<br />
where A(m,d) is a subset of <math>[1,3]^m</math> for which any two points differ from each other in at least d places.<br />
<br />
Mathworld’s entry on error-correcting codes suggests it might be NP-complete to find the maximum size of A(m,d) in general. However, the size of A(m,d) can be bounded by sphere-packing arguments. For example, points in A(m,3) are surrounded by non-intersecting spheres of Hamming radius 1, and points in A(m,5) are surrounded by non-intersecting spheres of Hamming radius 2.<br />
<br />
* <math>|A(m,1)| = 2^m</math> because it includes all points in <math>[1,3]^m</math><br />
* <math>|A(m,2)| = 2^{m-1}</math> because it can include all points in <math>[1,3]^m</math> with an odd number of ones.<br />
* <math>|A(m,3)| \le 2^m/(m+1)</math> because the size of a Hamming sphere is m+1.<br />
<br />
The integer programming routine from Maple 12 was used to obtain upper bounds for <math>c'_6</math> and <math>c'_7</math>. A large number of linear inequalities, such as those described above in sections (n=3) and (n=4), were combined. The details are in [[Maple calculations]]. The results were that <math>c'_6 \le 356</math> and <math>c'_7 \le 1041</math>.<br />
<br />
A [[genetic algorithm]] has provided the following examples:<br />
<br />
* <math>c'_6 \geq 353</math> (26 examples; [http://twofoldgaze.wordpress.com/2009/03/10/353-element-solution/ here is one])<br />
* <math>c'_7 \geq 988</math> [http://twofoldgaze.wordpress.com/2009/03/10/978-element-solution/ Here is the example]<br />
<br />
One of these 353-point genetic results turned out to be built entirely from Gamma sets (Recall Gamma(a,b,c) is those points with a 1s, b 2s and c 3s.) The Gamma sets were (6,0,0),(5,0,1),(3,3,0), (3,2,1),(3,1,2),(2,2,2),(1,3,2),(0,3,3),(2,0,4),(0,2,4),(0,1,5). This led to a search for solutions built from complete Gamma sets, in the Fujimura-type problem described above.<br />
<br />
For general N, the following collection of Gamma sets <math>\Gamma_{a,b,c}</math> is roughly one third better than the A sets described above when n is large:<br />
<br />
* <math> \Gamma(a,q,N-a-q) </math> when <math>a\ne 2 (mod 3)</math><br />
* <math> \Gamma(a,q-1,N+1-a-q) </math> when <math>a\ne 1 (mod 3)</math><br />
* <math> \Gamma(a,q-2,N+2-a-q) </math> when <math>a = 0 (mod 3)</math><br />
* <math> \Gamma(a,q-3,N+3-a-q) </math> when <math>a = 2 (mod 3)</math><br />
<br />
Extra points may be found in the following way: Suppose one has an unused cell (a+r,b,c+r) which forms a degenerate isosceles triangle with another cell (a,b+2r,c) that is being fully used, but is otherwise not forming any triangles with existing cells. Then there is a little bit of room to add a few more points: namely, one can add any subset of to the set without creating lines as long as no two points in that subset are a Hamming distance of exactly 2r apart. For instance one can certainly add a single point to the Gamma example without difficulty.<br />
<br />
For N=5, one can start with the 122-point solution (0 0 5), (1 1 3), (0 2 3 ), (1 2 2 ), (2 1 2 ), (2 2 1 ),(3 2 0 ), (3 1 1), (5 0 0 ). Add a single point from each of (1,0,4) and (4,0,1), to give a maximal 124-point solution.<br />
<br />
For N=10, twelve points from Gamma(5,0,5) can be added to the 24786-point solution, to give a 24798-point solution. No two of these twelve points are Hamming distance 4 from each other, so no triangles are made with Gamma(3,4,3) which is part of the 24786-point solution.<br />
<br />
== Larger sides (k>3) ==<br />
<br />
The following set gives a lower bound for Moser’s cube <math>[4]^n</math> (values 1,2,3,4): Pick all points where q entries are 2 or 3; and also pick those where q-1 entries are 2 or 3 and an odd number of entries are 1. This is maximized when q is near n/2, giving a lower bound of<br />
<br />
<math>\binom{n}{\lfloor n/2\rfloor} 2^n + \binom{n}{\lfloor n/2\rfloor-1} 2^{n-1}</math><br />
<br />
The following set improves this bound to <math>(2+o(1))\binom{n}{\lfloor n/2\rfloor} 2^n</math>. It includes four half-layers.<br />
<br />
Pick all points with a 1s, b 2s, c 3s, d 4s, for which<br />
* a+d = q or q-1, a and b have same parity<br />
* a+d = q-2 or q-3, a and b have opposite parity<br />
<br />
which is comparable to <math>\sqrt{\frac{8}{n\pi}}4^n</math> by [[Stirling's formula]].<br />
<br />
For k=5: If A, B, C, D, and E denote the numbers of 1-s, 2-s, 3-s, 4-s and 5-s then the first three points of a geometric line form a 3-term arithmetic progression in A+E+2(B+D)+3C. So, for k=5 we have a similar lower bound for the Moser’s problem as for DHJ k=3, i.e. <math>5^{n - O(\sqrt{\log n})}</math>.<br />
<br />
The k=6 version of Moser implies DHJ(3). Indeed, any k=3 combinatorial line-free set can be "doubled up" into a k=6 geometric line-free set of the same density by pulling back the set from the map <math>\phi: [6]^n \to [3]^n</math> that maps 1, 2, 3, 4, 5, 6 to 1, 2, 3, 3, 2, 1 respectively; note that this map sends k=6 geometric lines to k=3 combinatorial lines.</div>Charles R Greathouse IVhttps://asone.ai/polymath/index.php?title=Upper_and_lower_boundsUpper and lower bounds2015-09-30T04:26:40Z<p>Charles R Greathouse IV: fix OEIS link</p>
<hr />
<div><center>'''Upper and lower bounds for <math>c_n</math> for small values of n.'''</center><br />
<br />
<math>c_n</math> is the size of the largest subset of <math>[3]^n</math> that does not contain a combinatorial line (OEIS [http://oeis.org/A156762 A156762]. A spreadsheet for all the latest bounds on <math>c_n</math> [http://spreadsheets.google.com/ccc?key=p5T0SktZY9DsU-uZ1tK7VEg can be found here]. In this page we record the proofs justifying these bounds. See also this page on [[higher-dimensional DHJ numbers]].<br />
<br />
<br />
{|<br />
| n || 0 || 1 || 2 || 3 || 4 || 5 || 6 || 7<br />
|-<br />
| <math>c_n</math> || 1 || 2 || 6 || 18 || 52 || 150 || 450 || [1302,1348]<br />
|}<br />
<br />
== Basic constructions ==<br />
<br />
For all <math>n \geq 1</math>, a basic example of a mostly line-free set is<br />
<br />
:<math>D_n := \{ (x_1,\ldots,x_n) \in [3]^n: \sum_{i=1}^n x_i \neq 0 \ \operatorname{mod}\ 3 \}</math>. (1)<br />
<br />
This has cardinality <math>|D_n| = 2 \times 3^{n-1}</math>. The only lines in <math>D_n</math> are those with<br />
<br />
# A number of wildcards equal to a multiple of three;<br />
# The number of 1s unequal to the number of 2s modulo 3.<br />
<br />
One way to construct line-free sets is to start with <math>D_n</math> and remove some additional points. We also have the variants <math>D_{n,0}=D_n, D_{n,1}, D_{n,2}</math> defined as<br />
<br />
:<math>D_{n,j} := \{ (x_1,\ldots,x_n) \in [3]^n: \sum_{i=1}^n x_i \neq j \ \operatorname{mod}\ 3 \}</math>. (1')<br />
<br />
When n is not a multiple of 3, then <math>D_{n,0}, D_{n,1}, D_{n,2}</math> are all cyclic permutations of each other; but when n is a multiple of 3, then <math>D_{n,0}</math> plays a special role (though <math>D_{n,1}, D_{n,2}</math> are still interchangeable).<br />
<br />
Another useful construction proceeds by using the slices <math>\Gamma_{a,b,c} \subset [3]^n</math> for <math>(a,b,c)</math> in the triangular grid<br />
<br />
:<math>\Delta_n := \{ (a,b,c) \in {\Bbb Z}_+^3: a+b+c = n \},</math>. (2)<br />
<br />
where <math>\Gamma_{a,b,c}</math> is defined as the strings in <math>[3]^n</math> with <math>a</math> 1s, <math>b</math> 2s, and <math>c</math> 3s. Note that<br />
<br />
:<math>|\Gamma_{a,b,c}| = \frac{n!}{a! b! c!}.</math> (3)<br />
<br />
Given any set <math>B \subset \Delta_n</math> that avoids equilateral triangles <math> (a+r,b,c), (a,b+r,c), (a,b,c+r)</math>, the set<br />
<br />
:<math>\Gamma_B := \bigcup_{(a,b,c) \in B} \Gamma_{a,b,c}</math> (4)<br />
<br />
is line-free and has cardinality<br />
<br />
:<math>|\Gamma_B| = \sum_{(a,b,c) \in B} \frac{n!}{a! b! c!},</math> (5)<br />
<br />
and thus provides a lower bound for <math>c_n</math>:<br />
<br />
:<math>c_n \geq \sum_{(a,b,c) \in B} \frac{n!}{a! b! c!}.</math> (6)<br />
<br />
All lower bounds on <math>c_n</math> have proceeded so far by choosing a good set of B and applying (6). Note that <math>D_n</math> is the same as <math>\Gamma_{B_n}</math>, where <math>B_n</math> consists of those triples <math>(a,b,c) \in \Delta_n</math> in which <math>a \neq b\ \operatorname{mod}\ 3</math>.<br />
<br />
Note that if one takes a line-free set and permutes the alphabet <math>\{1,2,3\}</math> in any fashion (e.g. replacing all 1s by 2s and vice versa), one also gets a line-free set. This potentially gives six examples from any given starting example of a line-free set, though in practice there is enough symmetry that the total number of examples produced this way is less than six. (These six examples also correspond to the six symmetries of the triangular grid <math>\Delta_n</math> formed by rotation and reflection.)<br />
<br />
Another symmetry comes from permuting the <math>n</math> indices in the strings of <math>[3]^n</math> (e.g. replacing every string by its reversal). But the sets <math>\Gamma_B</math> are automatically invariant under such permutations and thus do not produce new line-free sets via this symmetry.<br />
<br />
== The basic upper bound ==<br />
<br />
Because <math>[3]^{n+1}</math> can be expressed as the union of three copies of <math>[3]^n</math>, we have the basic upper bound<br />
<br />
:<math>c_{n+1} \leq 3 c_n.</math> (7)<br />
<br />
Note that equality only occurs if one can find an <math>n+1</math>-dimensional line-free set such that every n-dimensional slice has the maximum possible cardinality of <math>c_n</math>.<br />
<br />
== n=0 ==<br />
<br />
:<math>c_0=1</math>:<br />
<br />
This is clear.<br />
<br />
== n=1 ==<br />
<br />
:<math>c_1=2</math>:<br />
<br />
The three sets <math>D_1 = \{1,2\}</math>, <math>D_{1,1} = \{2,3\}</math>, and <math>D_{1,2} = \{1,3\}</math> are the only two-element sets which are line-free in <math>[3]^1</math>, and there are no three-element sets.<br />
<br />
== n=2 ==<br />
<br />
:<math>c_2=6</math>:<br />
<br />
