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G.f. A(x) satisfies: Sum_{n>=0} n!*x^n/A(x)^n = 1/(1-x).
3

%I #8 Aug 01 2015 14:19:13

%S 1,1,2,6,26,150,1066,8862,83506,874302,10035538,125082870,1680770250,

%T 24211249062,372151797498,6080329604238,105238649649762,

%U 1923853815102030,37047429233963170,749689860387252966

%N G.f. A(x) satisfies: Sum_{n>=0} n!*x^n/A(x)^n = 1/(1-x).

%H Vaclav Kotesovec, <a href="/A159311/b159311.txt">Table of n, a(n) for n = 0..446</a>

%F G.f. satisfies: A(x) = exp( Sum_{n>=1} [(n-1)*a(n) + 1]*x^n/n ).

%F G.f. satisfies: x*A'(x) = A(x)*(1/(1-x) - A(x))/(1 - A(x)).

%F G.f.: A(x) = x/Series_Reversion(G(x)) so that G(x/A(x)) = x where G(x) = g.f. of A003319 (indecomposable permutations).

%F a(n) ~ n*n!/exp(1) * (1 - 2/n - 3/n^2 - 49/(3*n^3) - 379/(3*n^4) - 18509/(15*n^5)). - _Vaclav Kotesovec_, Aug 01 2015

%e G.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 26*x^4 + 150*x^5 + 1066*x^6 +...

%e log(A(x)) = x + 3*x^2/2 + 13*x^3/3 + 79*x^4/4 + 601*x^5/5 + 5331*x^6/2 +...

%e where coefficients of log(A(x)) is given by (n-1)*a(n) + 1:

%e 3 = 1*2 + 1, 13 = 2*6 + 1, 79 = 3*26 + 1, 601 = 4*15 + 1, 5331 = 5*1066 + 1.

%e Let G(x) = g.f. of A003319, then G(x/A(x)) = x where:

%e G(x) = x + x^2 + 3*x^3 + 13*x^4 + 71*x^5 + 461*x^6 + 3447*x^7 +...

%e and G(x) = 1 - 1/(1 + x + 2!*x^2 + 3!*x^3 + 4!*x^4 +...).

%o (PARI) {a(n)=local(G003319=1-1/sum(k=0,n+1,k!*x^k+x^2*O(x^n)));polcoeff(x/serreverse(G003319),n)}

%Y Cf. A003319, A159312.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Apr 16 2009

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