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-   -   P(n+1)<(sqrt(P(n))+1)^2 (https://www.mersenneforum.org/showthread.php?t=4895)

Crook 2005-10-26 13:32

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.

R.D. Silverman 2005-10-26 13:49

[QUOTE=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.[/QUOTE]

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.

maxal 2005-10-26 21:11

[QUOTE=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?[/QUOTE]
There is a famous [url=http://mathworld.wolfram.com/LandausProblems.html]Legendre's conjecture AKA the 3rd Landau's problem[/url] 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.

Citrix 2005-10-26 21:29

[url]http://www.primepuzzles.net/conjectures/conj_008.htm[/url]

Is my favorite conjecture related to distribution of primes.


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