# 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 $S\leq 15-6=9$.

## 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 $S\leq 21-9=12$.

## 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 $\overline{c}^\mu_n \geq n^2 \exp(-O(\sqrt{\log n}))$.