The function defined by $$ F[N,M]=\sum_{m=0}^{N-1}\frac{(-1)^{N-1-m}(m+1)^M}{m!(N-1-m)!} $$ where $N,M$ are positive integers. I want to show $$ F[N,N-1]=1,\ F[N,M]=0 $$ for $N>2$ and $M<N-1$. Any suggestion will be helpful.
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3$\begingroup$ In standard hypergeometric notation, your sum is $$\frac{(-1)^{N-1}}{(N-1)!}\,{}_{m+1}F_m\left(\begin{matrix}1-N,2,\dots,2\\1,\dots,1\end{matrix}\right).$$ Your identities are special cases of the Karlsson-Minton identities (Minton, J. Math. Phys. 11 (1970) 1375-1376; Karlsson, J. Math. Phys. 12 (1971) 270-271). There are many ways to prove these, e.g. using integral representations, partial fraction identities or difference operators. $\endgroup$– Hjalmar RosengrenCommented Mar 15 at 5:37
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1$\begingroup$ For $f(x)=(x+1)^M$ the number $(N-1)!F[N,M]$ is the $(N-1)$-st finite difference at point 0, thus the result $\endgroup$– Fedor PetrovCommented Mar 16 at 15:50
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We can derive an explicit formula: \begin{split} F(N,M) &= [y^M]\frac{M!}{(N-1)!}\sum_{m=0}^{N-1} \binom{N-1}m (-1)^{N-1-m} e^{(m+1)y} \\ &= [y^M]\frac{M!}{(N-1)!} e^y (e^y-1)^{N-1} \\ &=[y^M]\frac{M!}{(N-1)!} \left( (e^y-1)^N + (e^y-1)^{N-1} \right) \\ &=N\cdot S(M,N) + S(M,N-1) \\ &=S(M+1,N) \end{split} where $S(\cdot,\cdot)$ are Stirling numbers of the second kind. Particular values $F(N,N-1)=1$ and $F(M,N)=0$ for $M<N-1$ easily follow from their definition.
answered Mar 16 at 15:13