A simple looking problem in partitions that became increasingly complex I began with problem which looked simple in the beginning but became increasingly complex as I dug deeper. 
Main questions: Find the number of solutions $s(n)$ of the equation
$$
n = \frac{k_1}{1} + \frac{k_2}{2} + \ldots + \frac{k_n}{n}
$$
where $k_i \ge 0$ is a non-negative integer. This is my main questions. After tying different approaches, the one that I found most optimistic is as follows. But soon even this turned out to be devil (as we shall see why). 
Let $l_n$ be the LCM of the first $n$ natural numbers We know that $\log l_n  =\psi(n)$. Multiplying both sides by $l_n$ we obtain
$$
n l_n = \frac{k_1 l_n}{1} + \frac{k_2 l_n}{2} + \ldots + \frac{k_n l_n}{n}
$$
Each term on the RHS is a positive integer thus our question is equivalent to finding the number of partitions of $nl_n$ in which each part satisfy some criteria.
Criteria 1: How small can a part be? Assume that there is a solution with $k_n = 1$ then the smallest term in the above sum will be the $n$-th term which is $l_n / n$. Hence each term in our partition is $\ge l_n/n$.
Criteria 2: How many prime factors can each part contain? If my calculation is correct then for $n \ge 2, 2 \le r \le n$, the minimum number of prime factors that $l_n /r$ can contain is $\pi(n)-1$. With these two selection criterion we have:
$s(n) \le $ No. of partitions of $n l_n$ into at most $n$ parts such that each part is greater than $l_n / n$ and has at least $\pi(n) - 1$ different prime factors.
May be we can narrow down further by adding sharper selection criterions but I thought it was already complicated enough for the time being. The asymptotics of the number of partitions of $n$ into $k$ parts $p(n,k)$ is well known, but I have not found in literature any asymptotics for the number of partitions of $n$ into $k$ parts such that each part is at least $m$, let alone the case when each part has a certain minimum number of prime factors. I am looking for any suggestions, reference materials that would help in these intermediate questions that would ultimately help in answering the main question.
 A: I've got the following counts (which agrees with Brendan's):
1:  1
2:  3
3:  10
4:  55
5:  196
6:  2730
7:  10032
8:  108999
9:  973258
10: 20780331
11: 79309308
12: 2614200602
13: 10073335754
14: 288845706742
15: 11805287917646
16: 254331289285523
A: This may or may not be useful to you; I didn't get a complete answer from it.
If you multiply the original equation by $n!$ on both sides, you get $$n \cdot n! = k_1 n! + k_2 \frac{n!}{2} + \dots + k_n \frac{n!}{n} .$$
In the factorial-base expansion $n = a_1 1! + a_2 2! + a_3 3! + \dots$, this is then partitioning $00\dots0n$ into parts 
$00\dots0001 = (n-1)! = \frac{n!}{n}$ , 
$00\dots0011 = (n-1)!+(n-2)! = \frac{n!}{n-1}$ , 
$00\dots0221 = \frac{n!}{n-3}$ , $00\dots6631$ , ... , 
$00\dots000\frac{n}{2}$ , 
$00\dots00001 = n!$.
The leading digits obey the obvious distribution, starting with $0\dots x1$, then $0\dots x2$, with the $x$ increasing at increasing rates.  Now, partition problems don't necessarily behave well under small changes in the allowed parts, but if you can prove some sort of well-behavedness in the vicinity of these summands -- say, just taking the $0\dots x j$ parts -- perhaps poking at the factorial-base expansion will give you some sense of the asymptotics?
(Interestingly, the very largest parts converge to a constant form with trailing zeros, but only about log of them have frozen at any $n$.)
