%I #26 Mar 27 2021 08:13:26
%S 1,2,1026,118100,62563330,20019531252,11393421713604,5550455033938152,
%T 3431955863873102850,2052124795850957537060,1367610300690018553312276,
%U 916694195766256069610158152,649630217578404016288230718276,467800319852823195772146025385000
%N a(n) = Sum_{k = 0..n} C(n,k)^10.
%H Vincenzo Librandi, <a href="/A182447/b182447.txt">Table of n, a(n) for n = 0..200</a>
%H Vaclav Kotesovec, <a href="/A182447/a182447_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 = 10): 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)^10*C(n,k)^10 = C(n,r)^10*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!)^10 = (Sum_{n>=0} x^n / (n!)^10)^2. - _Ilya Gutkovskiy_, Jul 17 2020
%p a := n -> hypergeom([seq(-n, i=1..10)],[seq(1, i=1..9)],1):
%p seq(simplify(a(n)),n=0..13); # _Peter Luschny_, Jul 27 2016
%t Table[Sum[Binomial[n, k]^10, {k, 0, n}], {n, 0, 25}]
%o (PARI) a(n) = sum(k=0, n, binomial(n,k)^10); \\ _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