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The Fibonacci numbers $F_n$ can be given by $$\sum_{k\geq0}F_kx^k=\frac{x}{1-x-x^2}.$$ Among many many properties of this sequence, consider the following two results:

(1) the coefficients of the infinite product $\prod_{i=2}^{\infty}(1-x^{F_i})$ are $-1, 0$ or $+1$;

(2) the coefficients of the finite product $\prod_{i=2}^N(1-x^{F_i})$ are $-1, 0$ or $+1$.

It is almost irresistible to look for more. To this end, consider for example, the tribonacci sequences given by $$\sum_{k\geq0}T_kx^k=\frac{x^2}{1-x-x^2-x^3}.$$

QUESTION. Are the coefficients of the finite or infinite products $$\prod_{i=3}^N(1-x^{T_i}) \qquad \text{and} \qquad \prod_{i=3}^{\infty}(1-x^{T_i})$$ bounded? What is the best bound?

Remark. An affirmative answer lends itself further generalization of this assessment.

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  • $\begingroup$ Do you know an example of a coefficient that exceeds 2 in absolute value? Have you tried to use the technique of Ardila at fardila.com/Articles/fibonacci.pdf? $\endgroup$
    – Richard Stanley
    Commented Jun 8, 2023 at 0:20
  • $\begingroup$ @RichardStanley: that's a good point. I did not find anything bigger than 2 in absolute value. I look at Ardila's but did not quite see how to navigate with it. $\endgroup$
    – T. Amdeberhan
    Commented Jun 8, 2023 at 2:02
  • $\begingroup$ Let $A_1,A_2,\dots$ be positive integers. One could conjecture that $\prod_{i=1}^n(1-x^{A_i})$ is "nice" (e.g., has bounded coefficients, with bound independent of $n$) if $\sum A_i t^i$ is a rational function whose denominator is the minimal polynomial (normalized to have constant term 1) of the reciprocal of a PV number. $\endgroup$
    – Richard Stanley
    Commented Jun 8, 2023 at 2:15
  • $\begingroup$ I might have made an error earlier. I just find that the coefficient of $x^{44}$ is $-3$. $\endgroup$
    – T. Amdeberhan
    Commented Jun 8, 2023 at 2:56
  • $\begingroup$ For the finite product, from $N=3$, the sequence of maximum absolute coefficient begins 1, 1, 1, 2, 2, 2, 3, 4, 4, 5, 6, 8, 8, 10, 12, 16, 16, 20, 24, 32, 32, 40, 48, 64, 64, 80, 96, 128, ... It doesn't look likely to be bounded. $\endgroup$
    – Peter Taylor
    Commented Jun 8, 2023 at 8:22

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Boundedness holds for $n$-bonacci numbers with even $n$, but not for odd $n$.

There is an analogue of the Zeckendorf representation: if you have the sequence of $n$-bonacci numbers, then each positive integer $k$ has a unique representation into a sum of some of those, with no $n$ consecutive terms.

Each expansion of $k$ as a sum of distinct $n$-bonacci numbers can be obtained from the $n$-Zeckendorf representation by a suitable sequence of replacements of a number by a sum of $n$ lots of $n$-bonacci numbers. It can even be done in a way that all representations obtained in the process contain only distinct numbers (in line with the product in current MO question).

For an odd $n$ (in particular, for $n=3$), all such representations are counted in the product under consideration with the same sign, so it counts just the number of such representations which is indeed unbounded. To this end, consider the modified product (where $y$ registers number of products contributing to $x^k$) \begin{align*} \prod_{i\geq3}(1+y\,x^{T_i}) &=1+yx+yx^2+\cdots+(y^5+y^3+y)x^{81}+(y^6+y^4+y^2)x^{82}+\cdots \end{align*} in which either the powers of $y$ are all of the same parity (e.g. coefficients of $x^{81}$ or $x^{82}$).

For even $n$, one can show that this process is similar to what happens with ordinary Fibonacci (to each $n$-Zeckendorf representation, one can put into correspondence a Zeckendorf representation of a different number to which just the same set of operations is applicable), so the resulting coefficients are still bounded by $1$. (I may fill the details later, if needed.) By contrast, take the case $n=4$ and let $J_k$ be given by $\sum_{k\geq0}J_kx^k=\frac{x^3}{1-x-x^2-x^3-x^4}$. The corresponding product \begin{align*} \prod_{i\geq4}(1+y\,x^{J_i}) &=1+yx+yx^2+\cdots+(y^7+y^4+y)x^{208}+(y^5+y^2)x^{209}+\cdots \end{align*} shows that the powers of $y$ can take up any parity.

Addendum. However, it seems that one can find products of such kind which will have bounded coefficients! E.g., $$ \prod_{n\geq 1}(1+x^{T_{4n-1}})(1-x^{T_{4n}})(1-x^{T_{4n+1}})(1-x^{T_{4n+2}}). $$

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  • $\begingroup$ The part that says "I may fill the details later, if needed", actually needs filling. $\endgroup$
    – T. Amdeberhan
    Commented Jun 9, 2023 at 20:57

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