ms said: I know the solution a[n] to 1-px = product(n = 1 to infinity) (1-x^n)^a[n] through a rather indirect route, but are there techniques for directly solving this equation? me: what's often called the Euler transform on sequences will do the trick this is from the "transforms" file that you can find on the OEIS web site: # EULER takes [a_1,a_2,a_3,...] and produces [b_1,b_2,b_3,...] with # 1 + Sum_{n=1..inf} b_n x^n = 1 / Product_{n=1..inf} (1-x^n)^a_n. here's the Maple code: EULER:=proc(a) local b,c,i,d: if whattype(a) <> list then RETURN([]); fi: c:=[]: for i to nops(a) do c:=[op(c), add( d*did(i,d)*a[d] , d=1..i)]: od: b:=[]: for i to nops(a) do b:=[op(b),(1/i)*(c[i]+ add( c[d]*b[i-d], d=1..i-1))]: od: RETURN(b); end: Here "did" is "did it divide": did:=proc(m,n) if irem(m,n) = 0 then 1 else 0: fi: end: Neil Sloane