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Let $(x+1)(x+2)\dotsb(x+n)=c(0,n)x^n+c(1,n)x^{n-1}+c(2,n)x^{n-2}\dotsb+c(n,n)$

By multiplying both sides by $(x+n+1)$ we obtain $c(k,n)=c(k,n-1)+nc(k-1,n-1)$ and it is clear that $c(n,n)=n!$ so $$c(k,n)-c(k,n-1)=nc(k-1,n-1)\implies\sum_{h=k+1}^n c(k,h)-c(k,h-1)=n\sum_{h=k+1}^n c(k-1,n-1)\implies c(k,n)=k!+\sum_{h=k+1}^nc(k-1,n-1)$$ which can be laboriously found with Faulhaber's formulae. The first ones are:

$c(1,n)=\frac{n(n+1)}{2}$

$c(2,n)=\frac{(n-1)n(n+1)(3n+2)}{24}$

$c(3,n)=\frac{5n^6+19n^5+15n^4-15n^3-20n^2-4n-7440}{240}$

It is not hard to show by induction that $c(k,n)$ is a polynomial in n of degree 2k. Is there a closed form for any k?

Sophie
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    They are just (apart index notation) the (unsigned) Stirling Numbers of 1st kind about which there is a vast literature and can be expressed in thousands of ways ! – G Cab Oct 20 '16 at 12:35
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    https://oeis.org/A094638 Next time just put in your integers for a nice $n$ (I took $n=5$) into OEIS, that often answers questions like this. – Dirk Oct 20 '16 at 12:37

1 Answers1

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Notice that

$$(x+1)(x+2)\dots(x+n)=\frac{\Gamma(x+n+1)}{\Gamma(x)}$$

Thus, by Taylor's theorem, we may directly extract coefficients.

You could also recall the unsigned Stirling Numbers of the $\it1$st kind.

$$(x+1)(x+2)\dots(x+n)=\sum_{k=0}^n\left[{n\atop k}\right]x^k$$

Simply Beautiful Art
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