Search Results

See Warren P. Johnson, Combinatorics of Higher Derivatives of Inverses, American Mathematical Monthly, Vol. 109, No. 3 (Mar., 2002), pp. 273-277, http://www.jstor.org/stable/2695356

A chain $$ A_k \subset A_{k-1} \subset\dots\subset A_1 \subset A_0 $$ can be represented by the ordered partition $(B_1, B_2, \dots, B_{k+1})$ of the set $A_0=\{1, 2, \dots, n\}$ where $B_1=A_k$, $B_2 …

The answer is the coefficient of $x_1^{c_1}\cdots x_n^{c_n}$ in $[(1+x_1)(1+x_2)\cdots(1+x_n)-1]^m$. Here $(1+x_1)\cdots (1+x_n) -1$ corresponds to the chips placed on each square; zero or one of eac …

If $\mathcal O$ is the set of all subsets of an $n$-element set $U$ then by inclusion-exclusion the number of subsets of $\mathcal O$ whose union is $U$ is $$\sum_{k=0}^n (-1)^{n-k} \binom nk 2^{2^k} …

If we write the right side of Robert's identity as $$(-1)^n\frac{(\nu-2)(\nu-4)\cdots(\nu - 2n+2)}{(\nu+2)(\nu+4)\cdots(\nu+2n)},$$ we see that the identity is a partial fraction expansion of a prope …

This formula is Proposition 3B in P. Flajolet, Combinatorial Aspects of Continued Fractions, Discrete Math 32 (1980), 125–161.

Since the terms aren't integers we can't find a combinatorial interpretation directly. If we multiply both sides by $(u+v+d)!$ and rearrange, we can rewrite the identity as $$ \sum_{i=0}^u (-1)^i \bi …

Here is a sketch of a proof. First I'll switch $q$ and $z$, since this is more consistent with standard “$q$-series” notation. Let $C_k(n)$ be the coefficient of $z^k$ in $\prod_{j=-n}^n (1+q^jz)$, …

Here's a more self-contained description of this module. For simplicity, I'll consider only the case $n=2m$. Consider the vector space $V$ spanned by the set $P$ of ordered partitions of $[2m]$ into …

Here is a proof of the first formula. We use the exponential generating function $\sum_{n=0}^\infty S(n,k) x^n/n! = (e^x-1)^k/k!$. Multiply the left side by $x^n/n!$ and sum on $n$ from 1 to $\infty$ …

Here's a sketch of the derivation of the exponential generating function. First consider the case $n$ even, $n=2m$. The problem is equivalent to counting partitions with $k$ blocks of the set $\{1,2,\ …

A standard approach to proving this kind of identity is to use differences of polynomials. First note that if change the limits on both sums to 0 and $n$, then we add two terms on the left and two te …

Let $g(t) = th(t^{p-1}/p)$. Then the functional equation for $g(t)$ gives $h(z) =1+zh(z)^p$. It is well known that the coefficients of $h(z)$ are given by $$h(z) = \sum_{j=0}^\infty \frac{1}{pj+1}\b …

Here are the details on proving these identities using Hadamard products of generating functions. (You can find explanations of how to compute Hadamard products of rational functions here.) I'll writ …

There are some papers on this topic from the point of view of rook theory by V. S. Shevelev.