李善兰恒等式

${\displaystyle {\binom {n+k}{k}}^{2}=\sum _{j=0}^{k}{\binom {k}{j}}^{2}{\binom {n+2k-j}{2k}}}$

其中${\displaystyle {\binom {k}{l}}={\frac {k!}{l!(k-l)!}}}$

李善兰恒等式是Saalschütz's theorem的一个整数特例。

${\displaystyle {}_{3}F_{2}(a,b,-n;c,1+a+b-c-n;1)={\frac {(c-a)_{n}(c-b)_{n}}{(c)_{n}(c-a-b)_{n}}}.}$[3] [4]

${\displaystyle \sum _{j=0}^{k}{\binom {k}{j}}^{2}{\binom {n+2k-j}{2k}}={\frac {(n+2k)!}{(2k)!n!}}\sum _{j=0}^{\infty }{\frac {(-k)^{(j)}(-k)^{(j)}(-n)^{(j)}}{(1)^{(j)}(-n-2k)^{(j)}j!}}={\frac {(n+2k)!}{(2k)!n!}}{}_{3}F_{2}(-k,-k,-n;1,-n-2k;1)}$

${\displaystyle ={\frac {(n+2k)!(1+k)_{n}(1+k)_{n}}{(2k)!n!(1)_{n}(1+2k)_{n}}}={\frac {(n+2k)!(n+k)!(n+k)!(2k)!}{(2k)!n!k!k!n!(n+2k)!}}={\binom {n+k}{k}}^{2}}$

• 组合数学
• 李善兰