hyperrecursion - Maple Help

sumtools

 hyperrecursion
 Zeilberger-Koepf's hyperrecursion algorithm

 Calling Sequence hyperrecursion(U, L, z, s(n))

Parameters

 U, L - lists of the upper and lower parameters z - evaluation point n - name, recurrence variable s - name, recurrence function

Description

 • This function is an implementation of Koepf's extension of Zeilberger's algorithm, calculating a (downward) recurrence equation for the sum

$\sum _{k}\mathrm{hyperterm}\left(U,L,k\right)$

 the sum to be taken over all integers k, with respect to n. Here, U and L denote the lists of upper and lower parameters, and z is the evaluation point. The arguments of U and L are assumed to be rational-linear with respect to n. The resulting expression equals zero.
 • The output is a recurrence which equals zero. The recurrence is output as a function of n, the recurrence variable, and $s\left(n\right),s\left(n-1\right),\mathrm{...}$.
 • The command with(sumtools,hyperrecursion) allows the use of the abbreviated form of this command.

Examples

 > $\mathrm{with}\left(\mathrm{sumtools}\right):$
 > $\mathrm{hyperrecursion}\left(\left[-n,a\right],\left[b\right],1,f\left(n\right)\right)$
 $\left({-}{n}{+}{a}{-}{b}{+}{1}\right){}{f}{}\left({n}{-}{1}\right){+}\left({b}{+}{n}{-}{1}\right){}{f}{}\left({n}\right)$ (1)

Dougall's identity

 > $\mathrm{hyperrecursion}\left(\left[a,1+\frac{a}{2},b,c,d,1+2a-b-c-d+n,-n\right],\left[\frac{a}{2},1+a-b,1+a-c,1+a-d,1+a-\left(1+2a-b-c-d+n\right),1+a+n\right],1,s\left(n\right)\right)$
 ${-}\left({a}{+}{n}\right){}\left({a}{-}{c}{-}{d}{+}{n}\right){}\left({a}{-}{b}{-}{d}{+}{n}\right){}\left({a}{-}{b}{-}{c}{+}{n}\right){}{s}{}\left({n}{-}{1}\right){+}\left({a}{-}{d}{+}{n}\right){}\left({a}{-}{c}{+}{n}\right){}\left({a}{-}{b}{+}{n}\right){}\left({a}{-}{b}{-}{c}{-}{d}{+}{n}\right){}{s}{}\left({n}\right)$ (2)
 > $\mathrm{hyperrecursion}\left(\left[a+\frac{1}{2},a,b,1-b,-n,\frac{2a+1}{3}+n,\frac{a}{2}+1\right],\left[\frac{1}{2},\frac{2a-b+3}{3},\frac{2a+b+2}{3},-3n,2a+1+3n,\frac{a}{2}\right],1,s\left(n\right)\right)$
 $\left({b}{-}{2}{+}{3}{}{n}\right){}\left({b}{+}{1}{-}{3}{}{n}\right){}\left({2}{}{a}{-}{1}{+}{3}{}{n}\right){}\left({2}{}{a}{+}{3}{}{n}\right){}{s}{}\left({n}{-}{1}\right){+}\left({3}{}{n}{-}{1}\right){}\left({3}{}{n}{-}{2}\right){}\left({3}{}{n}{-}{1}{+}{b}{+}{2}{}{a}\right){}\left({3}{}{n}{-}{b}{+}{2}{}{a}\right){}{s}{}\left({n}\right)$ (3)