There are four six-element sets in <math>[3]^2</math> which are line-free, which we denote <math>x = D_{2,2}</math>, <math>y=D_{2,1}</math>, <math>z=D_2</math>, and <math>w</math> and are displayed graphically as follows.<br />
<br />
13 .. 33 .. 23 33 13 23 .. 13 23 ..<br />
x = 12 22 .. y = 12 .. 32 z = .. 22 32 w = 12 .. 32<br />
.. 21 31 11 21 .. 11 .. 31 .. 21 31<br />
<br />
Combining this with the basic upper bound (7) we see that <math>c_2=6</math>.<br />
<br />
== n=3 ==<br />
<br />
:<math>c_3=18</math>:<br />
<br />
We describe a subset <math>A</math> of <math>[3]^3</math> as a string <math>abc</math>, where <math>a, b, c \subset [3]^2</math> correspond to strings of the form <math>1**</math>, <math>2**</math>, <math>3**</math> in <math>[3]^3</math> respectively. Thus for instance <math>D_3 = xyz</math>, and so from (7) we have <math>c_3=18</math>.<br />
<br />
'''Lemma 1.'''<br />
* The only 18-element line-free subset of <math>[3]^3</math> is <math>D_3 = xyz</math>.<br />
* The only 17-element line-free subsets of <math>[3]^3</math> are formed by removing a point from <math>D_3=xyz</math>, or by removing either 111, 222, or 333 from <math>D_{3,2} = yzx</math> or <math>D_{3,3}=zxy</math>.<br />
<br />
'''Proof'''. We prove the second claim. As <math>17=6+6+5</math>, and <math>c_2=6</math>, at least two of the slices of a 17-element line-free set must be from x, y, z, w, with the third slice having 5 points. If two of the slices are identical, the last slice can have only 3 points, a contradiction. If one of the slices is a w, then the 5-point slice will contain a diagonal, contradiction. By symmetry we may now assume that two of the slices are x and y, which force the last slice to be z with one point removed. Now one sees that the slices must be in the order xyz, yzx, or zxy, because any other combination has too many lines that need to be removed. The sets yzx, zxy contain the diagonal {111,222,333} and so one additional point needs to be removed. <br />
<br />
The first claim follows by a similar argument to the second.<br />
<math>\Box</math><br />
<br />
== n=4 ==<br />
<br />
:<math>c_4=52</math>:<br />
<br />
Indeed, divide a line-free set in <math>[3]^4</math> into three blocks <math>1***, 2***, 3***</math> of <math>[3]^3</math>. If two of them are of size 18, then they must both be xyz, and the third block can have at most 6 elements, leading to an inferior bound of 42. So the best one can do is <math>18+17+17=52</math> which can be attained by deleting the diagonal {1111,2222,3333} from <math>D_{4,1} = xyz\ yzx\ xzy</math>, <math>D_4 = yzx\ zxy\ xyz</math>, or <math>D_{4,2} = zxy\ xyz\ yzx</math>. In fact,<br />
<br />
'''Lemma 2.'''<br />
<br />
* The only 52-element line-free sets in <math>[3]^4</math> are formed by removing the diagonal {1111,2222,3333} from <math>D_{4,j}</math> for some j=0,1,2.<br />
* The only 51-element line-free sets in <math>[3]^4</math> are formed by removing the diagonal and one further point from <math>D_{4,j}</math> for some j=0,1,2.<br />
* The only 50-element line-free sets in <math>[3]^4</math> are formed by removing the diagonal and two further points from <math>D_{4,j}</math> for some j=0,1,2 OR is equal to one of the three permutations of the set <math>X := \Gamma_{3,1,0} \cup \Gamma_{3,0,1} \cup \Gamma_{2,2,0} \cup \Gamma_{2,0,2} \cup \Gamma_{1,1,2} \cup \Gamma_{1,2,1} \cup \Gamma_{0,2,2}</math>.<br />
<br />
'''Proof''' It suffices to prove the third claim. In fact it suffices to show that every 50-point line-free set is either contained in the 54-point set <math>D_{4,j}</math> for some j=0,1,2, or is some permutation of the set X. Indeed, if a 50-point line-free set is contained in, say, <math>D_4</math>, then it cannot contain 2222, since otherwise it must omit one point from each of the four pairs formed from {2333, 2111} by permuting the indices, and must also omit one of {1111, 1222, 1333}, leading to at most 49 points in all; similarly, it cannot contain 1111, and so omits the entire diagonal {1111,2222,3333}, with two more points to be omitted. Similarly when <math>D_4</math> is replaced by one of the other <math>D_{4,j}</math><br />
<br />
Next, observe that every three-dimensional slice of a line-free set can have at most <math>c_3=18</math> points; thus when one partitions a 50-point line-free set into three such slices, it must divide either as 18+16+16, 18+17+15, 17+17+16, or some permutation of these.<br />
<br />
Suppose that we can slice the set into two slices of 17 points and one slice of 16 points. By the various symmetries, we may assume that the 1*** slice and 2*** slices have 17 points, and the 3*** slice has 16 points. By Lemma 1, the 1-slice is <math>\{1\} \times D_{3,j}</math> with one point removed, and the 2-slice is <math>\{2\} \times D_{3,k}</math> with one point removed, for some <math>j,k \in \{0,1,2\}</math>.<br />
<br />
If j=k, then the 1-slice and 2-slice have at least 15 points in common, so the 3-slice can have at most <math>27-15=12</math> points, a contradiction. If jk = 01, 12, or 20, then observe that from Lemma 1 the *1**, *2**, *3** slices cannot equal a 17-point or 18-point line-free set, so each have at most 16 points, leading to only 48 points in all, a contradiction. Thus we must have jk = 10, 21, or 02.<br />
<br />
Let's first suppose that jk=02. Then by Lemma 1, the 2*** slice contains the nine points formed from {2211, 2322, 2331} and permuting the last three indices, while the 1*** slice contains at least eight of the nine points formed from {1211, 1322, 1311} and permuting the last three indices. Thus the 3*** slice can contain at most one of the nine points formed from {3211, 3322, 3311} and permuting the last three indices. If it does contain one of these points, say 3211, then it must omit one point from each of the four pairs {3222, 3233}, {3212, 3213}, {3221, 3231}, {3111, 3311}, leading to at most 15 points on this slice, a contradiction. So the 3*** slice must omit all nine points, and is therefore contained in <math>\{3\} \times D_{4,1}</math>, and so the 50-point set is contained in <math>D_{4,1}</math>, and we are done by the discussion at the beginning of the proof.<br />
<br />
The case jk=10 is similar to the jk=02 case (indeed one can get from one case to the other by swapping the 1 and 2 indices). Now suppose instead that jk=12. Then by Lemma 1, the 1*** slice contains the six points from permuting the last three indices of 1123, and similarly the 2*** slice contains the six points from permuting the last three indices of 2123. Thus the 3*** slice must avoid all six points formed by permuting the last three indices of 3123. Similarly, as 1133 lies in the 1*** slice and 2233 lies in the 2*** slice, 3333 must be avoided in the 3*** slice.<br />
<br />
Now we claim that 3111 must be avoided also; for if 3111 was in the set, then one point from each of the six pairs formed from {3311, 3211}, {3331, 3221} and permuting the last three indices must lie outside the 3*** slice, which reduces the size of that slice to at most <math>27-6-1-6=14</math>, which is too small. Similarly, 3222 must be avoided, which puts the 3*** slice inside <math>\{3\} \times D_3</math> and then places the 50-point set inside <math>D_4</math>, and we are done by the discussion at the beginning of the proof.<br />
<br />
We have handled the case in which at least one of the slicings of the 50-point set is of the form 50=17+17+16. The only remaining case is when all slicings of the 50-point set are of the form 18+17+15 or 18+16+16 (or a permutation thereof). By the symmetries of the situation, we may assume that the 1*** slice has 18 points, and thus by Lemma 1 takes the form <math>\{1\} \times D_3</math>. Inspecting the *1**, *2**, *3** slices, we then see (from Lemma 1) that only the *1** slice can have 18 points; since we are assuming that this slicing is some permutation of 50=18+17+16, we conclude that the *1** slice must have exactly 18 points, and is thus described precisely by Lemma 1. Similarly for the **1* and ***1 slices. Indeed, by Lemma 1, we see that the 50-point set must agree exactly with <math>D_{4,1}</math> on any of these slices. In particular, on the remaining portion <math>\{2,3\}^4</math> of the cube, there are exactly 6 points of the 50-point set in <math>\{2,3\}^4</math>.<br />
<br />
Suppose that 3333 was in the set; then since all permutations of 3311, 3331 are known to lie in the set, then 3322, 3332 must lie outside the set. Also, as 1222 lies in the set, at least one of 2222, 3222 lie outside the set. This leaves only 5 points in <math>\{2,3\}^4</math>, a contradiction. Thus 3333 lies outside the set; similarly 2222 lies outside the set.<br />
<br />
Let a be the number of points in the 50-point set which are some permutation of 2233, thus <math>0 \leq a \leq 6</math>. If a=0 then the set lies in <math>D_{4,1}</math> and we are done. If a=6 then the set is exactly X and we are done. Now suppose a=1,2,3. By symmetry we may assume that 2233 lies in the set. Then (since 2133, 1233 2231, 2213 are known to lie in the set) 2333, 3233, 2223, 2232 lie outside the set, which leaves at most 5 points inside <math>\{2,3\}^4</math>, a contradiction.<br />
<br />
The remaining case is when a=4,5. Then one of the three pairs {2233, 3322}, {2323, 3232}, {2332, 3223} lie in the set. By symmetry we may assume that {2233, 3322} lie in the set. Then by arguing as before we see that all eight points formed by permuting 2333 or 3222 lie outside the set, leading to at most 5 points inside <math>\{2,3\}^4</math>, a contradiction.<br />
<math>\Box</math><br />
<br />
== n=5 ==<br />
<br />
:<math>c_5=150</math>:<br />
<br />
'''Lemma 3'''. Any line-free subset of <math>D_{5,j}</math> can have at most 150 points.<br />
<br />
'''Proof'''. By rotation we may work with <math>D_5</math>. This set has 162 points. By looking at the triplets {10000, 11110, 12220} and cyclic permutations we must lose 5 points; similarly from the triplets {20000,22220, 21110} and cyclic permutations. Finally from {11000,11111,11222} and {22000,22222,22111} we lose two more points. <math>\Box</math><br />
<br />
Equality can be attained by removing <math>\Gamma_{0,4,1}, \Gamma_{0,5,0}, \Gamma_{4,0,1}, \Gamma_{5,0,0}</math> from <math>D_5</math>. Thus <math>c_5 \geq 150</math>.<br />
