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%I #4 Dec 17 2017 15:59:27
%S 1,1,4,34,482,10056,286372,10591372,491169996,27826318000,
%T 1887581200256,150885500428224,14028718134958936,1500672248541122944,
%U 182987661921689610000,25231215606822797450176
%N E.g.f.: A(x) = exp( Sum_{n>=1} n(n+1)/2 * a(n-1)*x^n/n! ) = Sum_{n>=0} a(n)*x^n/n! with a(0)=1.
%F a(n) = Sum_{k=1..n} k(k+1)/2 * C(n-1,k-1)*a(k-1)*a(n-k) for n>0, with a(0)=1.
%F E.g.f. satisifies: A(x) = exp( d/dx x^2*A(x)/2 ). - _Paul D. Hanna_, Dec 17 2017
%e E.g.f: A(x) = 1 + x + 4*x^2/2! + 34*x^3/3! + 482*x^4/4! + 10056*x^5/5! +...
%e log(A(x)) = x + 3*1*x^2/2! + 6*4*x^3/3! + 10*34*x^4/4! + 15*482*x^5/5! +...
%e such that log(A(x)) = x*A(x) + x^2*A'(x)/2 = d/dx x^2*A(x)/2.
%o (PARI) {a(n) = if(n==0,1,n!*polcoeff(exp(sum(k=1,n,k*(k+1)/2*a(k-1)*x^k/k!)+x*O(x^n)),n))}
%o for(n=0,25,print1(a(n),", "))
%o (PARI) {a(n) = if(n==0,1,sum(k=1,n,k*(k+1)/2*binomial(n-1,k-1)*a(k-1)*a(n-k)))}
%o for(n=0,25,print1(a(n),", "))
%o (PARI) {a(n) = my(A=1); for(i=1,n, A = exp(deriv(x^2*A/2 +x^2*O(x^n)))); n!*polcoeff(A,n)}
%o for(n=0,25,print1(a(n),", ")) \\ _Paul D. Hanna_, Dec 17 2017
%Y Cf. A156326, A156327.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Feb 08 2009
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