I thought it was easy but not quite. If it can be shown it will give another proof of a formula I found. Here $m$ nad $n$ are non-negative integers. I would like the following to be (hopefully relatively simply) proven: $$\sum_k \left[ {m+1} \atop k \right]{ k+n \brace m}(-1)^{m+k} = 0$$ The case $n=0$ is easy since the only non-zero terms for this are when $k=m$ and $k=m+1$ $$\left[ m+1 \atop m\right] { m \brace m} - \left[ m+1 \atop m+1\right] { m+1 \brace m} = \left[ m+1 \atop m\right] \cdot 1 - 1\cdot { m+1 \brace m}$$ This is zero since it is well known that $$\left[ m+1 \atop m\right] = { m+1 \brace m}$$

Can the general case $n \ge 0$ be shown?

  • $\begingroup$ I tried the technique from the other post and it appears that it will produce the above formula as well, the algebra and the steps are the same, except that we need to start the generating function at $n=-1.$ The transformation from the EGF to the OGF also goes through since $n+k\ge 0.$ Alternatively we may replace $n$ by $n-1$ in the above and start the generating function at $n=0.$ $\endgroup$ Mar 29, 2014 at 0:55
  • $\begingroup$ Oh well, I guess I have not found a simpler proof then. For the other post that is. $\endgroup$
    – adam W
    Mar 29, 2014 at 0:59
  • $\begingroup$ These identities suggest that there should be a combinatorial proof, especially for the one that sums to $m^n.$ $\endgroup$ Mar 29, 2014 at 1:12
  • $\begingroup$ I think I can show it: use the identity $\left[ m+1 \atop k \right] = \left[ m+2 \atop k+1\right] - (m+1)\left[ m+1 \atop k+1\right]$ repeatedly until all terms are zero by the inversion formula. Anyone wants to show the details can get a free accept from me. $\endgroup$
    – adam W
    Mar 29, 2014 at 1:28


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