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a(n) = Sum_{k=0..n} binomial(5*n,k).
3

%I #17 Apr 09 2024 11:45:27

%S 1,6,56,576,6196,68406,768212,8731848,100146724,1156626990,

%T 13432735556,156713948672,1835237017324,21560768699762,

%U 253994850228896,2999267652451776,35490014668470052,420718526924212654,4995548847105422048,59402743684137281920

%N a(n) = Sum_{k=0..n} binomial(5*n,k).

%F a(n) = [x^n] 1/((1-2*x) * (1-x)^(4*n)).

%F a(n) ~ 5^(5*n + 1/2) / (3*sqrt(Pi*n) * 2^(8*n - 1/2)). - _Vaclav Kotesovec_, Apr 05 2024

%F a(n) = Sum_{k=0..floor(n/2)} binomial(5*n+1,n-2*k). - _Seiichi Manyama_, Apr 09 2024

%F a(n) = binomial(1+5*n, n)*hypergeom([1, (1-n)/2, -n/2], [1+2*n, 3/2+2*n], 1). - _Stefano Spezia_, Apr 09 2024

%t Table[32^n - Binomial[5*n, 1+n] * Hypergeometric2F1[1, 1 - 4*n, 2+n, -1], {n, 0, 20}] (* _Vaclav Kotesovec_, Apr 05 2024 *)

%o (PARI) a(n) = sum(k=0, n, binomial(5*n, k));

%Y Cf. A032443, A066380, A066381.

%K nonn,easy

%O 0,2

%A _Seiichi Manyama_, Apr 05 2024