Skip to main content
MathOverflow

Return to Answer

added 14 characters in body
Source Link
Max Alekseyev
Max Alekseyev
  • 34.4k
  • 5
  • 74
  • 152

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), \end{split}\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.

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), \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.

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.

Source Link
Max Alekseyev
Max Alekseyev
  • 34.4k
  • 5
  • 74
  • 152

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), \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.