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I'm calculating

$$\prod _{i=1} ^k\big( n(n-1)-2i \big)$$

for $n$ fixed, and I would like to express it as a sum, using a kind of Newton's formula. The problem is, the coefficients are a bit more complicated than the binomial ones, there is the $-2$ that is always there, but $i$ changes, so that I would need to be able to calculate each time the sum of all the possible products of $i$ distinct terms in $1,2, \dots, k$.

For example, for $k=4$, I would have :

$$\begin{eqnarray} i=1 &:& \quad 1 + 2 + 3 + 4 &=& 10 \\ i=2 &:& \quad 1 \cdot 2 + 1 \cdot 3 + 1 \cdot 4 + 2 \cdot 3 + 2 \cdot 4 + 3 \cdot 4 &=& 35 \\ i=3 &:& \quad 1 \cdot 2 \cdot 3 + 1 \cdot 2 \cdot 4 + 1 \cdot 3 \cdot 4 + 2 \cdot 3 \cdot 4 &=& 50 \\ i=4 &:& \quad 1 \cdot 2 \cdot 3 \cdot 4 &=& 24 \end{eqnarray}$$

Is there a known formula, or at least a name, for those kind of coefficients? Thank you in advance and sorry if it's not clear.

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    $\begingroup$ You can calculate your product directly: it equals $2^k\left(\binom{n}{2}-1\right)!/\left(\binom{n}{2}-k-1\right)!$. $\endgroup$
    – GH from MO
    Commented Feb 22, 2017 at 15:37
  • $\begingroup$ @GHfromMO I wanted to express it as a sum because it was allowing me to cancel other terms of my formula but I think this expression is actually better, it's more understandable. Thanks a lot :) $\endgroup$
    – Alice J.
    Commented Feb 22, 2017 at 16:22
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    $\begingroup$ Your $10,35,50,24$ are Stirling numbers of the first kind. See oeis.org/A094638 or oeis.org/A008276 or oeis.org/A054654 $\endgroup$
    – Gerry Myerson
    Commented Feb 22, 2017 at 22:15
  • $\begingroup$ @GerryMyerson Thank you, that's exactly what I needed ! $\endgroup$
    – Alice J.
    Commented Feb 23, 2017 at 9:02
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    $\begingroup$ Welcome to mathOverflow! By the way, whenever I encounter any integer sequence, the first thing I do is type it into oeis.org. That tells me more about the thing roughly 90% of the time (and when the thing isn't in there, I'll add an entry about it!). That's a good resource to have in your back pocket. $\endgroup$
    – Pat Devlin
    Commented Feb 23, 2017 at 12:55

1 Answer 1

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They are the absolute value of Stirling numbers of the $1^{st}$ kind, wihch are given at OEIS in A008275.

The unsigned version is A094638.

A recurrence formula is: $$a_{n,k}=a_{n-1,k-1}+(n-1)a_{n-1,k}$$

As they are the coefficients of $x^k$ in:

$$\prod_\limits{i=1}^n (x+i)$$

letting $x=1$ means the coefficients sum to $(n+1)!$.

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  • $\begingroup$ happy to help :) $\endgroup$
    – JMP
    Commented Feb 23, 2017 at 9:41

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