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I have a real life problem in which I need to find the $n^\text{th}$ number in the sequence below:

1  --->   0
2  --->   2
3  --->   6
4  --->   20
5  --->   70
6  --->   252
7  --->   924
8  --->   3432
9  --->   12870

Beyond the right side being divisible by $n$, I can't see the pattern or formula of getting from $n$ to the right side.

I can compute the value on the right side given an integer of $n$, so I can verify if someone, for example, knows what the value for $n=10$ should be.

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  • $\begingroup$ "knows what the value for $n=10$ should be." Actually, there are several possibilities, see my answer. $\endgroup$ – Dietrich Burde Aug 17 at 18:42
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    $\begingroup$ No matter what the first nine or one thousand terms are, the answer to the question is “whatever”. $\endgroup$ – egreg Aug 17 at 18:45
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OEIS is often helpful: Central binomial coefficients

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  • $\begingroup$ omg that was quick! Did you just recognise the sequence or what? $\endgroup$ – piccolo Aug 17 at 18:06
  • $\begingroup$ Well $252$ seemed familiar, but actually I just searched OEIS :D $\endgroup$ – ploosu2 Aug 17 at 18:07
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This is probably the central binomial coefficient $$\binom{2n}{n} = \frac{(2n)!}{(n!)^2}. $$ On the other hand, it can also be a polynomial, namely $$ f(x)=\frac{79x^8}{2880}−\frac{4583x^7}{5040}+\frac{3739x^6}{288} −\frac{7409x^5}{72}+\frac{1413631x^4}{2880} $$

$$ −\frac{1028681x^3}{720} +\frac{117655x^2}{48} −\frac{62509x}{28}+810, $$ which has values $f(1),f(2),\ldots ,f(9)$ as above. Here $f(10)=45480$, whereas $\binom{20}{10}=48620$.

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  • $\begingroup$ Nice and clean!! (+1) $\endgroup$ – Mostafa Ayaz Aug 17 at 18:09
  • $\begingroup$ Oh, thank you!! $\endgroup$ – Dietrich Burde Aug 17 at 18:10

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