3
$\begingroup$

The present quest emanates from this study by R. Stanley, including his recent MO question. Define the product (polynomials after full expansion) $$I_n(x)=\prod_{i=1}^n(1+x^{F_{i+1}})$$ based on the Fibonacci numbers $F_n$ with $F_1=F_2=1$. My focus is on the sequence $c_p(n)$, which enumerates the number of monomials, in $I_n(x)$, having non-zero coefficients modulo a prime $p$.

For example, if $p=2$ then $c_2(n)$ starts with $2, 4, 6, 8, 12, 20, 32, 48, 72, 112, \dots$. One may readily translate this problem to counting monomials in the product $\prod_{i=1}^n(1-x^{F_{i+1}})$ (see OEIS A104767) whose coefficients are $-1, 0, 1$ (see this paper).

QUESTION. Given a prime $p$, is it true that the generating function $\sum_{n\geq1}c_p(n)y^n$ is a rational function?

Examples. We have \begin{align} \sum_{n\geq1}c_2(n)y^n &=\frac{2y+2y^3}{1-2y+2y^2-2y^3}, \\ \sum_{n\geq1}c_3(n)y^n &=\frac{2y+3y^3+4y^5}{1-2y+2y^2-3y^3+4y^4-4y^5}, \\ \sum_{n\geq1}c_5(n)y^n &=\frac{2y+y^3+3y^5+4y^7}{1-2y+y^2-y^3+2y^4-3y^5+4y^6-4y^7}. \end{align}

$\endgroup$
2
  • 4
    $\begingroup$ An affirmative answer follows from Theorem 6.1 of arxiv.org/pdf/2101.02131.pdf. For some related open problems, see Conjectures 6.2 and 6.3. $\endgroup$ Commented Sep 24, 2022 at 22:27
  • $\begingroup$ Yes, indeed. Thank you. $\endgroup$ Commented Sep 25, 2022 at 13:50

0

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.