q-polynomials in terms of a basis

Question

Consider the polynomials $$f_n(q)=\prod_{j=1}^n(1+q^j) \qquad \text{and} \qquad g_m(q)=1+q+q^2+\cdots+q^m.$$ I'll list a few examples to motivate my question. Direct calculations show that $$f_1=g_1, \qquad f_2=g_3, \qquad f_3=g_6+q^3g_0, \qquad f_4=g_{10}+q^3g_4, \qquad f_5=g_{15}+q^3g_9+q^5g_5, \qquad etc.$$

I would then ask:

QUESTION 1. Are there (numerical) coefficients $a_k\in\mathbb{Z}$ such that the following holds? $$f_n(q)=\sum_{k\geq0}a_k\cdot q^k\cdot g_{\binom{n+1}2-2k}(q).$$

QUESTION 2. If so, are $a_k\geq0$?

Postscript. Richard Stanley responded with a confirmation to the above questions. If it is alright, I like to change them to the following: can you find the coefficients $a_k$ explicitly?

Remark. If Question 2 is answered positively, then we get a new proof of $f_n$ being symmetric and unimodal.

Answer 1

I'm convinced that an "explicit" form for $a_k$ might be ambitious. So, here is a sketchy argument for $a_k\geq0$ by induction. One can easily find a base case.

Denote $\Delta_n=\binom{n+1}2$ and $\Delta^*_n=\lfloor\frac12\Delta_n\rfloor$. Suppose $f_n(q)=\sum_{k=0}^{\Delta^*_n}a_kq^kg_{\Delta_n-2k}(q)$ where each $a_k\geq0$. By definition, we know $f_{n+1}(q)=(1+q^{n+1})f_n(q)$ and hence \begin{align} f_{n+1}(q):&=\sum_{k=0}^{\Delta^*_n}a_kq^kg_{\Delta_n-2k}(q) +\sum_{k=0}^{\Delta^*_n}a_kq^{n+1+k}g_{\Delta_n-2k}(q) \\ &=\sum_{k=0}^{\Delta^*_n}a_k(q^k+\cdots+{\color{red}{q^{\Delta_n-k}}}) +\sum_{k=0}^{\Delta^*_n}a_k(q^{n+1+k}+\cdots+{\color{blue}{q^{\Delta_{n+1}-k}}}) \\ &=\sum_{k_1=0}^{\Delta^*_n}a_{k_1}(q^{k_1}+\cdots+{\color{blue}{q^{\Delta_{n+1}-k_1}}}) +\sum_{k_2=0}^{\Delta^*_n}a_{k_2}(q^{n+1+k_2}+\cdots+{\color{red}{q^{\Delta_n-k_2}}}) \\ &=\sum_{k_1=0}^{\Delta^*_n}a_{k_1}q^{k_1}(1+\cdots+q^{\Delta_{n+1}-2k_1}) +\sum_{k_2=0}^{\Delta^*_n}a_{k_2}q^{n+1+k_2}(1+\cdots+q^{\Delta_{n+1}-2(n+1+k_2)}) \\ &=\sum_{k_1=0}^{\Delta^*_n}a_{k_1}q^kg_{\Delta_{n+1}-2k_1}(q) +\sum_{k_2=0}^{\Delta^*_n}a_{k_2}q^{n+1+k_2}g_{\Delta_{n+1}-2(n+1+k_2)}(q). \end{align} Apart from rearrangement and collapsing like-terms, the coefficients can only be non-negative again.