This is a follow up on [my earlier MO question][0]. 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 $2$. [0]: https://mathoverflow.net/questions/353715/an-identity-for-polynomials-over-partitions [1]: http://mathworld.wolfram.com/ConjugatePartition.html