<br />
Another pattern of 150 points is this: Take the 450 points<br />
in <math>{}[3]^6</math> which are (1,2,3), (0,2,4) and permutations,<br />
then select the 150 whose final coordinate is 1. That gives<br />
this many points in each cube:<br />
<br />
17 18 17<br />
<br />
17 17 18<br />
<br />
12 17 17<br />
<br />
'''Lemma 4'''. A line-free subset of <math>[3]^5</math> with over 150 points cannot have two parallel <math>[3]^4</math> slices, each of which contain at least 51 points.<br />
<br />
'''Proof'''. Suppose not. By symmetry, we may assume that the 1**** and 2**** slices have at least 51 points, and that the whole set has at least 151 points, which force the third slice to have at least <math>151-2c_4 = 47</math> points.<br />
<br />
By Lemma 2, the 1**** slice takes the form <math>\{1\} \times D_{4,j}</math> for some <math>j=0,1,2</math> with the diagonal {11111,12222,13333} and possibly one more point removed, and similarly the 2**** slice takes the form <math>\{2\} \times D_{4,k}</math> for some <math>k=0,1,2</math> with the diagonal {21111,22222,23333} and possibly one more point removed.<br />
<br />
Suppose first that j=k. Then the 1-slice and 2-slice have at least 50 points in common, leaving at most 31 points for the 3-slice, a contradiction. Next, suppose that jk=01. Then observe that the *i*** slice cannot look like any of the configurations in Lemma 2 and so must have at most 50 points for i=1,2,3, leading to 150 points in all, a contradiction. Similarly if jk=12 or 20. Thus we must have jk equal to 10, 21, or 02.<br />
<br />
Let's suppose first that jk=10. The first slice then is equal to <math>\{1\} \times D_{4,1}</math> with the diagonal and possibly one more point removed, while the second slice is equal to <math>\{2\} \times D_{4,0}</math> with the diagonal and possibly one more point removed. Superimposing these slices, we thus see that the third slice is contained in <math>\{3\} \times D_{4,2}</math> except possibly for two additional points, together with the one point 32222 of the diagonal that lies outside of <math>\{3\} \times D_{4,2}</math>.<br />
<br />
The lines x12xx, x13xx (plus permutations of the last four digits) must each contain one point outside the set. The first two slices can only absorb two of these, and so at least 14 of the 16 points formed by permuting the last four digits of 31233, 31333 must lie outside the set. These points all lie in <math>\{3\} \times D_{4,2}</math>, and so the 3**** slice can have at most <math>|D_{4,2}|-14+3=43</math> points, a contradiction.<br />
<br />
The case jk=02 is similar to the case jk=10 (indeed one can obtain one from the other by swapping 1 and 2). Now we turn to the case jk=21. Arguing as before we see that the third slice is contained in <math>\{3\} \times D_4</math> except possibly for two points, together with 33333. <br />
<br />
If 33333 was in the set, then each of the lines xx333, xxx33 (and permutations of the last four digits) must have a point missing from the first two slices, which cannot be absorbed by the two points we are permitted to remove; thus 33333 is not in the set. For similar reasons, 33331 is not in the set, as can be seen by looking at xxx31 and permutations of the last four digits. Indeed, any string containing four threes does not lie in the set; this means that at least 8 points are missing from <math>\{3\} \times D_4</math>, leaving only at most 46 points inside that set. Furthermore, any point in the 3**** slice outside of <math>\{3\} \times D_4</math> can only be created by removing a point from the first two slices, so the total cardinality is at most <math>46+52+52 = 150</math>, a contradiction.<math>\Box</math><br />
<br />
'''Corollary'''. <math>c_5 \leq 152</math><br />
<br />
'''Proof'''. By Lemma 4 and the bound <math>c_4=52</math>, any line-free set with over 150 points can have one slice of cardinality 52, but then the other two slices can have at most 50 points. <math>\Box</math><br />
<br />
<br />
'''Lemma 5''' Any solution with 151 or more points has a slice with at most 49 points.<br />
<br />
'''Proof''' Suppose we have 151 points without a line, and each of three slices has at least 50 points.<br />
<br />
Using earlier notation, we split subsets of <math>[3]^4</math> into nine subsets of <math>[3]^2</math>. <br />
So we think of x,y,z,a,b and c as subsets of a square. Each slice is one of the following.<br />
*<math>D_4 = y'zx,zx'y,xyz</math> (with one or two points removed)<br />
*<math>D_{4,2} = z'xy,xyz,yzx'</math> (with one or two points removed)<br />
*<math>D_{4,1} = xyz,yz'x,zxy'</math> (with one or two points removed)<br />
*<math>X = xyz, ybw, zwc</math><br />
*<math>Y = axw, xyz, wzc</math><br />
*<math>Z = awx, wby, xyz</math><br />
<br />
where a, b and c have four points each.<br />
<br />
.. 32 33 31 .. 33 .. .. ..<br />
a = .. 22 23 b = .. .. .. c = 21 22 ..<br />
.. .. .. 11 .. 13 11 12 ..<br />
<br />
x', y' and z' are subsets of x, y and z respectively, and have five points each.<br />
<br />
Suppose all three slices are subsets of <math>D_{4,j}</math>. <br />
We can remove at most five points from the full set of three D_{4,j}. <br />
Consider columns 2,3,4,6,7,8. At most two of these columns contain xyz, so one point must be removed from the other four.<br />
This uses up all but one of the removals.<br />
So the slices must be <math>D_{4,2},D_{4,1},D_{4,0}</math> or a cyclic permutation of that.<br />
Then the cube, which contains the first square of slice 1; the fifth square of slice 2; <br />
and the ninth square of slice 3, contains three copies of the same square. <br />
It takes more than one point removed to remove all lines from that cube.<br />
So we can't have all three slices subsets of <math>D_{4,j}</math>.<br />
<br />
Suppose one slice is X,Y or Z, and two others are subsets of <math>D_{4,j}</math>. <br />
We can remove at most three points from the full <math>D_{4,j}</math><br />
By symmetry, suppose one slice is X. Consider columns 2,3,4 and 7. They must be cyclic permutations of x,y,z,<br />
and two of them are not xyz, so must lose a point. <br />
Columns 6 and 8 must both lose a point, and we only have 150 points left.<br />
So if one slice is X,Y or Z, the full set contains a line.<br />
<br />
Suppose two slices are from X,Y and Z, and the other is a subset of <math>D_{4,j}</math>. <br />
By symmetry, suppose two slices are X and Y. Columns 3,6,7 and 8 all contain w, and therefore at most 16 points each.<br />
Columns 1,5 and 9 contain a,b, or c, and therefore at most 16 points. <br />
So the total number of points is at most 7*16+2*18 = 148. This contradicts the assumption of 151 points.<br />
<math>\Box</math><br />
<br />
'''Corollary''' <math>c_5 \leq 151 </math><br />
<br />
'''Proof''' By Lemmas 2 and 4, the maximum number of points is 52+50+49=151. <math>\Box</math><br />
<br />
'''Lemma 5.1''' No solution with 151 points contains as a slice the X defined in Lemma 2<br />
<br />
'''Proof''' Suppose one row is X. Another row is <math>D_{4,j}</math>.<br />
<br />
Suppose X is in the first row. Label the other rows with letters from the alphabet.<br />
<br />
xyz ybw zwc<br />
<br />
mno pqr stu<br />
<br />
def ghi jkl<br />
<br />
Reslice the array into a left nine, middle nine and right nine. One of these squares<br />
contains 52 points, and it can only be the left nine. One of its three columns contains<br />
18 points, and it can only be its left-hand column, xmd. So m=y and d=z. But none of the {math>D_{4,j}</math> begins with y or z, which is a contradiction. So X is not in the first row.<br />
<br />
So X is in the second or third row. By symmetry, suppose it is in the second row<br />
<br />
def ghi jkl<br />
<br />
xyz ybw zwc<br />
<br />
mno pqr stu<br />
<br />
Again, the left-hand nine must contain 52 points, so it is <math>D_{4,2}</math>.<br />
So either the first row is <math>D_{4,2}</math> or the third row is <math>D_{4,0}</math>.<br />
If the first row is <math>D_{4,2}</math> then the only way to have 50 points in the middle or right-hand nine is if the middle nine is X<br />
<br />
z'xy xyz yzx'<br />
<br />
xyz ybw zwc<br />
<br />
yzx' zwc stu<br />
<br />
In the seventh column, s contains 5 points and in the eighth column, t contains 4 points.<br />
The final row can now contain at most 48 points, and the whole array contains only 52+50+48 = 150 points.<br />
<br />
If the third row is <math>D_{4,0}</math>, then neither the middle nine nor the right-hand nine contains 50 points, by the classification of Lemma 4 and the formulas at the start of Lemma 5.<br />
Again, only 52+49+49 = 150 points are possible.<br />
<br />
A similar argument is possible if X is in the third row; or if X is replaced by Y or Z.<br />
<br />
So when a 151-point set is sliced into three, one slice is <math>D_{4,j}</math> and another slice is 50 points contained in <math>D_{4,k}</math>. <math>\Box</math><br />
<br />
'''Lemma 5.2''' There is no 151-point solution<br />
<br />
'''Proof''' Assume by symmetry that the first row contains 52 points and the second row contains 50.<br />
<br />
If <math>D_{4,1}</math> is in the first row, then the second row must be contained in <math>D_{4,0}</math>. <br />
<br />
xyz yz'x zxy'<br />
<br />
y'zx zx'y xyz<br />
<br />
def ghi jkl<br />
<br />
But then none of the left nine, middle nine or right nine can contain 52 points, which contradicts the corollary to Lemma 5.<br />
<br />
Suppose the first row contains D_{4,0}. Then the second row is contained in <math>D_{4,2}</math>, otherwise the cubes formed from the nine columns of the diagram would need to remove too many points.<br />
<br />
y'zx zx'y xyz<br />
<br />
z'xy xyz yzx'<br />
<br />
def ghi jkl<br />
<br />
But then neither the left nine, middle nine or right nine contains 52 points.<br />
<br />
So the first row contains <math>D_{4,2}</math>, and the second row is contained in <math>D_{4,1}</math>. Two points may be removed from the second row of this diagram.<br />
<br />
z'xy xyz yzx'<br />
<br />
xyz yz'x zxy'<br />
<br />
