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Feb 17, 2017 at 20:03 vote accept Stanley Yao Xiao
Feb 16, 2017 at 18:06 answer added Jan-Christoph Schlage-Puchta timeline score: 5
Feb 16, 2017 at 17:34 comment added Greg Martin That's not accurate. The weight is not 1 over the log of the interval length, but rather 1 over the log of the size of the integers in the interval. And that difference is precisely related to my initial comments (long intervals with no primes at all), and equally related to the fact that no such asymptotic formula can exist.
Feb 16, 2017 at 16:50 comment added Stanley Yao Xiao However one has the prime number theorem, which basically says that the number of primes in an interval is roughly equal to the 'weighted length' of the interval, where the weight is $1/\log x$.
Feb 16, 2017 at 9:16 comment added Greg Martin No, not even in one dimension. Fixing Vol$(R)$, there are intervals of that length with no primes whatsoever, and intervals of that length with lots of primes.
Feb 15, 2017 at 19:07 history asked Stanley Yao Xiao CC BY-SA 3.0