Integrality and leading-coefficient conjecture for colored-triangle coefficient polynomials

Proof for every coefficient order. Write

$$P_n(q)=2^{1-n}T_2(n;q)=\sum_{k\ge0}a_k(n)q^k.$$

We prove that, for every $k\ge0$, the eventual polynomial $k!a_k(n)$ belongs to $\mathbb Z[n]$ and has leading coefficient $5^k$.

Quotient the depth-two triangles by the $n-1$ independent energy-preserving interface involutions of source Corollary 2.3 and Proposition 3.4(iii). The normalized polynomial $P_n(q)$ then counts canonical orbit representatives. Every such representative decomposes uniquely into its ordered direct-sum-irreducible blocks, and the energy is additive across blocks. Hence

$$F(z,q):=\sum_{n\ge0}P_n(q)z^n =\frac1{1-C(z,q)}, \tag{1}$$

where $C$ counts nonempty irreducible canonical triangles.

We first establish that each fixed-energy part of $C$ is a genuine polynomial in $z$. Let $b_j$ be the $j$-th bottom color, and let $A_j$ be the active color set immediately before it. Interlacing gives $|A_j|=j$ and $b_j\in A_j$. Define

$$e_j=\#\{c\in A_j:c>b_j\}, \quad r_j=\#(A_j\setminus[j]), \quad h_i=\#\{j\le i:b_j>i\}, \quad E=\sum_j e_j.$$

The displacement identity and the active-set bound give

$$D:=\sum_i h_i =\sum_j(j-b_j)_+ =\sum_j(b_j-j)_+ \le E,$$

and

$$r_j\le e_j+(b_j-j)_+, \qquad \sum_jr_j\le2E.$$

Whenever $h_i=r_i=r_{i+1}=0$, the first $i$ bottom colors are exactly $[i]$, the adjacent active sets are $[i]$ and $[i+1]$, and the canonical interface separates the two ordered direct-sum blocks. Thus every noncut is either counted by some positive $h_i$, or is adjacent to some positive $r_j$. Consequently,

$$\#\{\text{noncuts}\} \le D+2\sum_jr_j \le5E.$$

An irreducible size-$s$ configuration has $s-1$ noncuts. Therefore $s\le5E+1$, and

$$C(z,q)=z+\sum_{j\ge1}c_j(z)q^j, \qquad c_j(z)\in\mathbb Z[z], \qquad \deg c_j\le5j+1. \tag{2}$$

Set $Q(z,q)=C(z,q)-z$. Expanding (1) gives

$$F(z,q) =\sum_{\ell\ge0} \frac{Q(z,q)^\ell}{(1-z)^{\ell+1}}.$$

Because $Q$ has no constant term in $q$,

$$[q^k]F(z,q) =\frac{N_k(z)}{(1-z)^{k+1}},$$

where

$$N_k(z)= \sum_{\ell=0}^k (1-z)^{k-\ell}[q^k]Q(z,q)^\ell \in\mathbb Z[z]. \tag{3}$$

Writing $N_k(z)=\sum_s d_s z^s$, we obtain for all sufficiently large $n$,

$$a_k(n) = \sum_s d_s\binom{n-s+k}{k},$$

and therefore

$$k!a_k(n) = \sum_s d_s \prod_{j=1}^k(n-s+j) \in\mathbb Z[n]. \tag{4}$$

Its leading coefficient is

$$\sum_s d_s=N_k(1).$$

At $z=1$, all summands in (3) vanish except $\ell=k$. Since $Q=c_1(z)q+O(q^2)$,

$$N_k(1)=c_1(1)^k. \tag{5}$$

Finally, source Proposition 3.5 gives $a_1(2)=1$ and $a_1(n)=5(n-2)$ for $n\ge3$. Comparing coefficients in (1),

$$[q]F(z,q)=\frac{c_1(z)}{(1-z)^2} =z^2+\frac{5z^3}{(1-z)^2},$$

so

$$c_1(z)=z^2+3z^3+z^4, \qquad c_1(1)=5.$$

Substituting into (5), the leading coefficient of $k!a_k(n)$ is exactly $5^k$ for every $k\ge0$. This proves both clauses of Conjecture 4.3.

Source: Blitvić and Petrov, Colored interlacing triangles and Genocchi medians, Conjecture 4.3, https://arxiv.org/abs/2602.04390.