def ghi jkl<br />
<br />
Slice it into the left nine, middle nine and right nine. Two of them are contained in <math>D_{4,j}</math><br />
so at least two of def, ghi, and jkl are contained in the corresponding slice of <math>D_{4,0}</math>.<br />
Slice along a different axis, and at least two of dgj,ehk,fil are contained in the corresponding slice of <br />
<math>D_{4,0}</math>. <br />
So eight of the nine squares in the bottom row are contained in the corresponding square of <math>D_{4,0}</math>.<br />
Indeed, slice along other axes, and all points except one are contained within <math>D_{4,0}</math>. <br />
This point is the intersection of all the 49-point slices. <br />
<br />
So, if there is a 151-point solution, then after removal of the specified point, <br />
there is a 150-point solution, within <math>D_{5,j}</math>, whose slices in each direction are 52+50+48.<br />
<br />
z'xy xyz yzx'<br />
<br />
xyz yz'x zxy'<br />
<br />
y'zx zx'y xyz<br />
<br />
One point must be lost from columns 3, 6, 7 and 8, and four more from the major diagonal z'z'z. That leaves 148 points instead of 150.<br />
<br />
So the 150-point solution does not exist with 52+50+48 slices; so the 151 point solution does not exist.<math>\Box</math><br />
<br />
<br />
An integer programming method has established the upper bound <math>c_5\leq 150</math>, with 12 extremal solutions.<br />
<br />
[http://abel.math.umu.se/~klasm/extremal-c5 This file] contains the extermisers. One point per line and different extermisers separated by a line with “—”<br />
<br />
[http://abel.math.umu.se/~klasm/linprog-d=5-t=3.lpt This is the linear program], readable by Gnu’s glpsol linear programing solver, which also quickly proves that 150 is the optimum.<br />
<br />
Each variable corresponds to a point in the cube, numbered according to their lexicografic ordering. If a variable is 1 then the point is in the set, if it is 0 then it is not in the set.<br />
There is one linear inequality for each combinatorial line, stating that at least one point must be missing from the line.<br />
<br />
== n=6 ==<br />
<br />
:<math>c_6=450</math>:<br />
<br />
The upper bound follows since <math>c_6 \leq 3 c_5</math>. The lower bound can be formed by gluing together all the [[slice]]s <math>\Gamma_{a,b,c}</math> where (a,b,c) is a permutation of (0,2,4) or (1,2,3).<br />
<br />
Computer verification, using the <math>c_5=150</math> extremals, has shown that there is exactly one extremiser for <math>c_6=450</math>.<br />
<br />
== n=7 ==<br />
<br />
:<math>1302 \leq c_7 \leq 1348</math>:<br />
<br />
To see the upper bound <math>c_7 \leq 3c_6-2</math>, observe that if two parallel six-dimensional slices had <math>c_6</math> points, then by uniqueness they are identical, and the third slice can have at most <math>3^6-c_6=279</math> points, far too few to get anywhere close to <math>1348</math>. Thus there can be at most one slice with <math>c_6</math> points, and the other two have at most <math>c_6-1</math>, giving the claim.<br />
<br />
The lower bound can be formed by removing 016,106,052,502,151,511,160,610 from <math>D_7</math>.<br />
<br />
'''Lemma 6''' Any line-free subset of <math>D_7</math> has at most 1302 points.<br />
<br />
'''Proof''' Start with the 1458 points of <math>D_7</math>. You must lose:<br />
<br />
* 42 points from (1,2,4),(1,5,1),(4,2,1)<br />
* 42 points from (2,1,4),(2,4,1),(5,1,1)<br />
* 21 points from (0,2,5),(0,5,2),(3,2,2)<br />
* 21 points from (2,0,5),(2,3,2),(5,0,2)<br />
* 15 points from (0,1,6),(0,4,3),(3,1,3),(0,7,0),(3,4,0),(6,1,0)<br />
* 15 points from (1,0,6),(1,3,3),(4,0,3),(7,0,0),(4,3,0),(1,6,0)<br />
<br />
where (a,b,c) is shorthand for the [[slice]] <math>\Gamma_{a,b,c}</math>.<br />
<math>\Box</math><br />
<br />
== Larger n ==<br />
<br />
The following construction gives lower bounds for the number of triangle-free points, <br />
There are of the order <math>2.7 \sqrt{log(N)/N}3^N</math> points for large N (N ~ 5000)<br />
<br />
It applies when N is a multiple of 3. <br />
* For N=3M-1, restrict the first digit of a 3M sequence to be 1. So this construction has exactly one-third as many points for N=3M-1 as it has for N=3M. <br />
* For N=3M-2, restrict the first two digits of a 3M sequence to be 12. This leaves roughly one ninth of the points for N=3M-2 as for N=3M.<br />
<br />
The current lower bounds for <math>c_{3m}</math> are built like this, with abc being shorthand for <math>\Gamma_{a,b,c}</math>:<br />
<br />
* <math>c_3</math> from (012) and permutations<br />
* <math>c_6</math> from (123,024) and perms<br />
* <math>c_9</math> from (234,135,045) and perms<br />
* <math>c_{12}</math> from (345,246,156,02A,057) and perms (A=10)<br />
* <math>c_{15}</math> from (456,357,267,13B,168,04B,078) and perms (B=11)<br />
<br />
To get the triples in each row, add 1 to the triples in the previous row; then include new triples that have a zero.<br />
<br />
A general formula for these points is given below. I think that they are triangle-free. (For N<21, ignore any triple with a negative entry.)<br />
<br />
* There are thirteen groups of points in the centre, formed from adding one of the following points, or its permutation, to (M,M,M), when N=3M:<br />
** (-7,-3,+10), (-7, 0,+7),(-7,+3,+4),(-6,-4,+10),(-6,-1,+7),(-6,+2,+4),(-5,-1,+6),(-5,+2,+3),(-4,-2,+6),(-4,+1,+3),(-3,+1,+2),(-2,0,+2),(-1,0,+1) <br />
* There are also eight string of points, stretching to the edges of the (abc) triangle:<br />
** For N = 3M<br />
*** M+(-8-2x,-6-2x,14+4x),M+(-8-2x,-3-2x,11+4x),M+(-8-2x,x,8+x),M+(-8-2x,3+x,5+x) and permutations (x>=0, M-8-2x>=0)<br />
*** M+(-9-2x,-5-2x,14+4x),M+(-9-2x,-2-2x,11+4x),M+(-9-2x,1+x,8+x),M+(-9-2x,4+x,5+x) and permutations (x>=0, M-9-2x>=0)<br />
<br />
<br />
An alternate construction:<br />
<br />
First define a sequence, of all positive numbers which, in base 3, do not contain a 1. Add 1 to all multiples of 3 in this sequence. This sequence does not contain a length-3 arithmetic progression.<br />
<br />
It starts 1,2,7,8,19,20,25,26,55, …<br />
<br />
Second, list all the (abc) triples for which the larger two differ by a number<br />
from the sequence, excluding the case when the smaller two differ by 1, but then including the case when (a,b,c) is a permutation of N/3+(-1,0,1)<br />
<br />
== Asymptotics ==<br />
<br />
DHJ(3) is equivalent to the upper bound<br />
<br />
:<math>c_n \leq o(3^n)</math><br />
<br />
In the opposite direction, observe that if we take a set <math>S \subset [3n]</math> that contains no 3-term arithmetic progressions, then the set <math>\bigcup_{(a,b,c) \in \Delta_n: a+2b \in S} \Gamma_{a,b,c}</math> is line-free. From this and the Behrend construction it appears that we have the lower bound<br />
<br />
:<math>c_n \geq 3^{n-O(\sqrt{\log n})}.</math><br />
<br />
More precisely, we have<br />
<br />
:<math>c_n > C 3^{n - 4\sqrt{\log 2}\sqrt{\log n}+\frac 12 \log \log n}</math><br />
for some absolute constant C, and where all logarithms are base-3.<br />
<br />
'''Proof''' For convenience, let n be a multiple of 3. Elkin’s bound gives <math>r_3(\sqrt{n}) > C \sqrt{n} (\log n)^{1/4} \exp_2(-2 \sqrt{\log_2 n})</math>, and let <math>R</math> be a subset of <math>(-3\sqrt{n}/2,3\sqrt{n}/2)</math> without 3-term APs and with size <math>r_3(\sqrt{n})</math>, and with all elements being integer multiples of 3 (again as a matter of convenience). For each <math>r,s\in R</math>, let <math>a = (n-r-s)/3</math>. The set <math>A</math> is the union of all <math>\Gamma_{a,a+r,a+s}</math>. Since all of <math>a, a+r,a+s</math> are between <math>n/3-2\sqrt{n}</math> and <math>n/3+2\sqrt{n}</math>, the size of <math>\Gamma_{a,a+r,a+s}</math> is at least <math>C 3^n / n</math>. Since there are <math>r_3(\sqrt{n})^2</math> choices for r and s, we have a set with size at least<br />
<br />
:<math>C (\sqrt{n} (\log n)^{1/4} \exp_2(-2 \sqrt{\log_2 n}))^2 3^n / n</math>.<br />
<br />
This simplifies to <math>C \sqrt{\log n} \exp_3(n-\alpha \sqrt{\log_3(n)})</math>, where <math>\alpha=4 \sqrt{\log_3(2)}</math>.<br />
<br />
Now suppose that <math>x_i\in \Gamma_{a_i,a_i+r_i,a_i+s_i}</math> is a combinatorial line in the set A. Then <math>(a_i+s_i)-(a_i)=s_i</math> is a 3-term AP contained in R, so the <math>s_i</math> are all the same. Similarly, all of the <math>r_i</math> are the same, and therefore all of the <math>a_i</math> are the same, too. But this implies that the <math>x_i</math> sequence is constant, which means the line is degenerate. <math>\Box</math><br />
<br />
[http://terrytao.wordpress.com/2009/02/05/upper-and-lower-bounds-for-the-density-hales-jewett-problem/#comment-35652 Numerics suggest] that the first large n construction given above above give a lower bound of roughly <math>2.7 \sqrt{\log(n)/n} \times 3^n</math>, which would asymptotically be inferior to the Behrend bound.<br />
<br />
The second large n construction had numerical asymptotics for <math>\log(c_n/3^n)</math> close to <math>1.2-\sqrt{\log(n)}</math> between n=1000 and n=10000, consistent with the Behrend bound.<br />
<br />
== Numerical methods ==<br />
<br />
A greedy algorithm [http://thetangentspace.com/wiki/Hales-Jewett_Theorem was implemented here]. The results were sharp for <math>n \leq 3</math> but were slightly inferior to the constructions above for larger n.</div>Charles R Greathouse IVhttps://asone.ai/polymath/index.php?title=Fujimura%27s_problemFujimura's problem2015-09-30T04:25:07Z<p>Charles R Greathouse IV: fix OEIS link</p>
<hr />
<div>Let <math>\overline{c}^\mu_n</math> the largest subset of the triangular grid<br />
<br />
:<math>\Delta_n := \{ (a,b,c) \in {\Bbb Z}_+^3: a+b+c=n \}</math><br />
<br />
which contains no equilateral triangles <math>(a+r,b,c), (a,b+r,c), (a,b,c+r)</math> with <math>r > 0</math>; call such sets ''triangle-free''. (It is an interesting variant to also allow negative r, thus allowing "upside-down" triangles, but this does not seem to be as closely connected to DHJ(3).) Fujimura's problem is to compute <math>\overline{c}^\mu_n</math> ([http://oeis.org/A157795 OEIS A157795]). This quantity is relevant to a certain [[hyper-optimistic conjecture]].<br />
