20051026, 13:32  #1 
Nov 2004
2^{4} Posts 
P(n+1)<(sqrt(P(n))+1)^2
Let P(n) denote the nth prime number. Then, does anybody have an idea why P(n+1)<(sqrt(P(n))+1)^2 is true? This would be a lower bound than the Tchebycheff result that there is always a prime between n and 2n. Regards.

20051026, 13:49  #2  
Nov 2003
2^{2}·5·373 Posts 
Quote:
true. The best that has been achieved, when last I looked was that there is always a prime between x and x + x^(11/20 + epsilon), for epsilon depending on x as x >oo. The fraction 11/20 may have been improved. Note that even R.H does not yield the result you want. R.H. would imply there is always a prime between x and x + sqrt(x)log x for sufficiently large x. You want one between x and 2sqrt(x)+1. 

20051026, 21:11  #3  
Feb 2005
2^{2}·3^{2}·7 Posts 
Quote:


20051026, 21:29  #4 
Jun 2003
1,579 Posts 
http://www.primepuzzles.net/conjectures/conj_008.htm
Is my favorite conjecture related to distribution of primes. 
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