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The motivation for my current question arises from this MO post by R. Stanley. 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?

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1 Answer 1

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Question 2 follows from Theorem 6.1 of arXiv:2101.02131. (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.

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  • $\begingroup$ Thank you for quick response. $\endgroup$ Commented May 18, 2023 at 18:58

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