Higher-degree ULC failure conjecture for the polynomials $f_m$ and $h_m$

Complete proof of the higher-degree ultra-log-concavity failure conjecture.

Consider Conjecture 3.9 of Stephanie Chen's arXiv:2503.16296v2. The associated paper appeared in Journal of Geometry and Physics 218 (2025), article 105666, doi:10.1016/j.geomphys.2025.105666; conjecture numbering here refers specifically to the accessible arXiv version. The paper already proves that the relevant polynomials are not ultra-log-concave overall. We prove its additional open assertion that every prescribed higher-degree inequality fails strictly.

Let

$$f_m(x)=\frac{(1+x)^m-(-1)^m}{x+2}, \qquad h_m(x)=f_m(x)+(-1)^m,$$

and write $a_j=[x^j]f_m(x)$. Both have degree $d=m-1$. We show that for every $4\le j\le m-2$,

$$\boxed{ \left(\frac{a_j}{\binom{m-1}{j}}\right)^2 < \frac{a_{j-1}}{\binom{m-1}{j-1}} \frac{a_{j+1}}{\binom{m-1}{j+1}}. } \tag{1}$$

Since $h_m-f_m$ is constant, the same strict inequality holds for $h_m$.

Pairing adjacent powers in the geometric quotient gives, for $j\ge1$,

$$a_j=\sum_{r\ge0}\binom{m-2r-2}{j-1}.$$

Set

$$b_j=\frac{a_j}{\binom{m-1}{j}},\qquad t=m-j.$$

These are probabilities that the maximum of a uniformly chosen $j$-subset of $\{1,\ldots,m-1\}$ has the same parity as $m-1$; hence $0<b_j\le1$. Comparing coefficients in

$$(x+2)f_m(x)=(1+x)^m-(-1)^m$$

gives

$$2(m-j)b_j+j b_{j-1}=m.$$

Eliminating the neighboring coefficients yields

$$b_{j-1}b_{j+1}-b_j^2 =\frac{j+t}{2j(t-1)}F_{j,t}(b_j),$$

where

$$F_{j,t}(z)=2z^2-(j+2t+1)z+j+t.$$

Thus it suffices to prove $F_{j,t}(b_j)>0$ for all $j\ge4,\ t\ge2$.

Fix $j$, and denote the normalized coefficient at $m=j+t$ by $b_{j,t}$. Its binomial formula gives the parity-step recurrence

$$b_{j,t+2}=T_{j,t}(b_{j,t}), \qquad T_{j,t}(z)= \frac{t(t+1)z+j(j+t)}{(j+t)(j+t+1)}.$$

Let $r_{j,t}$ be the smaller root of $F_{j,t}$. Since

$$F_{j,t}(0)=j+t>0,\qquad F_{j,t}(1)=1-t,$$

we have $r_{j,1}=1$ and $0<r_{j,t}<1$ when $t>1$.

We establish the strict barrier transport

$$T_{j,t}(r_{j,t})<r_{j,t+2} \qquad(j\ge4,\ t\ge1). \tag{2}$$

Define

$$\begin{aligned} L&=j^3+4j^2t+2j^2+4jt^2+8jt+5j+4t^2+4t,\\ J&=j^2t+2jt^2+3jt+2j+2t^2+2t,\\ \theta&=\frac{(j+t)J}{tL}. \end{aligned}$$

Direct polynomial identities give

$$\begin{aligned} &(j+t)^2(j+t+1)^2F_{j,t+2}(T_{j,t}(z))\\ &\qquad=t^2(t+1)^2F_{j,t}(z) +(t+1)\bigl((j+t)J-tLz\bigr), \end{aligned}$$

and

$$F_{j,t}(\theta) =-\frac{2j(j+t)^2}{t^2L^2}K_{j,t},$$

where, writing $u=j-4\ge0$,

$$K_{j,t} =u^3t+u^2(t^2+11t) +u(6t^2+35t-4)+(5t^2+25t-16)>0$$

for every $t\ge1$. Therefore $F_{j,t}(\theta)<0$, so $r_{j,t}<\theta$. Substitution of $z=r_{j,t}$ into the first identity yields

$$F_{j,t+2}\!\left(T_{j,t}(r_{j,t})\right)>0.$$

Since $T_{j,t}$ is increasing and $0<T_{j,t}(r_{j,t})<1$, while $F_{j,t+2}(1)<0$, this proves (2).

For odd $t$, start from

$$b_{j,1}=1=r_{j,1};$$

the strict transport (2) proves $b_{j,t}<r_{j,t}$ for every odd $t\ge3$. For even $t$, start from

$$b_{j,2}=\frac{j}{j+1}, \qquad F_{j,2}\!\left(\frac{j}{j+1}\right) =\frac{2}{(j+1)^2}>0;$$

again (2) propagates $b_{j,t}<r_{j,t}$ to every even $t\ge2$. Hence

$$F_{j,t}(b_{j,t})>0 \qquad\text{for every }j\ge4,\ t\ge2.$$

The determinant identity proves (1) at every $4\le j\le m-2$ for both $f_m$ and $h_m$, exactly as conjectured.

The source's earlier global non-ultra-log-concavity result is prior; the new conclusion is strict failure at every individual degree in the full conjectured range.