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.

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