TITLE: Partitions comprised only of divisors
QUESTION [6 upvotes]: How many of the partitions of a natural number $n$ are comprised only of its divisors? That is, if $$p(n)=\sum_{\sum_{1}^n kj_k=n:j_k\geq 0} 1_{\[j_1,j_2,...\]},$$
is the ordinary partition function (i.e. the total number of partitions of $n$), then I want to know something about the counting function $$s(n)=\sum_{\sum_{d|n}dj_d=n:j_d\geq 0}1_{[j_1,...,j_d,...]}.$$
I would be happy to hear of anything that is known about this function, but I am particularly interested in (a) its generating functions, and (b) a bijection between this restricted partition and another (hopefully more intuitive to count) restricted partition. Any insights would be welcome. 
I should add that google finds a number of papers that study "partitions of $n$ into divisors of $m$", e.g. Gupta, 1970s, but those methods reduce to rather vacuous statements when evaluated at $m=n$.
Thanks!

REPLY [8 votes]: Bounds for this partition function were given by the editors of The American Mathematical Monthly, Paul Erdos, and Andrew Odlyzko in the March 1992 issue, p. 277, as a solution to Advance Problem number 6640. The bounds they prove are:
$$
({\tau(n)}/{2}-1)(\log n + O({\log n}/{\log \log n}))
\leq 
\log s(n)
\leq
({\tau (n)}/{2})\log n + O(\log \log n),
$$
where $\tau(n)$ is the number of divisors of $n$.