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Let $c(n,k)$ be the $k^{th}$ unsigned Stirling number of the first kind. There are formulas for big $k$'s, namely, $c(n,n-1)= \binom{n}{2}$ and $c(n,n-2) = \frac{3n-1}{4} \binom{n}{3}$.

I wonder whethere there are formulas for small $k$'s, e.g. $c(n,1), c(n,2)$?

metamorphy
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Ninja
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    Does your convention imply that $c\left(n,k\right)$ counts the number of permutations of $\left{1,2,\ldots,n\right}$ having exactly $k$ cycles? In that case, $c\left(n,1\right)$ is simply the number of $n$-cycles, which is $\left(n-1\right)!$. As for $c\left(n,2\right)$, there is the formula $c\left(n,2\right) = \left(n-1\right)! H_{n-1}$, where $H_j$ denotes the $j$-th harmonic number $1/1 + 1/2 + \cdots + 1/j$. Barring a good formula for $H_j$, I don't think a good one exists for $c\left(n,2\right)$. – darij grinberg Sep 30 '19 at 16:00
  • Yes, it is. Thus, I wonder some closed form for $c(n,3)$ and the rest. – Ninja Sep 30 '19 at 16:22
  • OEIS for $c(n,2)$: https://oeis.org/A000254. You can also find $c(n,3)$ there, though if $c(n,2)$ is hard to get a nice formula for, I can't imagine $c(n,3)$ would be much easier. – Kevin Long Sep 30 '19 at 19:24
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    See the end of this section. – metamorphy Sep 30 '19 at 19:24
  • @metamorphy, thank you for your valuable comment. I checked the related part of the generalization. However, there is 'Pochhammer symbol', $(m)_{k}$ in the formula but it is not clear how it is used. The page of the symbol says that it may be used for either the rising or the falling factorial. Lastly, Wiki says that Pochhammer himself used it as $\binom{m}{k}$. Do you have any idea on this symbol and how it is used here? – Ninja Oct 01 '19 at 22:21
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    Well, it has to be defined wherever it is used ;) Throughout that article, it means the falling factorial (as stated at the very beginning; one can confirm it by trying a few small $k$). – metamorphy Oct 02 '19 at 04:59
  • Just answered this older question (asking basically the same). – metamorphy Jan 30 '21 at 16:31

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