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a(n) = Sum_{k=0..n} n^k * binomial(k+n,k).
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%I #9 Dec 27 2023 07:52:19

%S 1,3,31,643,20421,873806,46994011,3042431715,230249448841,

%T 19940350062394,1944516598602711,210829412453667998,

%U 25156743053019602701,3275876521195372322892,462262670054775645538099,70264375447526610838701091

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

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

%F a(n) ~ 4^n * n^(n - 1/2) / sqrt(Pi). - _Vaclav Kotesovec_, Dec 27 2023

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

%Y Cf. A001700, A026641, A368488.

%K nonn,easy

%O 0,2

%A _Seiichi Manyama_, Dec 27 2023