The IsProperHypergeometricTerm(T,n,k) command returns true if $T\left(n\,k\right)$ is a proper hypergeometric term. Otherwise, it returns false.

The IsHolonomic(T,n,k) command returns true if the bivariate hypergeometric term $T\left(n\,k\right)$ is holonomic. Otherwise, it returns false.

A bivariate hypergeometric term $T\left(n\,k\right)$ is proper if it can be written as $T\left(n\,k\right)=P\left(n\,k\right)\mathrm{Tp}\left(n\,k\right)$ where $P\left(n\,k\right)$ is a polynomial of n and k, and $\mathrm{Tp}\left(n\,k\right)=\frac{u^n{v}^k\left(\prod _{i=1}^l\left(b_{i}k+a_{i}n+g_i\right)!\right)}{\prod _{i=1}^m\left({\mathrm{ap}}_i+{\mathrm{bp}}_i+{\mathrm{gp}}_i\right)!}$, $a_i,b_i,{\mathrm{ap}}_i,{\mathrm{bp}}_i$ are integers, and $l,m$ are non-negative integers, $g_i,{\mathrm{gp}}_i,u,v$ are complex numbers.

It can be shown that $T\left(n\,k\right)$ is proper if and only if it is holonomic.

Note: If $T\left(n\,k\right)$ is a proper hypergeometric term, the termination of Zeilberger's algorithm is guaranteed.

$\mathrm{with}\left(\mathrm{SumTools}\left[\mathrm{Hypergeometric}\right]\right)\:$

$T≔\left(-\frac{4-k-4n}{-1-k-4n}+\frac{3-k-4n}{-2-k-4n}\right){\left(-1\right)}^k\mathrm{binomial}\left(n+1\,k\right)\mathrm{binomial}\left(2n-2k+1\,n\right)$

$T≔\left(-\frac{4-k-4n}{-1-k-4n}+\frac{3-k-4n}{-2-k-4n}\right){\left(\mathrm{-1}\right)}^k\left(\genfrac{}{}{0ex}{}{n+1}{k}\right)\left(\genfrac{}{}{0ex}{}{2n-2k+1}{n}\right)$

$\mathrm{ConjugateRTerm}\left[1\right]\left(T\,n\,k\,'\mathrm{listform}'\right)$

$\left[-\frac{2}{\sqrt{\mathrm{\pi }}}\,\frac{\left(n-k\right)!\left(k+4n\right)!{\left(-\frac{1}{4}\right)}^k{4}^n\left(n-k+\frac{1}{2}\right)!\left(n+1\right)!}{\left(n+1-k\right)!\left(2+k+4n\right)!k!\left(n-2k+1\right)!n!}\right]$

$\mathrm{IsProperHypergeometricTerm}\left(T\,n\,k\right)$

$\mathrm{true}$

$T≔\frac{-48-94n+10k}{{\left(5-5n+k{9}^{\frac{1}{3}}\right)}^3}{\left(-1\right)}^k\mathrm{binomial}\left(n+1\,k\right)\mathrm{binomial}\left(2n-2k+1\,n\right)$

$T≔\frac{\left(-48-94n+10k\right){\left(\mathrm{-1}\right)}^k\left(\genfrac{}{}{0ex}{}{n+1}{k}\right)\left(\genfrac{}{}{0ex}{}{2n-2k+1}{n}\right)}{{\left(5-5n+k{9}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^3}$

$\mathrm{ConjugateRTerm}\left[1\right]\left(T\,n\,k\,'\mathrm{listform}'\right)$

$\left[-\frac{1458}{5\sqrt{\mathrm{\pi }}{\left(59^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}n-59^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}-9k\right)}^3}\,\frac{\left(24+47n-5k\right)!\left(n-k\right)!{\left(-\frac{1}{4}\right)}^k{4}^n\left(n-k+\frac{1}{2}\right)!\left(n+1\right)!}{\left(23+47n-5k\right)!\left(n+1-k\right)!k!\left(n-2k+1\right)!n!}\right]$

$\mathrm{IsProperHypergeometricTerm}\left(T\,n\,k\right)$

$\mathrm{false}$

Abramov, S.A., and Petkovsek, M. "Proof of a Conjecture of Wilf and Zeilberger." Preprint series. Vol. 39. (2001): 748. University of Ljubljana, ISSN 1318--4865.