Stirling Number Identities

Examples

The Stirling package provides a command for computing a multivariate recurrence equation satisfied by a given sum whose summand involves Stirling numbers.

<<RISC`Stirling`

Package Stirling version 0.4
written by Manuel Kauers
Copyright 2007-2008, Research Institute for Symbolic Computation (RISC),
Johannes Kepler University, Linz, Austria

The following command delivers a recurrence equation in n and m for the sum (∑_(k=0)^m)(({{m}, {k}})) (Subscript[𝒮, n]^((k))):

FindRec[Binomial[m,k]StirlingS2[n,k],k,n,m]

Indeed:

Table[%,{n,0,5},{m,0,5}]/.SUM[n_,m_]->Sum[Binomial[m,k]StirlingS2[n,k],{k,0,n}]

You may also ask for a recurrence equation in one variable only, but then the summation algorithm may not terminate:

FindRec[x^k StirlingS1[n,k],k,n]

FindRec[x^k StirlingS2[n,k],k,n]

The summand must be of the form (Subscript[h, n,k,m])S(c+b k+a n,f+e k+d n)  where (Subscript[h, n,k,m]) is a (proper) hypergeometric term (viz. a product of binomials, factorials, exponentials, and polynomials), S is one of StirlingS1,StirlingS2,EulerianE1,EulerianE2, and a,b,c,d are such that a c-b d=±1.

FindRec[Binomial[m,k]EulerianE2[n+k,k],k,n,m]

FindRec[(-1)^k Binomial[m+n,k]EulerianE1[n+k,k],k,n,m]

etc.

FindRec[(-1)^k/(k+1)Binomial[m,k]Binomial[2k,k]StirlingS2[n+k,k],k,n,m]//Timing

If you have questions or if you find bugs, please send them to Manuel Kauers (mkauers@risc.uni-linz.ac.at)