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a(n) = Sum_{k=0..n} binomial(2*n, n-k) * k^n.
1

%I #25 Oct 03 2021 09:09:24

%S 1,1,8,90,1408,28350,697344,20264244,679313408,25805186550,

%T 1095482736640,51397070440716,2640925289349120,147491783753286700,

%U 8895880971425939456,576279075821454657000,39905347440408027725824,2941534126495441574472870,229966392623413457628168192

%N a(n) = Sum_{k=0..n} binomial(2*n, n-k) * k^n.

%H Seiichi Manyama, <a href="/A345876/b345876.txt">Table of n, a(n) for n = 0..360</a>

%F a(n) ~ 2^(2*n + 1/2) * r^(n+1) * n^n / (sqrt(1 + r^2) * exp(n) * (1 - r^2)^n), where r = 0.647918229029602749602061258113970414114660380467168496836586... is the positive root of the equation (1 + r) = (1 - r)*exp(1/r).

%t Join[{1}, Table[Sum[Binomial[2*n, n-k]*k^n, {k, 0, n}], {n, 1, 20}]]

%o (PARI) a(n) = sum(k=0, n, binomial(2*n, n-k) * k^n); \\ _Michel Marcus_, Oct 03 2021

%Y Cf. A072034, A298851.

%K nonn

%O 0,3

%A _Vaclav Kotesovec_, Oct 03 2021