Consider: $$f(x) = \frac{1}{e}\sum_{n=0}^{\infty}\frac{n^x}{n!}$$ When $x$ is a whole number f(x) is also a whole number. Is there a way to resolve this summation?
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1You're sure it's $n^x$ in the numerator and not $x^n$, right? – Brian Tung Sep 14 '18 at 22:25
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$f(x)$ represents the $x^{th}$ Bell number via Dobinski's formula. A search for the Bell numbers and related will provide further information. – Leucippus Sep 14 '18 at 22:33
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As suggested by Leucippus https://en.wikipedia.org/wiki/Dobi%C5%84ski%27s_formula – user Sep 14 '18 at 22:40
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It looks similar, but the mention of Bell numbers is what the answer I wanted. The other question had a different answer to it. – Nimish Sep 14 '18 at 22:58
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One method of examination is to obtain a generating function. For this is can be determined that: \begin{align} \sum_{n=0}^{\infty} f(n) \, \frac{t^n}{n!} &= \sum_{n=0}^{\infty} \frac{t^{n}}{n!} \, \frac{1}{e} \, \sum_{k=0}^{\infty} \frac{k^{n}}{k!} \\ &= \frac{1}{e} \, \sum_{k=0}^{\infty} \frac{1}{k!} \, \sum_{n=0}^{\infty} \frac{(k t)^{n}}{n!} \\ &= \frac{1}{e} \, \sum_{k=0}^{\infty} \frac{e^{k t}}{k!} \\ &= e^{e^{t} - 1}. \end{align} This is the generating function for the Bell numbers, $B_{n}$. This implies that $f(n) = B_{n}$.
Leucippus
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