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a(n) = Sum_{k = 0..n} binomial(n,k)^11.
12

%I #21 Aug 04 2022 03:28:17

%S 1,2,2050,354296,371185666,200097656252,222100237312864,

%T 193798873701831680,231719476114879600642,257097895846251291074612,

%U 330463219813679264204224300,419460465362069257397304825200,573863850341313751827291703127200

%N a(n) = Sum_{k = 0..n} binomial(n,k)^11.

%H Seiichi Manyama, <a href="/A342294/b342294.txt">Table of n, a(n) for n = 0..306</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 a(n) ~ 2^(p*n)/sqrt(p) * (2/(Pi*n))^((p-1)/2) * (1 - (p-1)^2/(4*p*n)), set p=11. - _Vaclav Kotesovec_, Aug 04 2022

%o (PARI) a(n) = sum(k=0, n, binomial(n, k)^11); \\ _Michel Marcus_, Mar 27 2021

%Y Column 11 of A309010.

%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

%O 0,2

%A _N. J. A. Sloane_, Mar 27 2021