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Wed Oct 29 10:20:21 CET 2008
gf(A000041)*gf(A010815) = 1..................(3)
where the two generating functions are defined by
gf(A000041) = sum_{i=0..infinity} A000041(i)*x^i ;
gf(A010815) = sum_{j=0..infinity} A010815(j)*x^j ;
so the long write-up of (3) is
sum_{i=0..infinity} A000041(i)*x^i * sum_{j=0..infinity} A010815(j)*x^j = 1
therefore by division
1/ sum_{j=0..infinity} A010815(j)*x^j = sum_{i=0..infinity} A000041(i)*x^i
and by the substitution x->x^n
1/ sum_{j=0..infinity} A010815(j)*x^(j*n) = sum_{i=0..infinity} A000041(i)*x^(i*n)
Insertion into equation (2) is
gf(JA) = 1+sum_{n>=1} x^n * sum_{i=0..infinity} A000041(i)*x^(i*n)
= 1+sum_{n>=1} sum_{i=0..infinity} A000041(i)*x^((i+1)*n)
= 1+sum_{n>=1} sum_{l=1..infinity} A000041(l-1)*x^(l*n)
Interchange of the two summations and insertion of the geometric series standard
sum_{n>=1} x^(l*n) = x^l/(1-x^l)
yields
gf(JA) = 1+ sum_{l=1..infinity} A000041(l-1)*x^l/(1-x^l)
which is (1), the result from Gary's reference.
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