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. Fujimura's problem is to compute [math]\displaystyle{ \overline{c}^\mu_n }[/math]. This quantity is relevant to a certain "hyper-optimistic" version of DHJ(3).

n=0

It is clear that [math]\displaystyle{ \overline{c}^\mu_0 = 1 }[/math].

n=1

It is clear that [math]\displaystyle{ \overline{c}^\mu_1 = 2 }[/math].

n=2

It is clear that [math]\displaystyle{ \overline{c}^\mu_2 = 4 }[/math] (e.g. remove (0,2,0) and (1,0,1) from [math]\displaystyle{ \Delta_2 }[/math]).

n=3

Deleting (0,3,0), (0,2,1), (2,1,0), (1,0,2) from [math]\displaystyle{ \Delta_3 }[/math] shows that [math]\displaystyle{ \overline{c}^\mu_3 \geq 6 }[/math]. In fact [math]\displaystyle{ \overline{c}^\mu_3 = 6 }[/math]: just note (3,0,0) or something symmetrical has to be removed, leaving 3 triangles which do not intersect, so 3 more removals are required.

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 doesn’t contain any equilateral triangles. It does contain the equilateral triangle (2,2,0),(2,0,2),(0,2,2), however, and it is not resolved whether this triangle refutes this solution.

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

[Cleanup required here]

A 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.

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].