From endomorphisms of rank 1 of the full transformation semigroup $[n]^{[n]}$ or idempotents in $[n]^{[n]}$, we have $$c_n:=\sum_{m=1}^n\binom{n}mm^{n-m}.$$
QUESTION. Is it true that $n$ divides $c_n - 1$?
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asked Dec 27, 2020 at 16:17
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2$\begingroup$ Yes, it is. See Exercise 1 (c) in my UMN Fall 2017 Math 4990 homework set #6. The proof is a simple application of the absorption formula $m \dbinom{n}{m} = n \dbinom{n-1}{m-1}$. If I am not mistaken, it is easy to make this a bijective proof. $\endgroup$– darij grinbergCommented Dec 27, 2020 at 16:42
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