<br />
We are also exploring issues raised by [[higher-dimensional Fujimura]].<br />
<br />
The following table was formed mostly by computer searches for optimal solutions. We also found human proofs for most of them (see below).<br />
<br />
{|<br />
| n || 0 || 1 || 2 || 3 || 4 || 5 || 6 || 7 || 8 || 9 || 10 || 11 || 12 || 13<br />
|-<br />
| <math>\overline{c}^\mu_n</math> || 1 || 2 || 4 || 6 || 9 || 12 || 15 || 18 || 22 || 26 || 31 || 35 || 40 || 46<br />
|}<br />
<br />
== n=0 ==<br />
<br />
<math>\overline{c}^\mu_0 = 1</math>:<br />
<br />
This is clear.<br />
<br />
== n=1 ==<br />
<br />
<math>\overline{c}^\mu_1 = 2</math>:<br />
<br />
This is clear.<br />
<br />
== n=2 ==<br />
<br />
<math>\overline{c}^\mu_2 = 4</math>:<br />
<br />
This is clear (e.g. remove (0,2,0) and (1,0,1) from <math>\Delta_2</math>).<br />
<br />
== n=3 ==<br />
<br />
:<math>\overline{c}^\mu_3 = 6</math>:<br />
<br />
For the lower bound, delete (0,3,0), (0,2,1), (2,1,0), (1,0,2) from <math>\Delta_3</math>.<br />
<br />
For the upper bound: observe that with only three removals each of these (non-overlapping) triangles must have one removal:<br />
<br />
* set A: (0,3,0) (0,2,1) (1,2,0)<br />
* set B: (0,1,2) (0,0,3) (1,0,2)<br />
* set C: (2,1,0) (2,0,1) (3,0,0)<br />
<br />
Consider choices from set A:<br />
<br />
* (0,3,0) leaves triangle (0,2,1) (1,2,0) (1,1,1)<br />
* (0,2,1) forces a second removal at (2,1,0) [otherwise there is triangle at (1,2,0) (1,1,1) (2,1,0)] but then none of the choices for third removal work<br />
* (1,2,0) is symmetrical with (0,2,1)<br />
<br />
== n=4 ==<br />
<br />
:<math>\overline{c}^\mu_4=9</math>:<br />
<br />
The set of all <math>(a,b,c)</math> in <math>\Delta_4</math> with exactly one of a,b,c =0, has 9 elements and is triangle-free.<br />
(Note that it does contain the equilateral triangle (2,2,0),(2,0,2),(0,2,2), so would not qualify for the generalised version of Fujimura's problem in which <math>r</math> is allowed to be negative.)<br />
<br />
Let <math>S\subset \Delta_4</math> be a set without equilateral triangles. If <math>(0,0,4)\in S</math>, there can only be one of <math>(0,x,4-x)</math> and <math>(x,0,4-x)</math> in S for <math>x=1,2,3,4</math>. Thus there can only be 5 elements in S with <math>a=0</math> or <math>b=0</math>. The set of elements with <math>a,b>0</math> is isomorphic to <math>\Delta_2</math>, so S can at most have 4 elements in this set. So <math>|S|\leq 4+5=9</math>. Similar if S contain (0,4,0) or (4,0,0). So if <math>|S|>9</math> S doesn’t contain any of these. Also, S can’t contain all of <math>(0,1,3), (0,3,1), (2,1,1)</math>. Similar for <math>(3,0,1), (1,0,3),(1,2,1)</math> and <math>(1,3,0), (3,1,0), (1,1,2)</math>. So now we have found 6 elements not in S, but <math>|\Delta_4|=15</math>, so <math>S\leq 15-6=9</math>.<br />
<br />
''Remark'': curiously, the best constructions for <math>c_4</math> uses only 7 points instead of 9.<br />
<br />
== n=5 ==<br />
<br />
:<math>\overline{c}^\mu_5=12</math>:<br />
<br />
The set of all (a,b,c) in <math>\Delta_5</math> with exactly one of a,b,c=0 has 12 elements and doesn’t contain any equilateral triangles.<br />
<br />
Let <math>S\subset \Delta_5</math> be a set without equilateral triangles. If <math>(0,0,5)\in S</math>, there can only be one of (0,x,5-x) and (x,0,5-x) in S for x=1,2,3,4,5. Thus there can only be 6 elements in S with a=0 or b=0. The set of element with a,b>0 is isomorphic to <math>\Delta_3</math>, so S can at most have 6 elements in this set. So <math>|S|\leq 6+6=12</math>. Similar if S contain (0,5,0) or (5,0,0). So if |S| >12 S doesn’t contain any of these. S can only contain 2 point in each of the following equilateral triangles:<br />
<br />
(3,1,1),(0,4,1),(0,1,4)<br />
<br />
(4,1,0),(1,4,0),(1,1,3)<br />
<br />
(4,0,1),(1,3,1),(1,0,4)<br />
<br />
(1,2,2),(0,3,2),(0,2,3)<br />
<br />
(3,2,0),(2,3,0),(2,2,1)<br />
<br />
(3,0,2),(2,1,2),(2,0,3)<br />
<br />
So now we have found 9 elements not in S, but <math>|\Delta_5|=21</math>, so <math>S\leq 21-9=12</math>.<br />
<br />
== n=6 ==<br />
<br />
:<math>\overline{c}^\mu_6 = 15</math>:<br />
<br />
<math>\overline{c}^\mu_6 \geq 15</math> from the bound for general n. <br />
<br />
Note that there are ten extremal solutions to <math>\overline{c}^\mu_3</math>:<br />
<br />
Solution I: remove 300, 020, 111, 003<br><br />
Solution II (and 2 rotations): remove 030, 111, 201, 102<br><br />
Solution III (and 2 rotations): remove 030, 021, 210, 102<br><br />
Solution III' (and 2 rotations): remove 030, 120, 012, 201<br><br />
<br />
Also consider the same triangular lattice with the point 020 removed, making a trapezoid. Solutions based on I-III are: <br />
<br />
Solution IV: remove 300, 111, 003<br><br />
Solution V: remove 201, 111, 102<br><br />
Solution VI: remove 210, 021, 102<br><br />
Solution VI': remove 120, 012, 201<br><br />
<br />
The on the 7x7x7 triangular lattice triangle 141-411-114 must have at least one point removed. Remove 141, noting by symmetry any logic that follows will also work for either of the other two points.<br />
<br />
Suppose we can remove all equilateral triangles on our 7×7x7 triangular lattice with only 12 removals.<br />
<br />
Here, "top triangle" means the top four rows of the lattice (with 060 at top) and "bottom trapezoid" means the bottom three rows.<br />
<br />
At least 4 of those removals must come from the top triangle (the solutions of <math> \overline{c}^\mu_3</math> mentioned above).<br />
<br />
The bottom trapezoid includes the overlapping trapezoids 600-420-321-303 and 303-123-024-006. If the solutions of these trapezoids come from V, VI, or VI', then 6 points have been removed. Suppose the trapezoid 600-420-321-303 uses the solution IV (by symmetry the same logic will work with the other trapezoid). Then there are 3 disjoint triangles 402-222-204, 213-123-114, and 105-015-006. Then 6 points have been removed. Therefore at least six removals must come from the bottom trapezoid.<br />
<br />
To make a total of 12 removals there must be either:<br><br />
Case A: 4 removals from the top triangle and 8 from the bottom trapezoid.<br><br />
Case B: 5 removals from the top triangle and 7 from the bottom trapezoid.<br><br />
Case C: 6 removals from the top triangle and 6 from the bottom trapezoid.<br><br />
<br />
* Suppose case A is true.<br />
<br />
Because 141 is already removed, the solution to the top triangle must remove either solution I (remove 060, 330, 033), solution II (remove 060, 231, 132), solution IIb (remove 033, 150, 240) or solution IIc (remove 330, 051, 042)<br />
<br />
Suppose I is the solution for the top triangle.<br />
<br />
Suppose 222 is open. Then 420, 321, 123, and 024 must all be removed. This leaves five disjoint triangles which require removals in the bottom trapezoid (150-600-105, 051-501-006, 222-402-204, 231-411-213, 132-312-114); therefore the bottom trapezoid needs at least 9 removals, but we can only make 8, therefore 222 is closed.<br />
<br />
Suppose 411 is open. Then 213 and 015 must be removed. This leaves five disjoint triangles such that each triangle must have exactly one removal (420-150-123, 321-051,024, 600-510-501, 402-312-303, 204-114-105), so the remaining point (006) must be open, forcing 501 to be removed. This makes 600 and 510 open, and based on the triangles 600-240-204 and 510-150-114 both 204 and 114 must both be removed, but 204 and 114 are on the same disjoint triangle, contradicting the statement that each triangle must have exactly one removal. So 411 is closed.<br />
<br />
This leaves six disjoint triangles each which must have at least one removal (420-123-150, 321-024-051, 510-213-240, 312-015-042, 501-204-231, 402-105-132). This forces 600 and 006 to be open. Based on the triangles 006-501-051 and 600-204-240, this forces 501 and 204 to be open. But then there are no removals from the triangle 501-204-231, which is a contradiction. Therefore the solution of the top triangle cannot be I.<br />
<br />
Suppose II is the solution for the top triangle.<br />
<br />
There are seven disjoint triangles (150-600-105, 051-501-006, 222-402-204, 240-510-213, 042-312-015, 330-420-321, 033-123-024), therefore of the three points remaining in the bottom trapezoid (411, 303, 114) exactly one must be removed.<br />
<br />
Suppose 411 is removed. Then 114 and 303 are open; 114 open forces 510 to be removed, forcing 213 to be open. 114 and 213 open force 123 to be closed, forcing 024 to be open. 024 open forces 222 to be closed, which forces 204 to be open, which leaves the triangle 213-303-204 open so we have a contradiction.<br />
<br />
By symmetry 114 also can't be removed. Therefore we must remove 303. This leaves 411 and 114 open, forcing 510 and 015 closed. 510 and 015 closed forces 312 and 213 open, forcing 222 closed. 222 closed forces 402 and 204 open. 402 and 204 open force 501 and 105 closed. 510 and 105 closed force 600 and 006 open, leaving equilateral triangles 600-204-240 and 402-006-042 open. Therefore 303 can't be removed, and so the solution of the top triangle can't be II.<br />
<br />
Suppose IIb is the solution for the top triangle.<br />
<br />
Suppose 024 is open. This forces 420, 321, and 222 closed. Five disjoint triangles remain (510-600-501, 231-411-213, 132-312-114, 123-303-105, 024-204-006) so each must have exactly one point removed, and the remaining points in the bottom trapezoid (402, 015) must be open. This forces 312, 411, 510, and 006 closed, which then force the the other points in their disjoint triangles open (600, 501, 231, 213, 132, 114, 204). The triangle 501-231-204 is therefore open, so we have a contradiction, therefore 024 is closed.<br />
<br />
Suppose 006 is open. This forces 600, 501 and 402 to be closed, and leaves five disjoint triangles (510-420-411, 321-231-222, 312-132-114, 303-213-204, 105-015-006) and so 123 must be open. This forces 222 closed, which forces 321 open, which forces 420 closed, which forced 510 and 411 open, which forces 213 closed, which forces 303 and 204 open, which forces 105 closed, which forces 015 and 006 to be open, leaving an open triangle at 411-015-051. Therefore we have a contradiction, so 006 is closed.<br />
