# $\sum_{p,q \text{ primes } p \le q} 1/(pq\log(pq))$

The sum $$\sum\limits_{p,q \text{ primes } p \le q} \frac{1}{pq\log(pq)}$$

is related to a conjecture of Erdős about primitive sequences.

It converges because the sequence is primitive. If my calculations are correct a lower bound is $1.062$.

Is it known to what is equal the sum?

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To the best of my knowledge, no, and I doubt very much it is a decent number. In general, if you cannot immediately express the sum over primes in terms of the $\zeta$-function using Euler's formula, it is hopeless because the explicit summation of whatever is not encoded in Euler requires knowledge of individual primes and that is well beyond the current state of art. Actually, even if you can express things in terms of $\zeta$, it is rather questionable whether it can be called "an explicit formula" because we don't know much about the values of $\zeta$ except at the even integers. – fedja Jun 4 '12 at 14:48
Thank you fedja, will accept this as answer. What do you mean by "decent number"? I suppose unconditional upper bound follows from the primitive sequence. – joro Jun 4 '12 at 15:07
Perhaps of interest: You can rewrite your sum as $$\frac{1}{2}\left(\int_{1}^{\infty}P(s)^{2}ds+\int_{1}^{\infty}P(2s)ds\right)$$ where $P(s)=\sum_{p} p^{-s}$ is the prime zeta function. – Eric Naslund Jun 4 '12 at 20:15
Thank you Eric. Can you give a lower bound? – joro Jun 5 '12 at 6:17