Inverse map for partition transform Let $(a_n)$, $n\in\mathbb{N}$, be a sequence of complex numbers, then formally one has 
(1)
$$\prod_{1}^{\infty}\left(1-a_nx^n\right)^{-1}=1+\sum_{1}^{\infty}\left(\sum_{j_1+2j_2+\cdots +nj_n=n}a_1^{j_1}a_2^{j_2}\cdots a_n^{j_n}\right)x^n=1+\sum_{1}^{\infty}b_n x^n,$$
say. I'm almost certain the answer to my question has been known for centuries, but I don't know where to find it: 
What is the inverse of the map defined by
(2)
$$b_n=\sum_{j_1+2j_2+\cdots +nj_n=n}a_1^{j_1}a_2^{j_2}\cdots a_n^{j_n},$$
and where can I find it in the literature? 
 A: Kevin Smith wrote:

I'm almost certain the answer to my question has been known for centuries, but I don't know where to find it...

You're right! It goes back to Euler. You're looking for a variation on Euler's infinite product representation algorithm (EIPRA). EIPRA takes
as input a sequence $b_n$ and outputs a sequence $a_n$ such that 
$$1+\sum_{n=1}^{\infty}b_n x^n = \prod_{n=1}^{\infty}\left(1-x^n\right)^{-a_n},$$
whereas you are looking for an algorithm that takes as input a sequence $b_n$ and outputs a sequence  $a_n$ such that
$$1+\sum_{n=1}^{\infty}b_n x^n = \prod_{n=1}^{\infty}\left(1-a_nx^n\right)^{-1}.$$
Fortunately, the key to both algorithms is to take the logarithmic derivative of both sides of the
equation, which transforms the infinite product into a Lambert series.  
[I see from the comments
that you already considered logarithmic differentiation, so  you must be looking for
something more explicit than the recurrence that follows.]
Taking the logarithmic derivative of both sides of your equation (1) (and multiplying by $x$),
$$
\sum_{n=1}^{\infty}\frac{a_n n x^n}{1-a_n x^n}=
\frac{\sum_{n=1}^{\infty}b_n n x^n}{ 1+\sum_{ n=1}^\infty b_n x^n } 
 :=  \sum_{n=1}^\infty c_n x^n .
$$ 
The sequence $c_n$ is easily determined by the sequence $b_n$, and the sequence $a_n$ is then determined from the sequence $c_n$ as in EIPRA.
Expanding $\frac{1}{1-a_n x^n}$ as a geometric series,
$$
\sum_{m=1}^\infty \sum_{j=1}^\infty m a_m^j x^{m j}
=  \sum_{n=1}^\infty c_n x^n ,
$$
hence, $a_n$ is defined recursively by
$$
n a_n 
=  c_n - \sum_{m|n,m\ne n} m a_m^{\frac n m}
.
$$  
A: To simplify things, one might consider the reciprocal generating function and expands the reciprocal of the product on the l.h.s. of (1) into 
(3)
    $$\prod_{1}^{\infty}(1-a_nx^n)=1+\sum_{1}^{\infty}c_nx^n,$$
where 
(4)
    $$c_n=\sum_{j_1+\cdots +nj_n=n:j_i\leq 1}(-a_1)^{j_1}\cdots (-a_n)^{j_n}$$
counts only partitions into distinct parts (i.e. $j_i\leq 1$). Noting that, by the multinomial expansion, one also has
(5)
$$c_n=\sum_{j_1+\cdots +nj_n=n}{{j_1+\cdots +j_n}\choose{j_1,...,j_n}}(-b_1)^{j_1}\cdots (-b_n)^{j_n}$$
 this reduces the problem to that of inverting the map defined by (4), and composition with the involution (5). To this end, recursive examination of (4) for the first 6 terms gives: 
    $$a_1=-c_1$$
    $$a_2=-c_2$$
    $$a_3=-c_3+c_1c_2$$
    $$a_4=-c_4+c_1c_3-c_1^2c_2$$
    $$a_5=-c_5+c_1c_4-c_1^2c_3+c_1^3c_2-c_1c_2^2+c_2c_3$$
    $$a_6=-c_6+c_1c_5-c_1^2c_4+c_1^3c_3-c_1^4c_2+c_1^2c_2^2-c_1c_2c_3+c_2c_4.$$
i.e. those partitions that are not of the form $n=dj_d$ for any $d < n$. So, one then supposes that 
(6)
    $$a_n=\sum_{j_1+\cdots +nj_n=n:dj_d\neq n:d < n}(-c_1)^{j_1}\cdots (-c_n)^{j_n}$$ 
is the inverse map. In an attempt to verify this conjecture, one may add in the missing terms to get 
    $$a_n+c_n+\sum _{d|n}(-c_d)^{n/d}=\sum_{j_1+\cdots +nj_n=n}(-c_1)^{j_1}\cdots (-c_n)^{j_n}.$$
UPDATE (02/04/12): The conjecture is incorrect - Martin Rubey has verified that the partitions are correct for $n\leq 15$, but the coefficients are not all unity.
