20190821, 23:22  #1  
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{3}×3×79 Posts 
Bertand's Postulate
Quote:
That does not sound right to me. Shouldn't that be something like: Quote:


20190821, 23:30  #2 
"Rashid Naimi"
Oct 2015
Remote to Here/There
11101101000_{2} Posts 
Follow up question:
* Is there a (known) way to formulate N as a function of k? Thanks in advance. 
20190822, 08:24  #3  
"Robert Gerbicz"
Oct 2005
Hungary
11·127 Posts 
Quote:
You can say stronger statement (following and modifying the proof) : there is c0>0 for that there is at least c0*n/log(n) primes in [n,2n]. Using this there is c1>0 for that for N=c1*n*log(n) there is at least n primes in [N,2N]. 

20190824, 01:21  #4 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{3}×3×79 Posts 
Thank you very much for the confirmation and the formulation R. Gerbicz.

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