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*} for example:
- $A_{n,2}(z)=(1+z)^{n/2}$
- $A_{n,3}(z)=(1+3z)^{n/3}$
- $A_{n,4}(z)=(1+4z+3z^2)^{n/4}$
- $A_{n,5}(z)=(1+5z+10z^2)^{n/5}$
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 for arbitrary $w$, maybe do some asymptotic analysis?
for example: $$\sum_{k=0}^{\alpha n}[z^k]A_{n,2}(z)=\sum_{k=0}^{\alpha n}\binom{n/2}{k}\sim2^{\frac{1}{2}H_2(2\alpha)n}<2^{\frac{2}{3}H_4(3\alpha)n}\sim\sum_{k=0}^{\alpha n}\binom{n/3}{k}3^k=\sum_{k=0}^{\alpha n}[z^k]A_{n,3}(z)$$$$\sum_{k=0}^{\alpha n}[z^k]A_{n,2}(z)=\sum_{k=0}^{\alpha n}\binom{n/2}{k}\sim2^{\frac{1}{2}H_2(2\alpha)n}\leq2^{\frac{2}{3}H_4(3\alpha)n}\sim\sum_{k=0}^{\alpha n}\binom{n/3}{k}3^k=\sum_{k=0}^{\alpha n}[z^k]A_{n,3}(z)$$ where $H_q(x)=x\log_q(q-1)-x\log_q(x)-(1-x)\log_q(1-x)$.
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 $[z^k]A_{n,w}(z)=[z^k]((1+z)^w)^{n/w}=[z^k](1+z)^n=\binom{n}{k}$.