OFFSET
0,3
FORMULA
G.f.: A(x) = 1/Product_{n>=1} (1 - x^n)^A179325(n); i.e., this sequence equals the Euler transform of A179325.
Given g.f. A(x), let B(x) be the g.f. of A179325, then A(x) and B(x) satisfy:
_ B(x) = x*A(B(x)) = Series_Reversion(x/A(x)),
_ B(x) = x*exp( Sum_{n>=1} B(B(x)^n)/n ).
G.f. satisfies: A(x) = exp( Sum_{n>=1} Sum_{k>=1} x^(n*k)/(n*k) * {[y^(k-1)] A(y)^k} ), where [y^(k-1)] A(y)^k denotes the coefficient of y^(k-1) in A(y)^k. - Paul D. Hanna, Feb 13 2013
EXAMPLE
G.f.: A(x) = 1 + x + 2*x^2 + 5*x^3 + 18*x^4 + 80*x^5 + 430*x^6 +...
which satisfies:
log(A(x)) = B(x) + B(x^2)/2 + B(x^3)/3 + B(x^4)/4 +...
where B(x) = x*A(B(x)) = g.f. of A179325 begins:
B(x) = x + x^2 + 3*x^3 + 12*x^4 + 59*x^5 + 331*x^6 + 2062*x^7 +...
PROG
(PARI) a(n)=local(A=1+x, B=x); for(i=1, n, B=serreverse(x/(A+x*O(x^n))); A=exp(sum(m=1, n, subst(B, x, x^m+x*O(x^n))/m))); polcoeff(A, n)
(PARI) /* As the Euler transform of A179325: */
{a(n)=local(A=1+x, B=x); for(i=1, n, B=serreverse(x/(A+x*O(x^n))); A=1/prod(k=1, n, (1-x^k+x*O(x^n))^polcoeff(B, k))); polcoeff(A, n)}
(PARI) a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, sum(k=1, n, x^(m*k)/(m*k)*polcoeff(A^k+O(x^k), k-1)) +x*O(x^n)))); polcoeff(A, n)
for(n=0, 25, print1(a(n), ", ")) \\ Paul D. Hanna, Feb 13 2013
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jul 06 2011
STATUS
approved