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%I
%S 1,2,13,172,3809,126526,5874517,362848088,28744087297,2839192902874,
%T 341922922464701,49297062811573732,8380916229314577313,
%U 1658770724530766046422,378056469777362366873989,98286603829297813268996176,28907477297195536067142301697
%N a(n) = Sum_{k=0..n} (n!/k!)^2*binomial(n,k).
%H Vincenzo Librandi, <a href="/A119400/b119400.txt">Table of n, a(n) for n = 0..100</a>
%F Sum_{n>=0} a(n)*x^n/n!^2 = BesselI(0,2*sqrt(x/(1-x)))/(1-x).
%F Recurrence: a(n)=(3*n^2-3*n+2)*a(n-1)-3*(n-1)^4*a(n-2)+(n-2)^3*(n-1)^3*a(n-3). - _Vaclav Kotesovec_, Sep 30 2012
%F a(n) ~ 1/sqrt(3)*n^(2*n+2/3)/exp(2*n-3*n^(1/3)). - _Vaclav Kotesovec_, Sep 30 2012
%F E.g.f.: exp(x) * Sum_{n>=0} x^n/n!^3 = Sum_{n>=0} a(n)*x^n/n!^3. - _Paul D. Hanna_, Nov 27 2012
%t Table[Sum[(n!/k!)^2*Binomial[n, k], {k, 0, n}], {n, 0, 16}] - _Stefan Steinerberger_, Jun 17 2007
%o (PARI) {a(n)=n!^3*polcoeff(exp(x+x*O(x^n))*sum(m=0, n, x^m/m!^3), n)} \\ _Paul D. Hanna_, Nov 27 2012
%o for(n=0,25,print1(a(n),", "))
%Y Cf. A000522, A002720, A216831
%K easy,nonn,changed
%O 0,2
%A _Vladeta Jovovic_, Jul 25 2006
%E More terms from _Stefan Steinerberger_, Jun 17 2007
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