This question is a simple generalization of a math problem I saw somewhere.
Consider the family of sets $$ A_n \equiv \{(a_1, \ldots, a_n)\in \mathbb{Z}_{\geq 1}^n \, : \, a_1 \leq \cdots \leq a_n \text{ and } \sum_{k = 1}^n \frac{1}{a_k} = 1\} $$
I think there are many interesting questions that could be asked about these $A_n$. Here are the ones I'm curious about:
- Is there a paper or standard reference on these $A_n$ I should know about?
- How fast does $|A_n |$ grow?
- What's the best way to calculate the elements of $A_n$?
If Math SE etiquette requires pinning me down to one question, I'm most interested in reading about this problem elsewhere, assuming it has already been written about (which I'm sure is a safe assumption).
So far, I've come up with an algorithm for calculating the $A_n$. Suppose $a_1, \ldots, a_k$ are given with $k < n$. Then $$ \sum_{j = k + 1}^n \frac{1}{a_j} = 1 - \sum_{j = 1}^k \frac{1}{a_j} > 0 $$ (If the RHS is $\leq 0$, there is no choice of $a_{k + 1}$ that works, and we are done).
But $$ \sum_{j = k + 1}^n \frac{1}{a_j} \leq \frac{n- k}{a_{k + 1}} $$ since $1 \leq a_1 \leq \cdots \leq a_n$. Thus $$ \frac{n-k}{a_{k + 1}} \geq 1 - \sum_{j = 1}^k \frac{1}{a_j} $$ Therefore $$ a_{k + 1} \leq \frac{n - k}{1 - \sum_{j = 1}^k \frac{1}{a_j}} $$ That is $$ a_{k+1} \in \left\{n \in \mathbb{Z} : a_k \leq n \leq \left\lfloor \frac{n - k}{1 - \sum_{j = 1}^k \frac{1}{a_j}}\right\rfloor\right\} \equiv B_{k + 1} $$ where $B_{k + 1}$ is clearly finite (and depends on $a_1, \ldots, a_k$). We may then pick a value for $a_{k +1}$ from $B_{k + 1}$ and similarly constrain the values for $a_{k + 2}$. We continue in this way until we get $B_j = \emptyset$ for some $j$ or until we've checked every element of $B_n$. We then select a different value for $a_{k + 1}$ from $B_{k + 1}$ and repeat. Once we've exhausted all the possibilities in $B_{k + 1}$, we make a different choice for $a_k$ from $B_k$ and so on until we've checked every possibility. Note that we always have $a_1 \in \{1, \ldots, n\}$, so there are definitely only finitely many possibilities to check.
Computer science is not my specialty. I'm sure there's a more elegant and rigorous way to describe this algorithm, but hopefully you get the idea. In particular, I hope it's clear that $A_n$ is finite and can be calculated by performing a finite (though possibly very large per this algorithm) number of computations.
If you're curious, the first few $A_n$ are: \begin{align} A_1 &= \{(1)\} \\A_2 &= \{(2,2)\} \\A_3 &= \{(2, 3, 6), (2, 4, 4), (3,3,3)\} \end{align}
