# Fujimura's problem

Let $\overline{c}^\mu_n$ the largest subset of the triangular grid

$\Delta_n := \{ (a,b,c) \in {\Bbb Z}_+^3: a+b+c=n \}$

which contains no equilateral triangles $(a+r,b,c), (a,b+r,c), (a,b,c+r)$ with $r \gt 0$; 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 $\overline{c}^\mu_n$. This quantity is relevant to a certain hyper-optimistic conjecture.

 n 0 1 2 3 4 5 $\overline{c}^\mu_n$ 1 2 4 6 9 12

## n=0

$\overline{c}^\mu_0 = 1$:

This is clear.

## n=1

$\overline{c}^\mu_1 = 2$:

This is clear.

## n=2

$\overline{c}^\mu_2 = 4$:

This is clear (e.g. remove (0,2,0) and (1,0,1) from $\Delta_2$).

## n=3

$\overline{c}^\mu_3 \geq 6$:

For the lower bound, delete (0,3,0), (0,2,1), (2,1,0), (1,0,2) from $\Delta_3$.

For the upper bound: observe that with only three removals each of these (non-overlapping) triangles must have one removal:

• set A: (0,3,0) (0,2,1) (1,2,0)
• set B: (0,1,2) (0,0,3) (1,0,2)
• set C: (2,1,0) (2,0,1) (3,0,0)

Consider choices from set A:

• (0,3,0) leaves triangle (0,2,1) (1,2,0) (1,1,1)
• (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
• (1,2,0) is symmetrical with (0,2,1)

## n=4

$\overline{c}^\mu_4=9$:

The set of all $(a,b,c)$ in $\Delta_4$ with exactly one of a,b,c =0, has 9 elements and is triangle-free. (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 $r$ is allowed to be negative.)

Let $S\subset \Delta_4$ be a set without equilateral triangles. If $(0,0,4)\in S$, there can only be one of $(0,x,4-x)$ and $(x,0,4-x)$ in S for $x=1,2,3,4$. Thus there can only be 5 elements in S with $a=0$ or $b=0$. The set of elements with $a,b\gt0$ is isomorphic to $\Delta_2$, so S can at most have 4 elements in this set. So $|S|\leq 4+5=9$. Similar if S contain (0,4,0) or (4,0,0). So if $|S|\gt9$ S doesn’t contain any of these. Also, S can’t contain all of $(0,1,3), (0,3,1), (2,1,1)$. Similar for $(3,0,1), (1,0,3),(1,2,1)$ and $(1,3,0), (3,1,0), (1,1,2)$. So now we have found 6 elements not in S, but $|\Delta_4|=15$, so $S\leq 15-6=9$.

Remark: curiously, the best constructions for $c_4$ uses only 7 points instead of 9.

## n=5

$\overline{c}^\mu_5=12$:

The set of all (a,b,c) in $\Delta_5$ with exactly one of a,b,c=0 has 12 elements and doesn’t contain any equilateral triangles.

Let $S\subset \Delta_5$ be a set without equilateral triangles. If $(0,0,5)\in S$, 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 $\Delta_3$, so S can at most have 6 elements in this set. So $|S|\leq 6+6=12$. 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:

(3,1,1),(0,4,1),(0,1,4)

(4,1,0),(1,4,0),(1,1,3)

(4,0,1),(1,3,1),(1,0,4)

(1,2,2),(0,3,2),(0,2,3)

(3,2,0),(2,3,0),(2,2,1)

(3,0,2),(2,1,2),(2,0,3)

So now we have found 9 elements not in S, but $|\Delta_5|=21$, so $S\leq 21-9=12$.

## n=6

$15 \leq \overline{c}^\mu_6 \leq 17$:

$15 \leq \overline{c}^\mu_6$ from the bound for general n.

Note that there are eight extremal solutions to $\overline{c}^\mu_3$:

Solution I: remove 300, 020, 111, 003
Solution II: remove 030, 111, 201, 102
Solution III (and 2 rotations): remove 030, 021, 210, 102
Solution III' (and 2 rotations): remove 030, 120, 012, 201

Also consider the same triangular lattice with the point 020 removed, making a trapezoid. Solutions based on I-III are:

Solution IV: remove 300, 111, 003
Solution V: remove 201, 111, 102
Solution VI: remove 210, 021, 102
Solution VI': remove 120, 012, 201

Suppose we can remove all equilateral triangles on our 7×7x7 triangular lattice with only 10 removals.

The triangle 141-411-114 must have at least one point removed. Remove 141, and note because of symmetry any logic that follows also applies to 411 and 114.

There are three disjoint triangles 060-150-051, 240-231-330, 042-132-033, so each must have a point removed.

(Now only six removals remaining.)

The remainder of the triangle 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 the remaining six removals must all come from the bottom three rows of the lattice.

Note this means the "top triangle" 060-330-033 must have only four points removed so it must conform to solution either I or II, because of the removal of 141.

Suppose the solution of the trapezoid 600-420-321-303 is VI or VI'. Both solutions I and II on the "top triangle" leave 240 open, and hence the equilateral triangle 240-420-222 remains. So the trapezoid can't be VI or VI'.

Suppose the solution of the trapezoid 600-420-321-303 is V. This leaves an equilateral triangle 420-321-330 which forces the "top triangle" to be solution I. This leaves the equilateral triangle 201-321-222. So the trapezoid can't be V.

Therefore the solution of the trapezoid 600-420-321-303 is IV. Since the disjoint triangles 402-222-204, 213-123-114, and 105-015-006 must all have points removed, that means the remaining points in the bottom three rows (420, 321, 510, 501, 312, 024) must be left open. 420 and 321 force 330 to be removed, so the "top triangle" is solution I. This leaves triangle 321-024-051 open, and we have reached a contradiction.

## General n

A lower bound for $\overline{c}^\mu_n$ is 2n for $n \geq 1$, 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

$\overline{c}^\mu_{n+1} \geq \overline{c}^\mu_n + 2$

for $n \geq 1$, because we can take an example for $\overline{c}^\mu_n$ (which cannot be all of $\Delta_n$) and add two points on the bottom row, chosen so that the triangle they form has third vertex outside of the original example.

An asymptotically superior lower bound for $\overline{c}^\mu_n$ is 3(n-1), made of all points in $\Delta_n$ with exactly one coordinate equal to zero.

A trivial upper bound is

$\overline{c}^\mu_{n+1} \leq \overline{c}^\mu_n + n+2$

since deleting the bottom row of a equilateral-triangle-free-set gives another equilateral-triangle-free-set. We also have the asymptotically superior bound

$\overline{c}^\mu_{n+2} \leq \overline{c}^\mu_n + \frac{3n+2}{2}$

which comes from deleting two bottom rows of a triangle-free set and counting how many vertices are possible in those rows.

Another upper bound comes from counting the triangles. There are $\binom{n+2}{3}$ 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 $\overline{c}^\mu_n$.

## Asymptotics

The corners theorem tells us that $\overline{c}^\mu_n = o(n^2)$ as $n \to \infty$.

By looking at those triples (a,b,c) with a+2b inside a Behrend set, one can obtain the lower bound $\overline{c}^\mu_n \geq n^2 \exp(-O(\sqrt{\log n}))$.