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%I #17 Jun 07 2019 11:33:37
%S 1,33,1268,50600,1972128,75121312,2803732096,102885494016,
%T 3722920064000,133152625650176,4715897847097344,165643005814853632,
%U 5776871664703455232,200235592430802124800,6903358709034568712192,236882142098621090889728,8094539021386254685569024
%N a(n) = Sum_{m=0..n} (Sum_{k=0..m} binomial(n,k))^5.
%H Seiichi Manyama, <a href="/A294436/b294436.txt">Table of n, a(n) for n = 0..300</a>
%H N. J. Calkin, <a href="http://dx.doi.org/10.1016/0012-365X(94)90394-8">A curious binomial identity</a>, Discr. Math., 131 (1994), 335-337.
%H M. Hirschhorn, <a href="http://dx.doi.org/10.1016/0012-365X(95)00086-C">Calkin's binomial identity</a>, Discr. Math., 159 (1996), 273-278.
%F a(n) ~ n * 2^(5*n - 1). - _Vaclav Kotesovec_, Jun 07 2019
%p A:=proc(n,k) local j; add(binomial(n,j),j=0..k); end;
%p S:=proc(n,p) local i; global A; add(A(n,i)^p, i=0..n); end;
%p [seq(S(n,5),n=0..30)];
%t Table[Sum[Sum[Binomial[n,k], {k,0,m}]^5, {m,0,n}], {n,0,15}] (* _Vaclav Kotesovec_, Jun 07 2019 *)
%o (PARI) a(n) = sum(m=0, n, sum(k=0, m, binomial(n,k))^5); \\ _Michel Marcus_, Nov 18 2017
%Y Same expression with exponent b instead of 5: A001792 (b=1), A003583 (b=2), A007403 (b=3), A294435 (b=4).
%K nonn
%O 0,2
%A _N. J. A. Sloane_, Nov 17 2017
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