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.

When you write $\sum_{n}$, you mean $\sum_{n=0}^{+\infty}$ or $\sum_{k=0}^n$ ? — Noomkwah, Apr 21 '22 at 18:43
I meant a finite decomposition, $\sum_{n=1}^N \frac{1}{a_n}$. — Daniel Schepler, Apr 21 '22 at 18:44
This is a duplicate of this question where solutions with all denominators distinct were found The largest denominator could be as small as $30030$ — Ross Millikan, Apr 21 '22 at 20:17
@RossMillikan OK, I agree this is a duplicate. Should I jjust delete it, or should we vote to close? — Daniel Schepler, Apr 21 '22 at 21:02
@DanielSchepler: I think just close it as a duplicate. I started it. — Ross Millikan, Apr 21 '22 at 22:18

score 5 · Answer 1 · answered Apr 21 '22 at 18:55

5

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.

answered Apr 21 '22 at 18:55

Daniel Schepler

22,856

4

What if the $a_n$ are required to be distinct, as is usually required of Egyptian fractions? See https://en.wikipedia.org/wiki/Egyptian_fraction and https://mathworld.wolfram.com/EgyptianFraction.html#:~:text=An%20Egyptian%20fraction%20is%20a,between%205%20and%20101. That is what I assumed was meant in the original question, which made it more interesting. – Yuval Peres Apr 21 '22 at 19:03
Hmm, if the $a_n$ were required to be distinct, then that would probably push it into the realm again where my Python script would be unworkable. – Daniel Schepler Apr 21 '22 at 19:15
If uniqueness was required, one might find: https://www.jstor.org/stable/3597210?refreqid=excelsior%3A345b3a13637016edd76c6ff4779c8343&seq=1, to be relevant. – person Apr 21 '22 at 19:22
One potential way of doing it would be a greedy algorithm — start with some set of (probably relatively prime) numbers ${p, q, r}$, say, start with the sum $S= 1/p+1/q+1/r$, and at each stage choose the smallest $t$ such that $S+1/t\leq 2$ but none of the partial sums are equal to 1. This will get a little unmanageable as you get too many terms — since there are $2^n$ partial sums to be checked — but for $n$ in the range of 20-30 that's still very feasible, and several different ${p,q,r}$ sets could be tested to try and find examples. – Steven Stadnicki Apr 21 '22 at 19:22

QC_QAOA · Answer 2 · 2022-04-21T20:34:46.327

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.

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

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