<br />
Given 024 and 006 closed, now note there are six disjoint triangles (600-510-501, 402-312-303, 204-114-105, 420-330-321, 222-132-123, 411-231-213). Therefore the remaining point in the bottom trapezoid 015 must be open, forcing 510, 411, and 312 to be closed. Using the disjoint triangles this forces 600, 501, 402, 303 and 213 to be open, which then forces 420 and 321 to be closed. Both 420 and 321 are on the same disjoint triangle, therefore we have a contradiction, so IIb can't be solution.<br />
<br />
Note by symmetry, the same logic for IIb will apply for IIc. Therefore case A isn't true.<br />
<br />
* Suppose case B is true.<br />
<br />
The row 330-231-132-033 has ten possible solutions (excluding reflections, which by symmetry will be handled by the same logic):<br />
<br />
Case Q: open-open-open-open<br><br />
Case R: closed-closed-closed-closed<br><br />
Case S: closed-open-open-open<br><br />
Case T: closed-open-closed-closed<br><br />
Case U: open-closed-closed-closed<br><br />
Case V: closed-open-closed-closed<br><br />
Case W: closed-closed-open-open<br><br />
Case X: closed-open-closed-open<br><br />
Case Y: open-closed-closed-open<br><br />
Case Z: closed-open-open-closed<br><br />
<br />
Consider all 10 cases. Note in all these cases we are still assuming 141 is removed.<br />
<br />
(Case Q) 330, 231, 132, 033 open forces 060, 150, 051, 240, 141, 042 closed, but the top triangle only allows 5 removals, so case Q forms a contradiction.<br />
<br />
(Case R) 060-240-042 is left open, so case R forms a contradiction.<br />
<br />
(Case S) 231, 132, and 033 open force 031 and 042 removed, which means from 240, 150, 061 there must be exactly one removal.<br />
<br />
Suppose 213 is open. Then 411 and 015 are closed, and five disjoint triangles are left where each triangle must have exactly one removal (600-420-402, 501-321-303, 312-222-213, 204-105-114, 123-024-033). So the two remaining points in the bottom trapezoid (510 and 006) must be open. 510 and 006 open force 303 closed, which forces 321 and 501 open, which forces 204 closed, which forces 114 and 103 open, which forces 402 and 312 closed, forcing 222 open, leaving 321-231-222 as an open triangle, so we have a contradiction.<br />
<br />
Therefore 213 is closed. This leaves six disjoint triangles (600-420-402, 501-231-204, 303-033-006, 411-321-312, 222-132-123, 114-024-015) which each have exactly one removal. Therefore 510 from the bottom trapezoid is open. Note since one of 150 and 060 must be open, then one of 114 or 015 must be closed; therefore 024 is open. This forces 123 closed, forcing 222 to be open, forcing 321 to be closed, forcing 411 and 312 open, forcing 421 closed, forcing 600 and 402 open, forcing 303 closed, forcing 006 open, forcing 204 closed, forcing 501 open, leaving the triangle 600-510-501 open and a contradiction.<br />
<br />
(Case T) 231 is closed, and 330, 132, and 033 open force 060, 150, and 042 closed. Since 141 is already closed we have five removals from the top triangle, so 240 and 051 are open.<br />
<br />
Suppose 024 is open. This forces 321 and 123 closed, leaving five disjoint triangles (600-510-501, 402-312-303, 204-114-105, 420-240-222, 213-033-015). Therefore 006 and 411 remaining in the bottom trapezoid are open, forcing 501, 303 and 015 closed, forcing 600, 510, 312, 402, and 213 open, forcing 222 closed, forcing 420 open, leaving the open triangle 420-510-411, so we have a contradiction.<br />
<br />
Therefore 024 is closed. There are now six disjoint triangles (510-600-501, 330-420-321, 312-402-303, 132-222-123, 033-213-015, 114-204-105). This leaves 411 and 006 open, which forces 015 and 501 closed, which forces 510 and 213 open, leaving the triangle 240-510-213 open, so we have a contradiction.<br />
<br />
(Case U) Given the removals 231, 132, 033, and 141, the only possible removal to not leave any equilateral triangles in the top triangle is 060. So 330, 240, 150, 051, and 042 are open.<br />
<br />
Suppose 420 is open. This forces 321, 222, and 123 closed. This leaves five disjoint triangles (600-330-303, 510-420-411, 204-105-114, 501-051-006, and 312-042-015), so we have a contradiction.<br />
<br />
Therefore 420 is closed. This leaves six disjoint triangles (600-330-303, 501-051-006, 204-114-105, 510-240-213, 411-321-312, 222-042-024). 015 in the bottom trapezoid is therefore open, forcing 312 and 411 closed, but 312 and 411 are on the same disjoint triangle, so we have a contradiction.<br />
<br />
(Case V) Given the removals 330, 132, and 033, the only possible removal to not leave any equilateral triangles in the top triangle is 060. So 150, 051, 240, 042, and 231 are open.<br />
<br />
Suppose 420 is open. Then 222 and 123 are closed, and we are left with 6 disjoint triangles (150-600-105, 240-510-213, 231-501-204, 024-114-015, 042-402-006, 411-321-312). This exceeds our limit of 7 removals in the bottom trapezoid, so we have a contradiction.<br />
<br />
Therefore 420 is closed. Again we are left with the same 6 disjoint triangles (150-600-105, 240-510-213, 231-501-204, 024-114-015, 042-402-006, 411-321-312). Therefore 222, 123, and 303 are open. This forces 321, 024, and 105 to be closed, which then forces 411 and 015 to be open, forming an open triangle at 051-411-015, so we have a contradiction.<br />
<br />
(Case W) Given the removals 330 and 231 with 132 and 033 open, 132 and 033 open force 042 closed. Given 141 is already closed, that leaves one more removal on the top triangle, which must be one of 060-150-051, so 240 must be open.<br />
<br />
Suppose 024 is open. This forces 123 to be closed, and leaves 6 disjoint triangles (600-510-501, 420-240-222, 411-321-312, 402-132-105, 213-033-015, 204-024-006). So 303 and 114 in the bottom trapezoid are left open, forcing 312 and 005 to be closed, forcing 204 and 321 to be open, forcing 600 and 501 to be closed. 600 and 501 are on the same disjoint triangle so we have a contradiction.<br />
<br />
Therefore 024 is closed. Suppose 312 is open. This forces 114 to be closed, and leaevs 5 disjoint triangles (600-510-501, 411-321-312, 303-213-204, 222-132-123, 105-015-006). Therefore 402 in the bottom trapezoid is open, which forces 105 to be closed, which forces 015 and 006 to be open, which forces 213 and 303 to be closed, which are on the same disjoint triangle, so we have a contradiction.<br />
<br />
Therefore 312 is closed. This leaves five disjoint triangles (510-240-213, 501-411-402, 222-132-123, 204-114-105, 303-033-006), and leaves 600, 420, 321, and 015 in the bottom trapezoid open. This forces 402 and 213 to be closed, which forces 510 and 411 to be open, leaving an open triangle at 510-420-411, so we have a contradiction.<br />
<br />
(Case X) 330 and 132 closed and 231 and 033 open imply 051 is closed, and since 141 is closed and we have only one more removal in the top triangle it must be one of 240, 042, or 060. This leaves 150 open.<br />
<br />
Suppose 321 is open. This forces 222 to be closed and leaves six disjoint triangles (600-420-402, 510-150-114, 411-321-312, 303-213-204, 105-015-006, 123-033-024), leaving 501 in the bottom trapezoid open. This forces 204 to be closed, which forces 303 to be open, leaving an open triangle at 501-321-303.<br />
<br />
So 321 is closed. This leaves six disjoint triangles (600-420-402, 510-150-114, 501-231-204, 312-222-213, 123-033-024, 105-015-006) and leaves 303 and 411 in the bottom trapezoid open. This forces 213 and 006 to be closed, and forces 312, 222, 105, and 015 to be open. This forces 600 to be closed and 402 to be open, leaving an open triangle at 402-312-303. So we have a contradiction.<br />
<br />
(Case Y) 330 and 033 are open, 330 and 033 open force 060 to be closed. With 141 closed there can be only one more removal, so suppose 150 is open (and note that if 150 was closed, we can use symmetry considering 051 to be open).<br />
<br />
Suppose 600 is open. This forces 303 to be closed, and leaves six disjoint triangles (510-150-114, 420-330-321, 411-501-402, 222-312-213, 033-123-024, 015-105-006). So 204 in the bottom trapezoid is left open. 600 and 150 open implies 105 is closed, forcing 015 and 006 to be open, forcing 024 and 213 to be closed, forcing 312, 222, and 123 to be open, forcing 042 in the top triangle to be closed (so 051 and 240 are open). 051 and 015 open force 411 to be closed, which forces 501 to be open and leaves the open triangle 051-501-006.<br />
<br />
So 600 is closed. This leaves six disjoint triangles (510-150-114, 420-330-321, 411-501-402, 222-312-213, 033-123-024, 015-105-006). 015 is the bottom trapezoid is then left open, forcing 312 and 411 to be closed. However, 312 and 411 are on the same disjoint triangle, so we have a contradiction.<br />
<br />
(Case Z) 330, 141, and 033 are closed, and 231 and 132 are open. Suppose 051 is open (note that if this forms a contradiction, by symmetrical argument we can say both 150 and 051 are closed).<br />
<br />
Suppose 114 is closed. This leaves six disjoint triangles (600-510-501, 402-312-303, 105-015-006, 411-231-213, 222-132-123, 321-051-024) and so 420 and 204 are open. This forces 501 to be closed, which forces 510 and 600 to be open, which forces 411 and 402 to be closed, which forces 213 and 303 to be open, leaving an open triangle at 213-303-204.<br />
<br />
So 114 is open. Suppose 213 is closed. This leaves five disjoint triangles (600-420-402, 501-321-303, 411-051-015, 222-132-123, 204-024-006) and in the bottom trapezoid 510 and 105 are open. This forces 204 to be closed, forcing 024 and 006 to be open, forcing 015 to be closed, forcing 411 to be open, forcing 420 to be closed, forcing 402 to be open, leaving the open triangle 402-132-105.<br />
<br />
So both 114 and 213 are open. Therefore 411 and 312 are closed. This leaves five disjoint triangles (600-510-501, 303-213-204, 105-015-006, 222-132-123, 321-051-024). The remaining points in the bottom trapezoid (420, 402) are then open, forcing 105 to be closed, forcing 015 and 006 to be open, forcing 024 to be closed. Also note 114 and 213 open force 123 to be closed, which forces 222 to be open, which forces 321 to be closed. 321 and 024 are on the same disjoint triangle, so we have a contradiction.<br />
