Let \begin{equation*} A_{n,w}(z)=\left(\sum_{i=0}^{\lfloor\frac{w}{2}\rfloor-1}\binom{w}{i}z^i+\frac{1}{2^{(w+1)\bmod 2}}\binom{w}{\lfloor\frac{w}{2}\rfloor}z^{\lfloor\frac{w}{2}\rfloor}\right)^{n/w} \end{equation*} Note that $[z^k]A_{n,w}(z)$ is the number of ways to put $k$ balls into $n/w$ boxes with $w$ positions such that:

• each box contains at most $\lfloor\frac{w}{2}\rfloor$ balls,
• any box containing exactly $\lfloor\frac{w}{2}\rfloor$ balls must use its first position if $w$ is even (note that $\binom{w-1}{w/2-1}=\frac{1}{2}\binom{w}{w/2}$ when $w$ is even).

I conjecture that:

Given $n, w$ such that $w(w+1) \mid n$. For all $0\leq r\leq n$, we have $$\sum_{k=0}^{r}[z^k]A_{n,w}(z)\leq\sum_{k=0}^{r}[z^k]A_{n,w+1}(z)$$

But I don't have enough tools to show that (maybe do some asymptotic analysis?).

The following basic facts might be helpful:

• Let $K_{w}(z)=(A_{n,w}(z))^{w/n}$, then we have $K_{w}(z)+z^wK_{w}(\tfrac{1}{z})=(1+z)^w$.
• For all $0\leq k\leq\lfloor\frac{w-1}{2}\rfloor$ we have $[z^k]A_{n,w}(z)=[z^k]((1+z)^w)^{n/w}=[z^k](1+z)^n=\binom{n}{k}$.