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a(n) = Sum_{k = 0..n} C(n,k)^8.
12

%I #27 Mar 27 2021 08:14:49

%S 1,2,258,13124,1810690,200781252,30729140484,4579408029576,

%T 770670360699138,132018919625044100,23913739057463037508,

%U 4433505541977804821256,848185646293853978499844,165563367990287610967653512,32993144260428865295508700680

%N a(n) = Sum_{k = 0..n} C(n,k)^8.

%H Vincenzo Librandi, <a href="/A182422/b182422.txt">Table of n, a(n) for n = 0..200</a>

%H Vaclav Kotesovec, <a href="/A182422/a182422_1.txt">Recurrence (of order 4)</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 = 8): 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)^8*C(n,k)^8 = C(n,r)^8*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!)^8 = (Sum_{n>=0} x^n / (n!)^8)^2. - _Ilya Gutkovskiy_, Jul 17 2020

%p a := n -> hypergeom([seq(-n, i=1..8)],[seq(1, i=1..7)],1):

%p seq(simplify(a(n)),n=0..14); # _Peter Luschny_, Jul 27 2016

%t Table[Total[Binomial[n, Range[0, n]]^8], {n, 0, 20}] (* _T. D. Noe_, Apr 28 2012 *)

%o (PARI) a(n) = sum(k=0, n, binomial(n,k)^8); \\ _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 28 2012