Class of $p_i$ such that $\sum_{i=1}^n \frac{1}{p_i} = l$ for some $l$. Let $l$ be any integer and $p_i$ belong to natural numbers for $1 \leq i \leq n$. Suppose they satisfy the following equation $$\sum_{i=1}^n \frac{1}{p_i} = l$$ Suppose we are given $l$ and $n$ is fixed. What can we conculde about $p_i$? More specifically, can we say any class of $p_i$ which statisfies the above equation. Or Is there any we can say explicitly which $p_i$ satifies the equation.
Given $n,l$ enumerate all the $p_1\le p_2\le \ldots \le p_n$ such that $$l=\sum_{i=1}^n \frac1{p_i}$$


Due to the $\le$ restriction given $p_1\ldots p_{j-1}$ we must have $p_j\le \frac{n-j}{l-\sum_{i=1}^{j-1} \frac1{p_i}}$. This gives the naive algorithm : 


Starting with $j=1$
If $j\le n$ let $r=\sum_{i=1}^{j-1} \frac1{p_i}$


If $r=l$ then return
Otherwise for each $p_j\le \frac{n-j}{l-r}$, enumerate recursively the solutions $p_{j+1}\le \ldots \le p_n$

If $j=n+1$ 


if $\sum_{i=1}^n \frac1{p_i} \ne l$ then return
otherwise enumerate all the permutations of $p_1,\ldots,p_n$
Given $n,l$ enumerate all the $p_1\le p_2\le \ldots \le p_n$ such that $$l=\sum_{i=1}^n \frac1{p_i}$$


Due to the $\le$ restriction given $p_1\ldots p_{j-1}$ we must have $p_j\le \frac{n-j}{l-\sum_{i=1}^{j-1} \frac1{p_i}}$. This gives the naive algorithm : 


Starting with $j=1$
If $j\le n$ let $r=\sum_{i=1}^{j-1} \frac1{p_i}$


If $r=l$ then return
Otherwise for each $p_j\le \frac{n-j}{l-r}$, enumerate recursively the solutions $p_{j+1}\le \ldots \le p_n$

If $j=n+1$ 


if $\sum_{i=1}^n \frac1{p_i} \ne l$ then return
otherwise enumerate all the permutations of $p_1,\ldots,p_n$