$$\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{43}+\cdots \le 1 $$

This is an interesting infinite summation. This is very closely resembling my other problem with has to do with the distribution of composites. That other problem is here. The same rules of that other summation apply, but it's good to repeat them :).

Rules of this summation:

1) All terms of the summation must be reciprocal of some prime. For example, $\frac{1}{6}$ could never be a term because 6 is not a prime number.

2) The next term must be the largest reciprocal term that does not allow the entire sum to be $>1$. The next term also must be < the term before it.

3) The sum must also be permanently less than $1$ if you were to stop at a number $\lt \infty$. If you were to stop on term $\infty$, then the sum would amount to exactly $1$ by definition.

Now that you know the rules, we must rewrite the summation in a different way in order to propose my question in a reasonable manner. $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\cdots\le 1$$ As a mote of clarification, $a, b, c, d$, etc. represent $2, 3, 7, 43$, etc. respectively.

The question:

-Envision $ f(1)=a, f(2)=b, f(3)=c$, and so on. Keep in mind there are an infinite number of terms, thus an infinite number of variables to represent primes. My question is what function, $f(n)$, could hold this property and maintain true no matter what term you are at.

-One who is loving this must remember that $a, b, c, d$, etc. are integers rather than fractions in this case and therefor $f(1) \not= 1/a, f(1) \not= 1/b$, etc. The reciprocal however is the form in which the numbers have that interesting property.

-What is $f(n)$? (The primary question)

-When answering this question, do the following, please provide some form of evidence or explain your case. If I missed something obvious, please let me know what that is with a website in the comments.

Have a great day! Thanks!

  • 3
    $\begingroup$ I cannot answer this question, but if you are interested, the next few primes are 1811, 654149, 27082315109, 153694141992520880899, and 337110658273917297268061074384231117039. I can keep going, but they are getting very large. $\endgroup$
    – Alex S
    Jul 6 '15 at 23:00
  • 8
    $\begingroup$ OEIS has the sequence listed as "Slowest-growing sequence of primes whose reciprocals sum to 1". $\endgroup$
    – Marcus M
    Jul 6 '15 at 23:01
  • $\begingroup$ @MarcusM What a curious title OEIS has for the sequence! Since the number of digits roughly doubles every term, I think there are much slower-growing sequence with this reciprocal sum (even oeis.org/A225669 sort of qualifies since it takes more terms to reach 10-20 digits). It's certainly the lexicographically first sequence, though. $\endgroup$
    – Erick Wong
    Jul 6 '15 at 23:42
  • $\begingroup$ That's pretty cool, I never even knew it was on OEIS. $\endgroup$
    – AAron
    Jul 7 '15 at 1:05
  • $\begingroup$ @AAron: It's always best to check. ;-$)$ $\endgroup$
    – Lucian
    Jul 7 '15 at 2:13

This function can be defined recursively as $$f(a) = NP\Big{(}\big{(}1-\sum_{k=1}^{a-1}\frac{1}{f(k)}\big{)}^{-1}\Big{)}$$ where $NP(x)$ is the next prime function. In other words, you look for the closest prime higher than the minimal number you need to keep the sum less than one in order to find $f(a)$. This sum can therefore not be expressed explicitly due to the "next prime" function used in the recursion.


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