Fujimura's problem

Let [math]\displaystyle{ \overline{c}^\mu_n }[/math] the largest subset of the triangular grid

[math]\displaystyle{ \Delta_n := \{ (a,b,c) \in {\Bbb Z}_+^3: a+b+c=n \} }[/math]

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

n	0	1	2	3	4	5
[math]\displaystyle{ \overline{c}^\mu_0 }[/math]	1	2	4	6	9	12

n=0

[math]\displaystyle{ \overline{c}^\mu_0 = 1 }[/math]:

This is clear.

n=1

[math]\displaystyle{ \overline{c}^\mu_1 = 2 }[/math]:

This is clear.

n=2

[math]\displaystyle{ \overline{c}^\mu_2 = 4 }[/math]:

This is clear (e.g. remove (0,2,0) and (1,0,1) from [math]\displaystyle{ \Delta_2 }[/math]).

n=3

[math]\displaystyle{ \overline{c}^\mu_3 \geq 6 }[/math]:

For the lower bound, delete (0,3,0), (0,2,1), (2,1,0), (1,0,2) from [math]\displaystyle{ \Delta_3 }[/math].

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

[math]\displaystyle{ \overline{c}^\mu_4=9 }[/math]:

The set of all [math]\displaystyle{ (a,b,c) }[/math] in [math]\displaystyle{ \Delta_4 }[/math] 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 [math]\displaystyle{ r }[/math] is allowed to be negative.)

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

n=5

[math]\displaystyle{ \overline{c}^\mu_5=12 }[/math]:

The set of all (a,b,c) in [math]\displaystyle{ \Delta_5 }[/math] with exactly one of a,b,c=0 has 12 elements and doesn’t contain any equilateral triangles.

Let [math]\displaystyle{ S\subset \Delta_5 }[/math] be a set without equilateral triangles. If [math]\displaystyle{ (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]\displaystyle{ \Delta_3 }[/math], so S can at most have 6 elements in this set. So [math]\displaystyle{ |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:

(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 [math]\displaystyle{ |\Delta_5|=21 }[/math], so [math]\displaystyle{ S\leq 21-9=12 }[/math].

General n

A lower bound for [math]\displaystyle{ \overline{c}^\mu_n }[/math] is 2n for [math]\displaystyle{ 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

[math]\displaystyle{ \overline{c}^\mu_{n+1} \geq \overline{c}^\mu_n + 2 }[/math]

for [math]\displaystyle{ n \geq 1 }[/math], because we can take an example for [math]\displaystyle{ \overline{c}^\mu_n }[/math] (which cannot be all of [math]\displaystyle{ \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.

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

A trivial upper bound is

[math]\displaystyle{ \overline{c}^\mu_{n+1} \leq \overline{c}^\mu_n + n+2 }[/math]

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

[math]\displaystyle{ \overline{c}^\mu_{n+2} \leq \overline{c}^\mu_n + \frac{3n+2}{2} }[/math]

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 [math]\displaystyle{ \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]\displaystyle{ \overline{c}^\mu_n }[/math].

Asymptotics

The corners theorem tells us that [math]\displaystyle{ \overline{c}^\mu_n = o(n^2) }[/math] as [math]\displaystyle{ n \to \infty }[/math].

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