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Supposing n a positive integer (1,2,3,...), we call Eg(n) the smallest number of different reciprocals (1,$\frac{1}{2}$,$\frac{1}{3}$,...) that sum up to n. We call also $\delta(n)$ the number of ways we can write n with Eg(n) reciprocals. We can easily see that Eg(1)=1 (1=1), and that $\delta(1)$=1. Also, we can easily prove that Eg(2)=4 (2=1+$\frac{1}{2}$+$\frac{1}{3}$+$\frac{1}{6}$), and that $\delta(2)=1$. For n=3, I proved, also easily, that Eg(3)=13:

3=1+$\frac{1}{2}$+$\frac{1}{3}$+$\frac{1}{4}$+$\frac{1}{5}$+$\frac{1}{6}$+$\frac{1}{8}$+$\frac{1}{9}$+$\frac{1}{10}$+$\frac{1}{12}$+$\frac{1}{15}$+$\frac{1}{16}$+$\frac{1}{720}$

So, the question is what is $\delta(3)$? And what are Eg(n) and $\delta(n)$ for bigger integers (e.g. 4,5,...)? Is there any research about this topic?

Rami Rami
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  • We assume distinct denominators, correct? – coffeemath Apr 13 '19 at 02:54
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    Of course, reciprocals must be different @coffeemath – Rami Rami Apr 13 '19 at 08:52
  • for the sum 3, the twelve denominators before 720, are the first 12 divisors of 720, likewise for the sum 2, the three denominators before 6 are the divisors of 6. is this a pattern that continues?. – James Arathoon Apr 13 '19 at 08:56
  • No, 3=1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9+1/14+1/24+1/28+1/45 @JamesArathoon – Rami Rami Apr 13 '19 at 09:06
  • Some thoughts: sum of 1 over each divisor of 5! equals 3, but that is 15 terms which cannot be further reduced. Your terms for 3 are selected from 1 over each divisor of 6! and then in your comment 1 over each divisor of 7!. Both of these by default sum to greater than 3 allowing you to find (somehow?) the maximum number of terms that can be removed from each respective sum. The number of divisors of 8! is 96 and the divisor inverses add up to less than 4. I wonder if you can find a solution to 4 with 96 or less terms. – James Arathoon Apr 13 '19 at 10:55
  • By calculation, we see that: H(30)<4<H(31); where H(n) is the nth harmonic number. This means that Eg(4)>or=31. @Yuriy S noticed that (with some modification): 4=1+1/2+......+1/30+1/200+1/77706+1/16532869712+1/3230579689970657935732+1/36802906522516375115639735990520502954652700 (35 reciprocals); which ensures that Eg(4) is one of {31,32,33,34,35}<96. – Rami Rami Apr 13 '19 at 13:09
  • @JamesArathoon This is the proof of Eg(4) being less than 96 – Rami Rami Apr 15 '19 at 12:51
  • Yes significantly so. As you say for $n>1$, the easily provable low bound is $\text{Eg}(n)\ge(\text{Floor}[\exp(n-\gamma)])$. However can an improved low bound and a high bound be found. Based on the very little data you have here, an initial guess for the bounds for $\text{Eg}(n)$ might be something like $ (\text{Floor}[\exp(n-\gamma+\frac{1}{n^2})-2])<\text{Eg}(n)< (\text{Floor}[\exp(n-\gamma+\frac{1}{n})-1])$ where $\gamma$ is the Euler-Mascheroni Constant. – James Arathoon Apr 15 '19 at 14:33
  • Based on this guess: the bounds for $\text{Eg}(2)$ are 3 to 5; those for $\text{Eg}(3)$ are 10 to 14; those for $\text{Eg}(4)$ are 30 to 38 and those for $\text{Eg}(5)$ are 84 to 100. – James Arathoon Apr 15 '19 at 14:33
  • Yes exactly. But my question is: is finding Eg(n) (exactly) an open problem? or was it discussed significantly by any mathematician? @JamesArathoon – Rami Rami Apr 15 '19 at 15:20
  • This paper by Ernest Croot was referenced in Guy's "Unsolved Problems in Number Theory" https://doi.org/10.1112/S0025579300007828 Your precise question is not explicitly listed by Guy as an open problem, so I expect you have now made it so. – James Arathoon Apr 15 '19 at 15:54
  • $$4=\sum _{k=1}^{30} (\frac{1}{k})+\frac{1}{200}+\frac{1}{77706}+\frac{1}{16532874306}+\frac{1}{59497349338165200}$$ (34 Reciprocals) – James Arathoon Apr 15 '19 at 19:08
  • Thank you...now,we only have to prove 4 cannot be written with 31,32 nor 33 reciprocals then we get 1,4,13,34(?),.... @JamesArathoon – Rami Rami Apr 15 '19 at 21:14
  • See http://oeis.org/A306349 for the "greedy" algorithm – GEdgar May 27 '19 at 10:58
  • @GEdgar thank you a lot...so the sequence is known up to n=5...is there a sequence for delta(n)? – Rami Rami May 31 '19 at 21:07

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