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%I #27 Mar 27 2021 08:14:17
%S 1,2,514,39368,10601986,2003906252,588906874144,159219918144128,
%T 51207103076632066,16425660314368351892,5697191745563573732764,
%U 2010823973962863400708688,739753103704422167184400096,277511604090132008416695054272,106814999715696983804826836579584
%N a(n) = Sum_{k = 0..n} C(n,k)^9.
%H Vincenzo Librandi, <a href="/A182446/b182446.txt">Table of n, a(n) for n = 0..200</a>
%H Vaclav Kotesovec, <a href="/A182446/a182446_1.txt">Recurrence (of order 5)</a>
%H M. A. Perlstadt, <a href="http://dx.doi.org/10.1016/0022-314X(87)90069-2">Some Recurrences for Sums of Powers of Binomial Coefficients</a>, Journal of Number Theory 27 (1987), pp. 304-309.
%F Asymptotic (p = 9): a(n) ~ 2^(p*n)/sqrt(p)*(2/(Pi*n))^((p - 1)/2)*( 1 - (p - 1)^2/(4*p*n) + O(1/n^2) ).
%F For r a nonnegative integer, Sum_{k = r..n} C(k,r)^9*C(n,k)^9 = C(n,r)^9*a(n-r), where we take a(n) = 0 for n < 0. - _Peter Bala_, Jul 27 2016
%F Sum_{n>=0} a(n) * x^n / (n!)^9 = (Sum_{n>=0} x^n / (n!)^9)^2. - _Ilya Gutkovskiy_, Jul 17 2020
%p a := n -> hypergeom([seq(-n, i=1..9)],[seq(1, i=1..8)],-1):
%p seq(simplify(a(n)),n=0..14); # _Peter Luschny_, Jul 27 2016
%t Table[Sum[Binomial[n, k]^9, {k, 0, n}], {n, 0, 25}]
%o (PARI) a(n) = sum(k=0, n, binomial(n,k)^9); \\ _Michel Marcus_, Jul 17 2020
%Y Sum_{k = 0..n} C(n,k)^m for m = 1..12: A000079, A000984, A000172, A005260, A005261, A069865, A182421, A182422, A182446, A182447, A342294, A342295.
%K nonn,easy
%O 0,2
%A _Vaclav Kotesovec_, Apr 29 2012
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