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a(n) = Sum_{m=0..n} (Sum_{k=0..m} binomial(n,k))^5.
5

%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