<br />
So 150 and 051 are both closed. However, this leaves the triangle 240-042-060 open, so case Z is impossible.<br />
<br />
Since all ten solutions have been eliminated, case B is impossible.<br />
<br />
* Suppose case C is true.<br />
<br />
Suppose the trapezoid 600-420-321-303 used solution IV. There are three disjoint triangles 402-222-204, 213-123-114, and 105-015-006. The remainder of the points in the bottom trapezoid (420, 321, 510, 501, 402, 312, 024) must be left open. 024 being open forces either 114 or 015 to be removed. <br />
<br />
Suppose 114 is removed. Then 213 is open, and with 312 open that forces 222 to be removed. Then 204 is open, and with 024 that forces 006 to be removed. So the bottom trapezoid is a removal configuration of 600-411-303-222-114-006, and the rest of the points in the bottom trapezoid are open. All 10 points in the top triangle form equilateral triangles with bottom trapezoid points, hence 10 removals in the top triangle would be needed (more than the 6 allowed), so 114 being removed doesn't work. <br />
<br />
Suppose 015 is removed. Then 006-024 forces 204 to be removed. Regardless of where the removal in 123-213-114, the points 420, 321, 222, 024, 510, 312, 501, 402, 105, and 006 must be open. This forces top triangle removals at 330, 231, 042, 060, 051, 132, our remaining 6 removals on the top triangle. However, we have already removed 141, forcing one removal too many, so the trapezoid 600-420-321-303 doesn't use solution IV.<br />
<br />
Suppose the trapezoid 600-420-321-303 uses solution VI. The trapezoid 303-123-024-006 can't be IV (already eliminated by symmetry) or VI' (leaves the triangle 402-222-204). Suppose the trapezoid 303-123-024-006 is solution VI. The removals from the bottom trapezoid are then 420, 501, 312, 123, 204, and 015, leaving the remaining points in the bottom trapezoid open. The remaining open points is forces 10 top triangle removals, so the trapezoid 600-420-321-303 doesn't use solution VI. Therefore the trapezoid 303-123-024-006 is solution V. The removals from the bottom trapezoid are then 420, 510, 312, 204, 114, and 105. The remaining points in the bottom trapezoid are open, and force 9 top triangle removals, hence the trapezoid 303-123-024-006 can't be V, and the solution for 600-420-321-303 can't be VI. <br />
<br />
The solution VI' for the trapezoid 600-420-321-303 can be eliminated by the same logic by symmetry. <br />
<br />
Therefore it is impossible for the bottom trapezoid to use only 6 removals.<br />
<br />
We have determined cases A, B, and C to be impossible, therefore it impossible to form a triangle free configuration on the 7x7x7 lattice with only 12 removals. Therefore <math>\overline{c}^\mu_6 = 15</math>.<br />
<br />
== n = 7 ==<br />
<br />
<math>\overline{c}^\mu_{7} \leq 22</math>:<br />
<br />
Using the same ten extremal solutions to <math> \overline{c}^\mu_3 </math> as previous proofs:<br />
<br />
Solution I: remove 300, 020, 111, 003<br><br />
Solution II (and 2 rotations): remove 030, 111, 201, 102<br><br />
Solution III (and 2 rotations): remove 030, 021, 210, 102<br><br />
Solution III' (and 2 rotations): remove 030, 120, 012, 201<br />
<br />
Suppose the 8x8x8 lattice can be triangle-free with only 13 removals.<br />
<br />
Slice the lattice into region A (070-340-043) region B (430-700-403) and region C (034-304-007). Each region must have at least 4 points removed. Note there is an additional disjoint triangle 232-322-223 that also must have a point removed. Therefore the points 331, 133, and 313 are open. 331-313 open means 511 must be removed, 331-133 open means 151 must be removed, and 133-313 open means 115 must be removed. Based on the three removals, the solutions for regions A, B, and C must be either I or II. All possible combinations for the solutions leave several triangles open (for example 160-520-124). So we have a contradiction, and <math> \overline{c}^\mu_7 \leq 22 </math>.<br />
<br />
== n = 8 ==<br />
<br />
== n = 9 ==<br />
<br />
== n = 10 ==<br />
<br />
== Computer data ==<br />
<br />
From integer programming, we have<br />
<br />
* n=3, maximum 6 points, [http://abel.math.umu.se/~klasm/solutions-n=3-k=3-FUJ 10 solutions]<br />
* n=4, maximum 9 points, [http://abel.math.umu.se/~klasm/solutions-n=4-k=3-FUJ 1 solution]<br />
* n=5, maximum 12 points, [http://abel.math.umu.se/~klasm/solutions-n=5-k=3-FUJ 1 solution]<br />
* n=6, maximum 15 points, [http://abel.math.umu.se/~klasm/solutions-n=6-k=3-FUJ 4 solutions]<br />
* n=7, maximum 18 points, [http://abel.math.umu.se/~klasm/solutions-n=7-k=3-FUJ 85 solutions]<br />
* n=8, maximum 22 points, [http://abel.math.umu.se/~klasm/solutions-n=8-k=3-FUJ 72 solutions]<br />
* n=9, maximum 26 points, [http://abel.math.umu.se/~klasm/solutions-n=9-k=3-FUJ 183 solutions]<br />
* n=10, maximum 31 points, [http://abel.math.umu.se/~klasm/solutions-n=10-k=3-FUJ 6 solutions]<br />
* n=11, maximum 35 points, [http://abel.math.umu.se/~klasm/solutions-n=11-k=3-FUJ 576 solutions]<br />
* n=12, maximum 40 points, [http://abel.math.umu.se/~klasm/solutions-n=12-k=3-FUJ 876 solutions]<br />
<br />
== General n ==<br />
<br />
A lower bound for <math>\overline{c}^\mu_n</math> is 2n for <math>n \geq 1</math>, by removing (n,0,0), the triangle (n-2,1,1) (0,n-1,1) (0,1,n-1), and all points on the edges of and inside the same triangle. In a similar spirit, we have the lower bound<br />
<br />
:<math>\overline{c}^\mu_{n+1} \geq \overline{c}^\mu_n + 2</math><br />
<br />
for <math>n \geq 1</math>, because we can take an example for <math>\overline{c}^\mu_n</math> (which cannot be all of <math>\Delta_n</math>) and add two points on the bottom row, chosen so that the triangle they form has third vertex outside of the original example.<br />
<br />
An asymptotically superior lower bound for <math>\overline{c}^\mu_n</math> is 3(n-1), made of all points in <math>\Delta_n</math> with exactly one coordinate equal to zero.<br />
<br />
A trivial upper bound is <br />
<br />
:<math>\overline{c}^\mu_{n+1} \leq \overline{c}^\mu_n + n+2</math><br />
<br />
since deleting the bottom row of a equilateral-triangle-free-set gives another equilateral-triangle-free-set. We also have the asymptotically superior bound<br />
<br />
:<math>\overline{c}^\mu_{n+2} \leq \overline{c}^\mu_n + \frac{3n+2}{2}</math><br />
<br />
which comes from deleting two bottom rows of a triangle-free set and counting how many vertices are possible in those rows.<br />
<br />
Another upper bound comes from counting the triangles. There are <math>\binom{n+2}{3}</math> triangles, and each point belongs to n of them. So you must remove at least (n+2)(n+1)/6 points to remove all triangles, leaving (n+2)(n+1)/3 points as an upper bound for <math>\overline{c}^\mu_n</math>.<br />
<br />
== Asymptotics ==<br />
<br />
The [[corners theorem]] tells us that <math>\overline{c}^\mu_n = o(n^2)</math> as <math>n \to \infty</math>.<br />
<br />
For any equilateral triangle (a+r,b,c),(a,b+r,c) and (a,b,c+r), the value y+2z forms an arithmetic progression of length 3. A Behrend set is a finite set of integers with no arithmetic progression of length 3 (see [[http://arxiv.org/PS_cache/arxiv/pdf/0811/0811.3057v2.pdf this paper]]). By looking at those triples (a,b,c) with a+2b inside a Behrend set, one can obtain the lower bound <math>\overline{c}^\mu_n \geq n^2 \exp(-O(\sqrt{\log n}))</math>.</div>Charles R Greathouse IVhttps://asone.ai/polymath/index.php?title=Dickson-Hardy-Littlewood_theoremsDickson-Hardy-Littlewood theorems2013-07-16T23:00:23Z<p>Charles R Greathouse IV: fix typo</p>
<hr />
<div>For any integer <math>k_0 \geq 2</math>, let <math>DHL[k_0,2]</math> denote the assertion that given any admissible <math>k_0</math>-tuple <math>{\mathcal H}</math>, that infinitely many translates of <math>{\mathcal H}</math> contain at least two primes. Thus for instance <math>DHL[2,2]</math> would imply the twin prime conjecture. The acronym DHL stands for "Dickson-Hardy-Littlewood", and originates from [http://arxiv.org/abs/1305.6289 this paper of Pintz].<br />
<br />
It is known how to deduce results <math>DHL[k_0,2]</math> from three classes of estimates:<br />
<br />
* Elliott-Halberstam estimates <math>EH[\theta]</math> for some <math>1/2 < \theta < 1</math>.<br />
* Motohashi-Pintz-Zhang estimates <math>MPZ[\varpi,\delta]</math> for some <math>0 < \varpi < 1/4</math> and <math>0 < \delta < 1/4+\varpi</math>.<br />
* Motohashi-Pintz-Zhang estimates <math>MPZ'[\varpi,\delta]</math> for densely divisible moduli for some <math>0 < \varpi < 1/4</math> and <math>0 < \delta < 1/4+\varpi</math>.<br />
<br />
The Elliott-Halberstam estimates are the simplest to use, but unfortunately no estimate of the form <math>EH[\theta]</math> for any <math>\theta > 1/2</math> is known unconditionally at present. Zhang was the first to establish a result of the form <math>MPZ[\varpi,\theta]</math>, which is weaker than <math>EH[1/2+2\varpi+]</math>, for some <math>\varpi,\theta>0</math>. More recently, we have switched to using <math>MPZ'[\varpi,\theta]</math>, an estimate of intermediate strength between <math>MPZ[\varpi,\delta]</math> and <math>EH[1/2+2\varpi+]</math>, as the conversion of this estimate to a <math>DHL[k_0,2]</math> result is more efficient in the <math>\delta</math> parameter. The precise definition of the <math>MPZ</math>, <math>MPZ'</math> and <math>MPZ''</math> estimates can be found at the page on [[distribution of primes in smooth moduli]].<br />
<br />
== Converting EH to DHL ==<br />
<br />
In the [http://www.ams.org/mathscinet-getitem?mr=2552109 breakthrough paper of Goldston, Pintz, and Yildirim], it was shown that <math>EH[\theta]</math> implied <math>DHL[k_0,2]</math> whenever<br />
<br />
:<math>2\theta > (1 + \frac{1}{2l_0+1}) (1 + \frac{2l_0+1}{k_0})</math><br />
<br />
