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:


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?

  • $\begingroup$ We assume distinct denominators, correct? $\endgroup$ – coffeemath Apr 13 at 2:54
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    $\begingroup$ Of course, reciprocals must be different @coffeemath $\endgroup$ – Rami Rami Apr 13 at 8:52
  • $\begingroup$ 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?. $\endgroup$ – James Arathoon Apr 13 at 8:56
  • $\begingroup$ 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 $\endgroup$ – Rami Rami Apr 13 at 9:06
  • $\begingroup$ 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. $\endgroup$ – James Arathoon Apr 13 at 10:55

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