If $\sum_n \frac{1}{a_n} = 2$ where $a_n$ are positive integers, is there a subset such that $\sum_{n\in S} \frac{1}{a_n} = 1$?

Question

As the title says: I'm wondering, out of curiosity, whether any (weak) Egyptian fraction decomposition of 2 always splits into two Egyptian fraction decompositions of 1. By a "weak" Egyptian fraction decomposition, I mean one where the denominators are not necessarily required to be distinct. (The problem with normal Egyptian fraction decompositions, where $a_n$ are required to be distinct, would also be interesting.)

In trying to construct a counterexample -- for example by taking enough prime numbers so that $\sum_{k=1}^n \frac{1}{p_k} > 2$, removing the last one, and inserting enough remainder terms to make up the difference -- I run into a tension between having the denominators grow slowly enough for the sum to diverge, and on the other hand having the sum be close enough to 2 that the remainders can't bring the sums otherwise close to 1. I haven't yet found a way for such an approach to work.

From the other end, I've written a simple Python script to try to search for decompositions of 2, and see for each whether it's splittable into 2 decompositions of 1. At the current stage, I've been able to run it for lengths up to 7 and haven't found any counterexamples. But when I try to allow length of 8, the run time (as the script stands) appears to be prohibitive; and if I trace through the decompositions of length 7 or less that it's testing, it appears that length 7 is not yet enough to get any really interesting decompositions.

Answer 1

After making some optimizations to the Python script, I was eventually able to find a counterexample of length 8: $$\frac{1}{2} + \frac{1}{3} + \frac{1}{3} + \frac{1}{5} + \frac{1}{5} + \frac{1}{5} + \frac{1}{5} + \frac{1}{30} = 2.$$ To see this is a counterexample, suppose we had a splitting, and consider the part of the splitting that includes 30. Then the sum of the reciprocals of the other elements must be $\frac{29}{30}$. Therefore, that part of the splitting must also include at least one each of the 2, 3, 5 terms. However, then already, $\frac{1}{2} + \frac{1}{3} + \frac{1}{5} > 1$, giving a contradiction.

If there are no bugs in the Python script, then it appears this is the unique counterexample of length 8 or less, up to permutation.

Answer 2

Here's a counterexample with $19$ terms: $2=\sum_{n\in S}\frac{1}{n}$ where

$$S=\{2, 3, 4, 7, 8, 9, 11, 13, 14, 16, 17, 20, 21, 22, 30, 1547, 17017, 612612, 2450448\}$$

You can manually check that this sum is indeed $2$ but no subset of $S$ leads to a sum of $1$.

---

ADDENDUM: Since the answer is negative, this begets an interesting generalization: Let $r$ be a given positive rational and let $T$ be a finite set of rational numbers in $(0,r)$. Does there exist a finite set $S\subset \mathbb{N}$ such that

$$\sum_{n\in S}\frac{1}{n}=r$$

but

$$\sum_{n\in R}\frac{1}{n}\not\in T$$

for all $R\in P(S)$ (the powerset of $S$)? For example, in the original question $r=2$ and $T=\{1\}$. I would conjecture such an $S$ must exist for all $r$ and $T\subset (0,r)$ but I have no idea how one would even begin such a proof.