The nonnegative-coefficient conjecture for the polynomial h(t,a;r)

We prove that the polynomial has nonnegative joint coefficients in $t$ and $a$ for every integer $r\ge0$.

Introduce $x=1+at$, $u=t/x$, and

$$A=(1+3u)(1+2u),\qquad B=(1+u)(1+4u),$$ $$C=(1+u)(1+2u),\qquad D=1+3u.$$

Regrouping the four homogeneous products in the defining polynomial gives

$$h(t,a;r)=x^{2r+1}g_r(u),$$

where

$$g_r(u)=(1+2u)A^r-(1+4u)B^r+(1+2u)C^r-D^r.$$

The key identity is the following exact positive generating function:

$$\begin{aligned} \sum_{r\ge0}g_r(u)z^r &=\frac{1+2u}{1-Az} -\frac{1+4u}{1-Bz} +\frac{1+2u}{1-Cz} -\frac1{1-Dz}\\ &=\frac{8u^4z^2(1+2u)} {(1-Az)(1-Bz)(1-Cz)(1-Dz)}. \end{aligned}$$

The identity follows simply by taking a common denominator.

Each of $A,B,C,D$ belongs to $\mathbb N[u]$. Therefore every factor

$$\frac1{1-Ez}=\sum_{j\ge0}E(u)^jz^j, \qquad E\in\{A,B,C,D\},$$

has nonnegative coefficients in both $u$ and $z$. The numerator $8u^4z^2(1+2u)$ does also. Hence

$$g_r(u)=\sum_{j=0}^{2r+1}c_{r,j}u^j \qquad\text{with }c_{r,j}\in\mathbb N$$

for every $r\ge0$; the degree bound follows directly from the defining expression for $g_r$.

Undoing the substitution yields the manifestly positive decomposition

$$h(t,a;r) =\sum_{j=0}^{2r+1}c_{r,j}\,t^j(1+at)^{2r+1-j}.$$

Thus

$$[t^{j+\ell}a^\ell]h(t,a;r) =c_{r,j}\binom{2r+1-j}{\ell}\ge0$$

for all $j,\ell$. This proves the conjecture for every integer $r$. In fact $h(t,a;0)=h(t,a;1)=0$, and for every $r\ge2$ the polynomial lies in $8t^4\mathbb N[t,a]$.