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a(n) = n! * Sum_{k=0..n} binomial(2*n, 2*k) / binomial(n,k).
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%I #19 Apr 26 2023 11:21:30

%S 1,2,10,72,664,7440,98064,1486464,25476480,487192320,10284768000,

%T 237574149120,5960907832320,161440734873600,4694193123379200,

%U 145855192928256000,4822943651308339200,169104439543534387200,6266811206473703424000

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

%H Vincenzo Librandi, <a href="/A182525/b182525.txt">Table of n, a(n) for n = 0..200</a>

%H Sela Fried and Toufik Mansour, <a href="https://arxiv.org/abs/2304.05728">Graph labelings obtainable by random walks</a>, arXiv:2304.05728 [math.CO], 2023.

%F Asymptotic: a(n) ~ Pi*n^(n+1)*2^(n-1/2)/exp(n). [_Vaclav Kotesovec_, May 06 2012]

%F E.g.f.: ( x * arctan(x / sqrt(1 - 2*x)) + sqrt(1 - 2*x) ) / (sqrt(1 - 2*x))^3. -_Sela Fried_, Apr 26 2023

%e E.g.f.: A(x) = 1 + 2*x + 10*x^2/2! + 72*x^3/3! + 664*x^4/4! + 7440*x^5/5! +...

%e a(1) = 1!*(1/1 + 1/1) = 2;

%e a(2) = 2!*(1/1 + 6/2 + 1/1) = 10;

%e a(3) = 3!*(1/1 + 15/3 + 15/3 + 1/1) = 72;

%e a(4) = 4!*(1/1 + 28/4 + 70/6 + 28/4 + 1/1) = 664;

%e a(5) = 5!*(1/1 + 45/5 + 210/10 + 210/10 + 45/5 + 1/1) = 7440; ...

%t RecurrenceTable[{a[n]==3*n*a[n-1]-n*(2n-3)*a[n-2],a[0]==1,a[1]==2},a,{n,0,25}] (* _Vaclav Kotesovec_, May 06 2012 *)

%o (PARI) {a(n)=n!*sum(k=0,n,binomial(2*n, 2*k)/binomial(n,k))}

%K nonn

%O 0,2

%A _Paul D. Hanna_, May 03 2012