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2 votes
2 answers
424 views

"Squeezing" the primes?

The logical idea here is to map a curve that encodes the primes into the region $(0,1)^2$ and analyze the distribution there more easily and achieve tight bounds. To assess the distribution of primes, ...
4 votes
1 answer
395 views

Mertens formulas aren't enough for prime number theorem

For the primes it's true that $$ \sum_{p \le x}\frac{1}{p} = \ln\ln x + M + O(1/\ln x) $$ where, $M$ is suitable constant, and, moreover, the prime number theorem gives that $$ \lim_{x\to\infty}\frac{\...
5 votes
0 answers
137 views

Is finding positive integer solutions of $\zeta(a/b) = c$ equivalent to deciding the rationality of $\gamma$?

This question requires little bit of explanation of the background hence it is a bit lengthy. Note: The question was initially posted in MSE but did not get answers hence posting in MO. For every ...
13 votes
2 answers
2k views

Asymptotics of the n-th prime using the gamma function

In the paper http://rgmia.org/papers/v8n2/eepnt.pdf, the author proves that proves an explicit inequality on prime numbers using the gamma function and as a corollary, he showed that. $$ p_n = n \...
46 votes
4 answers
8k views

Why could Mertens not prove the prime number theorem?

We know that $$ \sum_{n \le x}\frac{1}{n\ln n} = \ln\ln x + c_1 + O(1/x) $$ where $c_1$ is a constant. Again Mertens' theorem says that the primes $p$ satisfy $$ \sum_{p \le x}\frac{1}{p} = \ln\ln ...