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 2005-10-26, 13:32 #1 Crook     Nov 2004 1016 Posts P(n+1)<(sqrt(P(n))+1)^2 Let P(n) denote the n-th 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.
2005-10-26, 13:49   #2
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

22×5×373 Posts

Quote:
 Originally Posted by Crook Let P(n) denote the n-th 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.
The short answer is no. Noone does. We have no proof that it is
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.

2005-10-26, 21:11   #3
maxal

Feb 2005

11·23 Posts

Quote:
 Originally Posted by Crook Let P(n) denote the n-th prime number. Then, does anybody have an idea why P(n+1)<(sqrt(P(n))+1)^2 is true?
There is a famous Legendre's conjecture AKA the 3rd Landau's problem that there is always a prime between n^2 and (n+1)^2 but I'm not sure if n is required to be integer. If n may be a positive real number then this conjecture directly implies your inequality. Otherwise, it implies a weaker inequality P(n+1)<(ceil(sqrt(P(n))+1)^2.

 2005-10-26, 21:29 #4 Citrix     Jun 2003 22×397 Posts http://www.primepuzzles.net/conjectures/conj_008.htm Is my favorite conjecture related to distribution of primes.

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