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A196304 G.f.: x = Sum_{n>=1} a(n)*x^n/(1 + n*(n+1)/2*x)^n. 2

%I #22 May 05 2017 04:52:24

%S 1,1,5,64,1587,65421,4071178,357962760,42379107165,6512954469625,

%T 1262574678261816,301690485704179584,87187147717429037215,

%U 29994563760476311689525,12119686846920536310216000,5685713204308826743851247936,3066004482905684870319668989977

%N G.f.: x = Sum_{n>=1} a(n)*x^n/(1 + n*(n+1)/2*x)^n.

%C Compare g.f. to: x = Sum_{n>=1} n^(n-2)*x^n/(1 + n*x)^n, which generates coefficients in the series reversion of x*exp(-x).

%H Alois P. Heinz, <a href="/A196304/b196304.txt">Table of n, a(n) for n = 1..200</a>

%F E.g.f.: x = Sum_{n>=1} a(n)*x^n/(n-1)! * exp(-n*(n+1)/2*x).

%F a(n) = A195737(n)/n for n>=1.

%F a(n) = Sum_{k=1..n-1} (-1)^(k-1)*binomial(n-1,k)*binomial(n+1-k,2)^k*a(n-k) for n>=2. - _Jonathan Noel_, May 05 2017

%e x = x/(1+x) + 1*x^2/(1+3*x)^2 + 5*x^3/(1+6*x)^3 + 64*x^4/(1+10*x)^4 + 1587*x^5/(1+15*x)^5 +...+ a(n)*x^n/(1+n*(n+1)/2*x)^n +...

%e The coefficients a(n) also satisfy:

%e x = x*exp(-x) + 1*x^2/1!*exp(-3*x) + 5*x^3/2!*exp(-6*x) + 64*x^4/3!*exp(-10*x) + 1587*x^5/4!*exp(-15*x) +...+ a(n)*x^n/(n-1)!*exp(-n*(n+1)/2*x) +...

%p p:= l-> (n-> n!*LinearAlgebra[Determinant](Matrix(n, (i, j)

%p -> (t->`if`(t<0, 0, l[i]^t/t!))(j-i+1))))(nops(l)):

%p a:= n-> p([i*(i+1)/2$i=1..n-1]):

%p seq(a(n), n=1..20); # _Alois P. Heinz_, Dec 03 2015

%o (PARI) {a(n)=if(n<1, 0, polcoeff(x-sum(m=1, n-1, a(m)*x^m/(1+m*(m+1)/2*x+x*O(x^n))^m), n))}

%o (PARI) {a(n)=if(n<1, 0, (n-1)!*polcoeff(x-sum(m=1, n-1, a(m)*x^m/(m-1)!*exp(-m*(m+1)/2*x+x*O(x^n))), n))}

%Y Cf. A195737, A082157.

%K nonn

%O 1,3

%A _Paul D. Hanna_, Sep 30 2011

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