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Stirling transform of (k-1)!

  1. Jul 3, 2015 #1
    While reading about combinatorial mathematics, I came across the Stirling transform. https://en.wikipedia.org/wiki/Stirling_transform

    So then, if I want to find the Stirling transform of, for instance, ##(k-1)!##, I have to compute this (using the explicit formula of the Stirling number of the second kind):

    ##\displaystyle\sum_{k=1}^{n}\left(\frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j}j^n(k-1)!\right)##

    This looks complicated and I don't know where to start. Mathematica gives the result ##(-1)^n\operatorname{Li}_{1-n}(2)##.

    Any insights or hints of how to arrive at that result will be appreciated
     
  2. jcsd
  3. Jul 8, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Jul 21, 2015 #3
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