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Borek

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## Main Question or Discussion Point

**reciprocals of primes summed to 1**

As the other thread about sum of primes started it reminded me about the idea I had long ago.

My starting point was Erathostenes sieve. It occured to me that multiples of 2 make half of all natural numbers, multiples of 3 make 1/3 of all natural numbers and so on. And as multiplies of prime numbers have to cover all natural numbers, correctly constructed sum of their reciprocals must equal 1. This is not as obvious as it seemed to me at first, as some multiplies of 2 and 3 will be calculated twice, so it has to be 1/2+1/3-1/(2*3) - but it still can be done.

Let P be set of all prime numbers. Let's define some more sets:

[tex]\forall {a, b} \in P, a < b \rightarrow ab \in P_2[/tex]

[tex]\forall {a, b, c} \in P, a < b < c \rightarrow abc \in P_3[/tex]

and so on P

_{4}, P

_{5}...

each of these sets has elements

[tex]p_{1i} \in P, p_{2i} \in P_2, p_{3i} \in P_3 ...[/tex]

when combined

[tex]\sum {\frac 1 p_{1i} } - \sum {\frac 1 p_{2i} } + \sum {\frac 1 p_{3i} } - \sum {\frac 1 p_{4i} } ... = 1[/tex]

(note: could be I am misusing notation, what I mean is "if a < b < c then abc is a member of set P

_{3}- that's a way of using each combination of three primes only once, could be it can be done much simpler; please remember I am a chemist )

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