The motivation for my current question arises from [this MO post by R. Stanley](https://mathoverflow.net/questions/430741/number-of-coefficients-equal-to-k-in-certain-fibonacci-polynomials). *Caveat.* There's a slight alteration.
With the convention $F\_1=F\_2=1$ for the Fibonacci numbers, define the polynomials $f\_n(x)=\prod\_{i=1}^n(1+x^{F\_i})$.


For a given prime $p$, expand $f\_n(x)$ and compute the coefficients modulo $p$. Denote the resulting polynomial by $g\_{n,p}(x)$ and its number of non-zero coefficients by $N(g\_{n,p})$.


**Example.** $g\_{1,2}=1+x, g\_{2,2}=1+x^2, g\_{3,2}=1+x^4, g\_{4,2}=1+x^3+x^4+x^7$, we find that
$N(g\_{1,2})=2, N(g\_{2,2})=2, N(g\_{3,2})=2, g\_{4,2}=4$. I like to propose two sets of questions.



> 
> **QUESTION 1.** Are any of these recurrences true?
> \begin{align}
> N(g\_{n,2})&=2N(g\_{n-1,2})-2N(g\_{n-2,2})+2N(g\_{n-3,2}), \\
> N(g\_{n,3})&=2N(g\_{n-1,3})-2N(g\_{n-2,3})+3N(g\_{n-3,3})-4N(g\_{n-4,3})+4N(g\_{n-5,3}), \\
> N(g\_{n,5})&=2N(g\_{n-1,5})-N(g\_{n-2,5})+N(g\_{n-3,5})-2N(g\_{n-4,5})+3N(g\_{n-5,5}) \\
> & \qquad \qquad \qquad-4N(g\_{n-6,5})+4N(g\_{n-7,3}).
> \end{align}
> 
> 
> 



> 
> **QUESTION 2.** For fixed prime $p$, does the sequence $\{N(g\_{n,p})\}\_{n\geq1}$ have a rational generating function?
> 
> 
> 



Fibonacci and product polynomials
Question 2 follows from Theorem 6.1 of [arXiv:2101.02131](https://arxiv.org/pdf/2101.02131.pdf). (In this reference, I consider $\prod\_{i=1}^n(1+x^{F\_{i+1}})$ rather than $\prod\_{i=1}^n(1+x^{F\_i})$, but the proof still works.) The result holds for any positive integer $p$, not just primes. The proof is constructive, so in principle one can obtain actual recurrences.


