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This is a follow up on my earlier MO question.

Given an integer partition $\lambda=(\lambda_1,\dots,\lambda_{\ell(\lambda)})$ of $n$ where $\ell(\lambda)$ is the length of $\lambda$, associate $\tilde\lambda'=\lambda,0$ found by appending one extra zero at the right end of $\lambda$. Further, define the following numerics $a(\lambda)_j=\tilde\lambda_j-\tilde\lambda_{j+1}$ for $j=1,2,\dots,\ell(\lambda)$.

For example, if $\lambda=(3,2,1,1)$ and $\tilde\lambda=(3,2,1,1,0)$ and $a(\lambda)=(1,1,0,1)$.

QUESTION. Is it true that the coefficients of the polynomial $A_n(q)$ are all in $\{-1,0,1,2\}$? $$A_n(q):=\sum_{\lambda\vdash n}q^{n-\lambda_1} \prod_{a(\lambda)_j\geq1}\frac{(q^{2a(\lambda)_j}-1)(q-1)}{q+1}.$$

REMARK. In fact, it appears that only coefficient of the middle-term can possibly be equal to $2$.

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    $\begingroup$ Since you're summing over all partitions, any reason why we need to take the conjugates first? The only reference to $\lambda$ itself seems to be the exponent of $q$ in the outer sum, and that can be replaced by $\lambda'_1$, right? $\endgroup$
    – user44191
    Commented Mar 13, 2020 at 5:34
  • $\begingroup$ @user44191: edited accordingly. $\endgroup$
    – T. Amdeberhan
    Commented Mar 13, 2020 at 14:26

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