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%I #8 Jan 22 2014 13:08:38
%S 1,1,3,21,171,1821,24123,373941,6693291,135897741,3081969243,
%T 77250233061,2120715880011,63277499072061,2039050439495163,
%U 70571948084252181,2610905715855178731,102824333281385113581
%N E.g.f. A(x) = series reversion of log(1+x)-x^3/3.
%F a(n)=(sum(k=0..n-1, (n+k-1)!*sum(j=0..k, (-1)^j/(k-j)!*sum(i=0..min(j,(n+j-1)/3),(1/i!)*(-1)^i* stirling1(n-3*i+j-1,j-i)/(3^i*(n-3*i+j-1)!))))), n>0.
%F a(n) ~ n^(n-1) / (sqrt(s*(2+s^3)) * exp(n) * r^(n-1/2)), where s = 1/3*(-1 + (25/2 - (3*sqrt(69))/2)^(1/3) + (1/2*(25 + 3*sqrt(69)))^(1/3)) = 0.75487766624669276... is the root of the equation s^2*(1+s) = 1 and r = log(1+s) - s^3/3 = 0.4190125789786... - _Vaclav Kotesovec_, Jan 22 2014
%t Rest[CoefficientList[InverseSeries[Series[Log[1+x]-x^3/3, {x, 0, 20}], x],x]*Range[0, 20]!] (* _Vaclav Kotesovec_, Jan 22 2014 *)
%o (Maxima) a(n):=(sum((n+k-1)!*sum((-1)^j/(k-j)!*sum((1/i!)*(-1)^i*stirling1(n-3*i+j-1,j-i)/(3^i*(n-3*i+j-1)!),i,0,min(j,(n+j-1)/3)),j,0,k),k,0,n-1));
%K nonn
%O 1,3
%A _Vladimir Kruchinin_, Feb 07 2012
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