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I'll try to state this formally, forgive me if I botch the notation. Let

$$R=\{n : n+1 \in\mathbb{P}\}.$$ (Where $\mathbb{P}$ is the set of primes)

Then the question is,

$$\exists S \subset (\mathbb{N} \setminus R)\text{ for }s\text{ of finite length, such that }\sum_{s\in S}{\frac{1}{s}}=1 ?$$

It is not hard to find finite subsets of $\mathbb{N}$ whose reciprocals sum to $1$, such as the trivial $\{1\}$, or $\{2,3,6\}$. I suspect that there are infinitely many finite subsets of $R$ with this property, such as $\{2,4,6,12\}$ or $\{2,4,6,18,36\}$.

However, I couldn't find any valid $S$ with a quick search through small numbers, and I'm wondering if it's known whether such a set exists.

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    $\begingroup$ Context for this question please? $\endgroup$
    – Mike
    Aug 23, 2023 at 18:54
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    $\begingroup$ No context other than mathematical curiosity, after looking again at the post I linked to. It seems potentially "interesting" if indeed no such set exists. $\endgroup$
    – Trevor
    Aug 23, 2023 at 18:56
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    $\begingroup$ That is mathematical curiosity is certainly fair, it is an interesting question. However, anyone spending energy on trying to come up with a solution deserves IMO to know e.g., whether or not this is a homework exercise or a contest question with a known solution. $\endgroup$
    – Mike
    Aug 23, 2023 at 18:59
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    $\begingroup$ Ahh, okay. Then to be clear, no, this was just my own supposition. It is not a contest question to the best of my knowledge, and I left homework behind many years ago. $\endgroup$
    – Trevor
    Aug 23, 2023 at 19:02
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    $\begingroup$ fair enough... +1 $\endgroup$
    – Mike
    Aug 23, 2023 at 19:18

3 Answers 3

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We can use the formula $$ \frac1n = \frac1{n+1} + \frac1{n(n+1)} $$ to obtain such a sum. Start with $$ 1 = \frac12 + \frac14 + \frac18 + \frac18 $$ Now, for $n=8$ we can rewrite the last term as $$ \frac18=\frac19+\frac{1}{72} $$ Since $72+1$ is a prime, we have to iterate this, i.e., $$ \frac18=\frac19+\frac{1}{73}+\frac{1}{5256}. $$ Similarly $$ \frac12=\frac13+\frac17+\frac{1}{43}+\frac{1}{1806} $$ and $$ \frac14=\frac15+\frac{1}{20}. $$ Together we obtain $$ 1=\left(\frac13+\frac17+\frac{1}{43}+\frac{1}{1806}\right)+\left(\frac15+\frac{1}{20}\right)+\frac18+\left(\frac19+\frac{1}{73}+\frac{1}{5256}\right). $$ All denominators plus one are not prime numbers.

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Your denominators can include any odd number greater than $1$.

Find an odd semiperfect number $n$ (some subset of its proper factors add to $n$). $945$ is the smallest such number. Write $945$ as a sum of (some) of its proper factors. Then divide by $945$.

$945=315+189+135+105+63+45+35+27+15+9+7$.

So $1=\frac{315+189+135+105+63+45+35+27+15+9+7}{945}$.

That is $$1=\frac13+\frac15+\frac17+\frac19+\frac{1}{15}+\frac{1}{21}+\frac{1}{27}+\frac{1}{35}+\frac{1}{63}+\frac{1}{105}+\frac{1}{135}$$

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Here's the greedy solution, obtained by repeatedly choosing the largest possible fraction whose denominator isn't one less than a prime.

$$ \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{13} + \frac{1}{93} + \frac{1}{44155} + \frac{1}{8968410360}$$

And note that $8968410361$ is $241 * 347 * 107243$, from Wolfram Alpha.

You can find new terms by editing the following very crude lisp program, which computes the smallest possible denominator for the next term (without checking the primality condition).

(ceiling (/ 1 (- (1- (+ 1/3 1/5 1/7 1/8 1/9)))))
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