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For example, $S = \{3,5,7,13\}$ gives $\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{12}=1$.

A few other such sets: $\{2\},\{3,5,7,19,37\},\{3,5,7,29,31,71\}$.

Are there infinitely many of these?


See also this post; it can be proven that there are infinitely many such sets where none of the elements are prime.

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    $\begingroup$ Cool idea, but like most prime number postulates, if it's not trivial, it is quite likely outside of our mathematical knowledge. $\endgroup$ Dec 10, 2019 at 4:40
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    $\begingroup$ Some experimentation (that supports the answer "yes"): codegolf.stackexchange.com/a/196966/56178 $\endgroup$ Dec 19, 2019 at 17:36
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    $\begingroup$ I realize this math.stackexchange.com/questions/4757620/… is a different question from what you are asking here. But it may be helpful and good for the forum if you were to link these two [good] questions with each other. $\endgroup$
    – Mike
    Aug 24, 2023 at 20:47

2 Answers 2

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I believe the answer to this is actually yes.

See Erdős–Graham problem.

This problem asked: if you partition all natural numbers larger than 1 into finitely many subsets, must the reciprocals of one of the subsets sum to 1? It was apparently proven in the affirmative in 2003.

By my reasoning, you could put all composite numbers in one subset, which already sums to more than 1 after taking the composites through 16, then divide up the remaining numbers (all primes) into any partitioning you like and be assured that at least one of those subsets will sum to 1. There are clearly infinitely many ways one could do that partitioning so as to obtain infinitely many different valid solution subsets.

Among other things, this would yield infinitely many distinct covering sets for Sierpinski numbers, which would therefore have infinitely many distinct possible coverage periods.


As pointed out below, I read this mistakenly so this isn't actually a solution. I'll leave this here as a potential jumping-off point for a solution, however; if one could assign the composites systematically into subsets in such a way that none of those subsets could contain a reciprocal sum to unity, then this would still work.

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    $\begingroup$ The property is rather "must the reciprocals of a subset of one of the subsets sum to $1$". $\endgroup$ Feb 25, 2022 at 13:35
  • $\begingroup$ Really? Crap... $\endgroup$
    – Trevor
    Feb 25, 2022 at 13:35
  • $\begingroup$ You have that wrong (obviously, if you think about it). Your link says that one of those subsets must itself have a finite subset, the sum of reciprocals of whose elements is equal to $1$ (a so-called Egyptian fraction) . And this fact can't be used to settle the OP's question. $\endgroup$
    – TonyK
    Feb 25, 2022 at 13:36
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Based on experimentation, it seems very likely the answer is yes.

In particular, by taking any existing valid set and selecting one of its elements $n$, then from $$\frac{1}{n}=\frac{1}{n+k}+\frac{k}{n(n+k)}\quad \text{where }k\mid n^2\text{ and }1\leq k\leq n,$$

it seems easy to recursively substitute until hitting on suitable values to generate a new set.

For example, I tried finding a set that didn't include $3$ and otherwise stayed about as compact as possible, and came up with $\{5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 61, 67, 71, 73, 109, 181, 241, 281, 421, 463, 541, 661, 2521\}$, a set of primes which indeed sums to $1$ (after subtracting $1$ and inverting each element).

On its own, this is obviously not a proof, but having played with this approach, it seems to strongly suggest the proposition is true. In fact, I suspect that given any finite lists of primes to be included or excluded, you could still find infinitely many valid lists subject to that constraint.

As per Greg Martin's comment above, see also https://codegolf.stackexchange.com/questions/196941/longest-prime-sums.

Barring a surprise proof, this seems likely to be as close to resolution as this question gets, so I'll consider it settled for now.

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