Search Results

Yes, these have been studied before. They were studied by MacMahon under the name "symmetric functions of several systems of quantities." Nowadays they are usually called MacMahon symmetric functions …

I am interested in polynomials $G_n(z)$ defined by the recurrence $$G_{n+1}(z) - 2G_n(z) + (1-nz)G_{n-1}(z)=0$$ for $n\ge1$ with the initial values $G_0(z) = 1$ and $G_1(z) = 1$. The next few values a …

Here is a proof that $a_1(n, p, q) = a_2(n,p,q)$. The generating function for the Stirling numbers of the second kind is $$\sum_{n=i}^\infty {n\brace i}\frac{x^n}{n!}= \frac{(e^x-1)^i}{i!}.$$ Also $$\ …

A closed form is \begin{equation*} \sum_{i=j+1}^{n} \binom{\binom{i}{j}}{2} =\sum_{k=1}^j \frac{1}{2}\binom{j}{k}\binom{j+k}{k}\binom{n+1}{j+k+1}.\tag{1}\label{474985_1} \end{equation*} For fixed $j$ …

This is a known result. To quote from Richard Stanley's Enumerative Combinatorics, Volume 2, second edition, solution to problem 8(e) of Chapter 5, page 115: This is equivalent to a conjecture of J. M …

Result 1 can be found in N. G. de Bruijn, Asymptotic Methods in Analysis, Dover, New York, 1981, p. 71. Result 2 follows from a formula of E. M. Wright, though he didn't state it in this form. A discu …

We proceed by "guessing" a generating function for $R(n,q)$ and verifying that it has the right properties. According to https://oeis.org/A143017, the generating function $G= \sum_{n=1}^\infty a(n) x^ …

Let \begin{equation*} A(x,q) = e^{qx}A(x)=e^{(q+1)x+(e^x-1)^2/2} \end{equation*} and define $a(n,q)$ by \begin{equation*} A(x,q) = \sum_{n=0}^\infty a(n,q) \frac{x^n}{n!}. \end{equation*} Then $a(n,0) …

Let $$ A(x,q)=\sum_{i=0}^{\infty}\frac{x^i}{\prod\limits_{j=0}^{i}(1-f(q+j)x)}, $$ so that $A(x) = A(x,0)$. Define $a(n,q)$ by $$A(x,q) = \sum_{n=0}^\infty a(n,q) x^n,$$ so that $a(n) = a(n,0)$. I wil …

Here's a sketch of a generating function proof. Recall that \begin{equation*} \sum_{m=0}^\infty \binom{2m+k}{m} x^m =\frac{c(x)^k}{\sqrt{1-4x}}, \end{equation*} where $c(x) = \sum_{n=0}^\infty C_n x^n …

No. It's easy to see that $B_n=4$ is impossible, and $B_n$ is never divisible by 8. This follows from the fact that $B_n$ is periodic modulo 8 with period 24. See, e.g., W. F. Lunnon, P. A. B. Pleasan …

Your $f_m(v,w)$ is a special case of Appell's $F_4$ hypergeometric function, $$f_m(v,w) = F_4(m+1;m+1;m+1,m+1;v,w).$$ Some information about analytic continuation of $F_4$ can be found in https://arxi …

A similar determinant, with $k^{i+j-2}$ instead of $k^{2i+2j-2}$ is evaluated in A. Zavrotsky, El Gesseliano (Spanish), Notas de Matematicas, no. 73, Universidad de los Andes, Facultad de Ciencias, De …

Algebraically, this identity is $$\sum_{n=0}^\infty C(n) x^n (1-x)^{n+1} = 1, $$ which is a consequence of the generating function $$\sum_{n=0}^\infty C(n) x^n = \frac{1-\sqrt{1-4x}}{2x}.$$

Here's an explanation of the combinatorial meaning of $T_{-n}(x)$. The combinatorial interpretation $T_n(x)$ is that it counts $n$-ary trees. More precisely, it counts ordered trees in which every ver …