# Find $n$-th number in this sequence

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.

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

• Well $252$ seemed familiar, but actually I just searched OEIS :D – ploosu2 Aug 17 at 18:07
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 $$\binom{20}{10}=48620$$.