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2 votes

What alternatives are there to the binomial poset theory of generating function families?

Let $\mathcal{E}$ be the poset of all idempotent matrices over $\mathbb{F}_q$ having only finitely many nonzero entries with the ordering $A \leq B$ iff $AB=BA=A$. Then $\mathcal{E}$ is a binomial ...
  • 121
3 votes

Number of coefficients equal to $k$ in certain "Fibonacci polynomials"

This is just a `formulas-free' version of the proof that the numbers $\nu_k(n)$ are indeed linear in $n$, for a fixed $k$ and large enough $n\geq n_0(k)$. The following lemma is the same as Max ...
5 votes
Accepted

Number of coefficients equal to $k$ in certain "Fibonacci polynomials"

I will use the set up of my answer to a previous question about "Fibonacci polynomials". The key observation is that the coefficient of $x^m$ in $P(x)$ equals the number of "unrollings&...
4 votes
Accepted

Congruences of binomial sums

I found the reference: Theorem 9.1 in Boris Adamczewski and Jason P. Bell, Diagonalization and Rationalization of algebraic Laurent series, Ann. Sci. Éc. Norm. Supér. 46 (2013), 963–1004. The authors ...
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3 votes

Reference for asymptotic estimates

Here's a guess at something to try. Write $R(x) = \frac{P(x)}{Q(x)}$. Your series $F(x)$ satisfies $$F(x) = \frac{R(x)}{F(x^2)}$$ so taking logarithms we get $$\log F(x) = \log R(x) - \log F(x^2).$$ ...
2 votes

Congruences of binomial sums

These binomial sequences, and diagonals, are holonomic, meaning that they satisfy recurrence relations with polynomial coefficients. If the leading coefficient has no roots modulo m, then the question ...
  • 294
1 vote
Accepted

If the coefficient of the polynomial positive

When $m=1$ and $k$ is odd, I compute that $\bar{S}(k) = (3k^5-20k^3+17k)/120$, which has a negative coefficient.

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