for some positive integer <math>l_0</math>. Actually (as noted [http://terrytao.wordpress.com/2013/06/03/the-prime-tuples-conjecture-sieve-theory-and-the-work-of-goldston-pintz-yildirim-motohashi-pintz-and-zhang/ here]), there is nothing preventing the argument for working for non-integer <math>l_0 > 0</math> as well, so we can optimise this condition as<br />
<br />
:<math>2\theta > (1 + \frac{1}{\sqrt{k_0}})^2</math>.<br />
<br />
Some further optimisation of this condition was performed in the paper of Goldston, Pintz, and Yildirim by working with general polynomial weights rather than monomial weights. In [http://www.renyi.hu/~revesz/ThreeCorr0grey.pdf this paper of Farkas, Pintz, and Revesz], the optimal weight was found (coming from a Bessel function), and the optimised condition<br />
<br />
:<math>2\theta > \frac{j_{k_0-2}^2}{k_0(k_0-1)}</math><br />
<br />
was obtained, where <math>j_{k_0-2}=j_{k_0-2,1}</math> is the first positive zero of the Bessel function <math>J_{k_0-2}</math>. See for instance [http://terrytao.wordpress.com/2013/06/03/the-prime-tuples-conjecture-sieve-theory-and-the-work-of-goldston-pintz-yildirim-motohashi-pintz-and-zhang/ this post] for details.<br />
<br />
== Converting MPZ to DHL ==<br />
<br />
The observation that <math>DHL[k_0,2]</math> could be deduced from <math>MPZ[\varpi,\delta]</math> if <math>k_0</math> was sufficiently large depending on <math>\varpi,\delta</math> was first made in the literature [http://www.ams.org/mathscinet-getitem?mr=2414788 by Motohashi and Pintz]. In the [http://annals.math.princeton.edu/wp-content/uploads/YitangZhang.pdf paper of Zhang], an explicit implication was established: <math>MPZ[\varpi,\varpi]</math> implies <math>DHL[k_0,2]</math> whenever there exists an integer <math>l_0>0</math> such that<br />
<br />
:<math> (1+4\varpi) (1-\kappa_2) > (1 + \frac{1}{2l_0+1}) (1 + \frac{2l_0+1}{k_0}) (1 + \kappa_1)</math><br />
<br />
where<br />
<br />
:<math> \kappa_1 := \delta_1( 1 + \delta_2^2 + k_0 \log(1+\frac{1}{4\varpi}) \binom{k_0+2l_0}{k_0}</math><br />
<br />
:<math> \kappa_2 := \delta_1 (1+4\varpi) ( 1 + \delta_2^2 + k_0 \log(1+\frac{1}{4\varpi}) \binom{k_0+2l_0+1}{k_0-1}</math><br />
<br />
:<math> \delta_1 := (1+4\varpi)^{-k_0}</math><br />
<br />
:<math> \delta_2 := \sum_{0 \leq j < 1+\frac{1}{4\varpi}} \frac{ \log(1+\frac{1}{4\varpi}) k_0)^j}{j!}.</math><br />
<br />
The value of <math>\delta_2</math> was lowered to <math>\prod_{0 \leq j < 1+\frac{1}{4\varpi}} (1 + k_0 \log(1+\frac{1}{j})</math> in [http://terrytao.files.wordpress.com/2013/05/bounds.pdf these notes]. Subsequently, the values of <math>\kappa_1,\kappa_2</math> were improved to<br />
<br />
:<math> \kappa_1 := (\delta_1 + \sum_{j=1}^{1/4\varpi} \delta_1^j \delta_{2,j}^2 + \delta_1 k_0 \log(1+\frac{1}{4\varpi})) \binom{k_0+2l_0}{k_0}</math><br />
<br />
:<math> \kappa_2 := (\delta_1 (1+4\varpi) + \sum_{j=1}^{1/4\varpi} \delta_1^j (1+4\varpi)^j \delta_{2,j}^2 + \delta_1 (1+4\varpi) k_0 \log(1+\frac{1}{4\varpi}) \binom{k_0+2l_0+1}{k_0-1}</math><br />
<br />
where<br />
<br />
:<math> \delta_{2,j} := \prod_{i=1}^j (1 + k_0 \log(1+\frac{1}{i}) )</math>;<br />
<br />
again, see [http://terrytao.files.wordpress.com/2013/05/bounds.pdf these notes]. As before, <math>l_0</math> can be taken to be non-integer.<br />
<br />
The constraint was then [http://terrytao.wordpress.com/2013/06/03/the-prime-tuples-conjecture-sieve-theory-and-the-work-of-goldston-pintz-yildirim-motohashi-pintz-and-zhang/ improved further in this post] to deduce <math>DHL[k_0,2]</math> from <math>MPZ[\varpi,\delta]</math> whenever<br />
<br />
:<math> (1+4\varpi) > (1 + \frac{1}{2l_0+1}) (1 + \frac{2l_0+1}{k_0}) (1 + \kappa)</math><br />
<br />
where<br />
<br />
:<math> \kappa = \sum_{1 \leq n \leq \frac{1+4\varpi}{2\delta}} (1 - \frac{2n\delta}{1+4\varpi})^{k_0/2+l_0} \prod_{j=1}^n (1+3k_0 \log(1+\frac{1}{j}))</math>.<br />
<br />
Using the optimal Bessel weight, this condition was improved to<br />
<br />
:<math> (1+4\varpi) > \frac{j^{2}_{k_0-2}}{k_0(k_0-1)} (1 + \kappa)</math>;<br />
<br />
again, see [http://terrytao.wordpress.com/2013/06/03/the-prime-tuples-conjecture-sieve-theory-and-the-work-of-goldston-pintz-yildirim-motohashi-pintz-and-zhang/ this post].<br />
<br />
A variant of this criterion was developed using the elementary Selberg sieve in [http://terrytao.wordpress.com/2013/06/08/the-elementary-selberg-sieve-and-bounded-prime-gaps/ this post], but never used. A subsequent refined criterion was established in [http://terrytao.wordpress.com/2013/06/11/further-analysis-of-the-truncated-gpy-sieve/ this post], namely that<br />
<br />
:<math> (1+4\varpi) (1-\kappa') > \frac{j^{2}_{k_0-2}}{k_0(k_0-1)} (1 + \kappa)</math><br />
<br />
where<br />
<br />
:<math> \kappa := \sum_{1 \leq n < \frac{1+4\varpi}{2\delta}} \frac{3^n+1}{2} \frac{k_0^n}{n!} (\int_{4\delta/(1+\varpi)}^1 (1-t)^{k_0/2} \frac{dt}{t})^n</math><br />
<br />
:<math> \kappa' := \sum_{2 \leq n < \frac{1+4\varpi}{2\delta}} \frac{3^n-1}{2} \frac{(k_0-1)^n}{n!} (\int_{4\delta/(1+\varpi)}^1 (1-t)^{(k_0-1)/2} \frac{dt}{t})^n.</math><br />
<br />
A slight refinement in [http://terrytao.wordpress.com/2013/06/11/further-analysis-of-the-truncated-gpy-sieve/#comment-234845 this comment] allows the condition <math>n \geq 2</math> in the definition of <math>\kappa'</math> to be raised to <math>n \geq 3</math>.<br />
<br />
An [http://terrytao.wordpress.com/2013/06/18/a-truncated-elementary-selberg-sieve-of-pintz argument of Pintz] yields the following improved values of <math>\kappa,\kappa'</math> in the above criterion:<br />
<br />
:<math> \kappa := 0 </math><br />
<br />
:<math> \kappa' := 2 \kappa_1 + 2 \kappa_2</math><br />
<br />
:<math> \kappa_1 := \int_{4\delta/(1+4\varpi)}^1 (1-t)^{(k_0-1)/2} \frac{dt}{t}</math><br />
<br />
:<math> \kappa_2 := (k_0-1) \int_{4\delta/(1+4\varpi)}^1 (1-t)^{k_0-1} \frac{dt}{t}</math>.<br />
<br />
== Converting MPZ' to DHL ==<br />
<br />
An efficient [http://terrytao.wordpress.com/2013/06/18/a-truncated-elementary-selberg-sieve-of-pintz argument of Pintz], based on the elementary Selberg sieve, allows one to deduce <math>DHL[k_0,2]</math> from <math>MPZ'[\varpi,\delta]</math> with almost no loss with respect to the <math>\delta</math> parameter. As currently optimised, the criterion takes the form<br />
<br />
:<math> (1+4\varpi) (1-2\kappa_1 - 2\kappa_2 - 2\kappa_3) > \frac{j^{2}_{k_0-2}}{k_0(k_0-1)}</math><br />
<br />
where<br />
<br />
:<math> \kappa_1 := \int_{\theta}^1 (1-t)^{(k_0-1)/2} \frac{dt}{t}</math><br />
<br />
:<math> \kappa_2 := (k_0-1) \int_{\theta}^1 (1-t)^{k_0-1} \frac{dt}{t}</math><br />
<br />
:<math> \kappa_3 := \tilde \theta \frac{J_{k_0-2}(\sqrt{\tilde \theta} j_{k_0-2})^2 - J_{k_0-3}(\sqrt{\tilde \theta} j_{k_0-2}) J_{k_0-1}(\sqrt{\tilde \theta} j_{k_0-2})}{ J_{k_0-3}(j_{k_0-2})^2 } <br />
\exp( A + (k_0-1) \int_{\tilde \delta}^\theta e^{-(A+2\alpha)t} \frac{dt}{t} )</math><br />
<br />
:<math> \alpha := \frac{j_{k_0-2}^2}{4(k_0-1)}</math><br />
:<math> \theta := \frac{\delta'}{1/4 + \varpi}</math><br />
:<math> \tilde \theta := \frac{(\delta' - \delta)/2 + \varpi}{1/4 + \varpi}</math><br />
:<math> \tilde \delta := \frac{\delta}{1/4 + \varpi}</math><br />
<br />
and <math>A>0</math> and <math>\delta \leq \delta' \leq \frac{1}{4} + \varpi</math> are parameters one is free to optimise over.<br />
<br />
Here is some simple Maple code to verify the above criterion for given choices of <math>k_0,\varpi,\delta,\delta',A</math>:<br />
<br />
k0 := [INSERT VALUE HERE];<br />
varpi := [INSERT VALUE HERE];<br />
delta := [INSERT VALUE HERE];<br />
deltap := [INSERT VALUE HERE]; <br />
A := [INSERT VALUE HERE];<br />
theta := deltap / (1/4 + varpi);<br />
thetat := ((deltap - delta)/2 + varpi) / (1/4 + varpi);<br />
deltat := delta / (1/4 + varpi);<br />
j := BesselJZeros(k0-2,1);<br />
eps := 1 - j^2 / (k0 * (k0-1) * (1+4*varpi));<br />
kappa1 := int( (1-t)^((k0-1)/2)/t, t = theta..1, numeric);<br />
kappa2 := (k0-1) * int( (1-t)^(k0-1)/t, t=theta..1, numeric);<br />
alpha := j^2 / (4 * (k0-1));<br />
e := exp( A + (k0-1) * int( exp(-(A+2*alpha)*t)/t, t=deltat..theta, numeric ) );<br />
gd := (j^2/2) * BesselJ(k0-3,j)^2;<br />
tn := sqrt(thetat)*j;<br />
gn := (tn^2/2) * (BesselJ(k0-2,tn)^2 - BesselJ(k0-3,tn)*BesselJ(k0-1,tn));<br />
kappa3 := (gn/gd) * e;<br />
eps2 := 2*(kappa1+kappa2+kappa3);<br />
# we win if eps2 < eps<br />
<br />
== Converting <math>MPZ''</math> to DHL ==<br />
<br />
There is a variant of <math>MPZ'</math> which we call <math>MPZ''</math> in which dense divisibility is replaced by the stronger condition of double dense divisibility; see [Distribution of primes in smooth moduli] for details. <math>MPZ''[\varpi,\delta]</math> is weaker than <math>MPZ'[\varpi,\delta]</math> but stronger than <math>MPZ[\varpi,\delta]</math>. It turns out (details [http://terrytao.wordpress.com/2013/06/30/bounded-gaps-between-primes-polymath8-a-progress-report/#comment-237306 here]) that one can also deduce <math>DHL[k_0,2]</math> from <math>MPZ''[\varpi,\delta]</math> with an almost identical numerology to the previous section, except that <math>\tilde \theta</math> is increased to<br />
<br />
:<math> \tilde \theta := \min( \frac{\delta' - \delta + \varpi}{1/4 + \varpi}, 1).</math><br />
<br />
So the Maple code is now changed slightly to the following:<br />
<br />
k0 := [INSERT VALUE HERE];<br />
varpi := [INSERT VALUE HERE];<br />
delta := [INSERT VALUE HERE];<br />
deltap := [INSERT VALUE HERE]; <br />
A := [INSERT VALUE HERE];<br />
theta := deltap / (1/4 + varpi);<br />
thetat := min( ((deltap - delta) + varpi) / (1/4 + varpi), 1);<br />
deltat := delta / (1/4 + varpi);<br />
j := BesselJZeros(k0-2,1);<br />
eps := 1 - j^2 / (k0 * (k0-1) * (1+4*varpi));<br />
kappa1 := int( (1-t)^((k0-1)/2)/t, t = theta..1, numeric);<br />
kappa2 := (k0-1) * int( (1-t)^(k0-1)/t, t=theta..1, numeric);<br />
alpha := j^2 / (4 * (k0-1));<br />
e := exp( A + (k0-1) * int( exp(-(A+2*alpha)*t)/t, t=deltat..theta, numeric ) );<br />
gd := (j^2/2) * BesselJ(k0-3,j)^2;<br />
tn := sqrt(thetat)*j;<br />
gn := (tn^2/2) * (BesselJ(k0-2,tn)^2 - BesselJ(k0-3,tn)*BesselJ(k0-1,tn));<br />
kappa3 := (gn/gd) * e;<br />
eps2 := 2*(kappa1+kappa2+kappa3);<br />
# we win if eps2 < eps</div>Charles R Greathouse IV