All Questions
8 questions
9
votes
2
answers
679
views
Сlosed formula for $(g\partial)^n$
The objective is to obtain a closed formula for:
$$
\boxed{A(n)=\big(g(z)\,\partial_z\big)^n,\qquad n=1,2,\dots}
$$
where $g(z)$ is smooth in $z$ and $\partial_z$ is a derivative with respect to $z$.
...
- 277
7
votes
0
answers
229
views
Has anyone seen these binary trees (Catalan-type related to the Gegenbauer polynomials and Motzkin paths)?
The OEIS entry A121448 enumerates binary trees with $n$ edges and $k$ vertices with outdegree 1.
Has anyone seen these trees?
The o.g.f. for this entry, $G(x,t)$, is essentially a discriminant ...
- 10.5k
5
votes
1
answer
345
views
Why does this "factorial sequence" appear in the OEIS?
For a reciprocal of a polynomial, $f = \frac{1}{p}$, we (presumably) may construct a sequence $(c_n)_{n=0}^\infty$ such that for all $N\ge 0$
$$f(k)k! = \sum_{n=0}^{N-1} c_n(k-n)! + O((k-N)!). $$
I ...
- 3,499
3
votes
1
answer
159
views
Proving that two sequences of polynomials defined over partitions are inverse to each other
For any fixed $c>0$ consider the polynomials
\begin{align*}
& p_n(X_1,X_2,\ldots) := \frac{n!}{c} \sum\limits_{b=1}^n \frac{c^b}{b!(n+1-b)!} \sum\limits_{\substack{l_1,\ldots,l_b \geq 1 \\ ...
- 1,295
3
votes
0
answers
264
views
Guises of the refined Eulerian numbers, generated by tangent vectors (OEIS A145271)
The Eulerian numbers (OEIS A008292, not to be confused with the Euler numbers) pop up in numerous scenarios in combinatorics and advanced analysis, one as the components of the h-vectors of the ...
- 10.5k
2
votes
1
answer
404
views
Relating face polytopes of permutohedra to integer partitions
The OEIS entries A019538, A049019, and A133314, relate a refinement of the face polynomials of the permutohedra (A049019) to partition polynomials (A133314) defined by multiplicative inversion of an ...
- 10.5k
2
votes
0
answers
121
views
Generalized identity with Stirling numbers of the second kind and falling factorials
It is known that Striling numbers of the second kind satisfy the relation
$$
\sum\limits_{k=0}^{n}{n \brace k}(x)_k = x^n.
$$
where $(x)_n$ is the falling factorials such that
$$
(x)_n = x(x-1)(x-2)\...
- 4,938
0
votes
0
answers
60
views
Algorithm for $q$-Bell numbers
Let $T(n,k)$ be A126347 (i.e., triangle, read by rows, with row polynomials $B(n, q)$). Here
$$
B(n, q) = \sum\limits_{k=0}^{n-1}\binom{n-1}{k}B(k, q)q^k, \\
B(0, q) = 1.
$$
Start with vector $\nu$ of ...
- 4,938