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a(n) = Sum_{k=0..n} n^k * binomial(n,k)^3.
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%I #16 Jul 11 2020 23:45:11

%S 2,21,352,8065,231876,7951069,314931968,14095941633,701590424500,

%T 38358147922501,2281458125531520,146469277526152321,

%U 10084388675810865248,740560093656498673965,57738578482070455269376,4760258648137662340202497,413561386818608994516491316

%N a(n) = Sum_{k=0..n} n^k * binomial(n,k)^3.

%H Vincenzo Librandi, <a href="/A241247/b241247.txt">Table of n, a(n) for n = 1..200</a>

%F a(n) ~ exp(1 - 3*n^(1/3)/2 + 3*n^(2/3)) * n^(n-2/3) / (2*Pi*sqrt(3)) * (1 + 5/(4*n^(1/3))).

%t Table[Sum[n^k*Binomial[n,k]^3,{k,0,n}],{n,1,20}]

%t Table[HypergeometricPFQ[{-n,-n,-n},{1,1},-n],{n,1,20}]

%o (PARI) a(n) = sum(k=0, n, n^k*binomial(n,k)^3); \\ _Michel Marcus_, Jul 11 2020

%Y Cf. A187021, A234971.

%K nonn,easy

%O 1,1

%A _Vaclav Kotesovec_, Apr 18 2014