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 …

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 $$\ …

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 …

A combinatorial proof of the matrix-tree theorem can be found in the paper by D. Zeilberger A combinatorial approach to matrix algebra, Discrete Math. 56 (1985), 61–72. The proof uses only the interp …

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 …

Speaking only for myself, the reason that I don't publish all my results is that I write very slowly. (It has nothing to do with the nature of combinatorics.) It takes a long time for me to arrange my …

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 …

There is a product formula for the order polynomial of an unusual poset in Solution of an Enumerative Problem Connected with Lattice Paths by Kreweras and Niederhausen. The poset is a product of a ch …

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 …