OFFSET
0,3
COMMENTS
Compare to e.g.f. G(x) of A052802, which satisfies: G(x) = 1/(1 + log(1 - x*G(x))).
FORMULA
E.g.f. satisfies: A(x*(1 - log(1+x))) = 1/(1 - log(1+x)).
E.g.f.: A(x) = (1/x)*Series_Reversion[x - x*log(1+x)].
a(n) = n!*[x^n] 1/(1 - log(1+x))^(n+1)/(n+1).
a(n) = Sum_{k=0..n} (binomial(n+k,n) * Sum_{j=0..k} (-1)^(j)*binomial(k,j) * (Sum_{i=0..j} (-1)^i*i!*binomial(j,i)*Stirling1(n,i)))/(n+1). - Vladimir Kruchinin, Feb 04 2012
a(n) ~ n^(n-1) / ((1-c)*sqrt(1+c) * exp(n) * (1/c+c-2)^n), where c = LambertW(1). - Vaclav Kotesovec, Dec 28 2013
a(n) = (1/(n+1)!) * Sum_{k=0..n} (n+k)! * Stirling1(n,k). - Seiichi Manyama, Nov 06 2023
EXAMPLE
E.g.f.: A(x) = 1 + x + 3*x^2/2! + 17*x^3/3! + 144*x^4/4! + 1634*x^5/5! + ...
where log(1 + x*A(x)) equals
1 - 1/A(x) = x + x^2/2! + 5*x^3/3! + 38*x^4/4! + 404*x^5/5! + 5514*x^6/6! + ...
MATHEMATICA
a[n_] := Sum[ Binomial[n+k, n]*Sum[ (-1)^(j)*Binomial[k, j]*Sum[ (-1)^i*i!*Binomial[j, i]*StirlingS1[n, i], {i, 0, j}], {j, 0, k}], {k, 0, n}]/(n+1); Table[a[n], {n, 0, 19}] (* Jean-François Alcover, Jun 24 2013, after Vladimir Kruchinin *)
CoefficientList[1/x*InverseSeries[Series[x-x*Log[1+x], {x, 0, 20}], x], x] * Range[0, 19]! (* Vaclav Kotesovec, Dec 28 2013 *)
PROG
(PARI) {a(n)=n!*polcoeff((1/x)*serreverse(x-x*log(1+x +x*O(x^n))), n)}
(PARI) {a(n)=n!*polcoeff(1/(1-log(1+x +x*O(x^n)))^(n+1)/(n+1), n)}
(PARI) /* by Vladimir Kruchinin's formula: */
{Stirling1(n, k)=n!*polcoeff(binomial(x, n), k)}
{a(n)=sum(k=0, n, binomial(n+k, n)*sum(j=0, k, (-1)^(j)*binomial(k, j)*sum(i=0, j, (-1)^i*i!*binomial(j, i)*Stirling1(n, i))))/(n+1)} \\ Paul D. Hanna
(Maxima) a(n):=sum(binomial(n+k, n)*sum((-1)^(j)*binomial(k, j)*sum((-1)^i*i!*binomial(j, i)* stirling1(n, i), i, 0, j), j, 0, k), k, 0, n))/(n+1); // Vladimir Kruchinin, Feb 04 2012
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Oct 30 2011
STATUS
approved