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Let $n$ be a natural number. Let $f(x)=\prod_{i=-n}^{n}(x-i)$. If $k$ is an even integer, then the coefficient of $x^k$ is zero. The coefficient of $x^{2n-1}$ is $-(1^2+\cdots+n^2)=-\dfrac{n(n+1)(2n+1)}{6}$. It is easy to see that the sum of all coefficients is zero. How about the rest of coefficients? Is there any explicit formula for them?

Let $g(x)=\dfrac{f(x)}{x-1}$. What are the coefficient of $g(x)$?

Moh514
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    The next coefficient is $\frac{1}{360} n (n^2 - 1) (4 n^2 - 1) (5 n + 6)$. This follows from it being equal to $\sum_{i>j}i^2j^2$ which can be evaluated from considering $(\sum i^2)^2 = 2\sum_{i>j} i^2j^2 + \sum i^4$ – Winther Jun 01 '16 at 01:17

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The coefficients of a polynomial given by its roots are given by Vieta's formulas.

These formulas express the coefficients as elementary symmetric functions of the roots.

In your case it may be simpler to write $f(x)=x\prod_{i=1}^{n}(x^2-i^2)=xh(x^2)$, where $h(x)=\prod_{i=1}^{n}(x-i^2)$, and apply Vieta's formulas to $h$.

The actual coefficients seem to be given by A008955.

lhf
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  • Actually I tried to use Vieta's formulas, but they don't help to find the coefficient explicitly, as far as I know. – Moh514 Jun 01 '16 at 01:15