20180128, 19:21  #12  
Aug 2006
3×1,993 Posts 
Quote:
R. K. Guy, C. B. Lacampagne and J. L. Selfridge, Primes at a glance, Math. Comp. 48 (1987), 183202. Agoh, Erdos, and Granville, Primes at a (somewhat lengthy) glance, The American Mathematical Monthly Vol. 104, No. 10, Dec., 1997, pages 943 to 945 

20180128, 20:03  #13  
"Rashid Naimi"
Oct 2015
Remote to Here/There
2×1,087 Posts 
Quote:
n!n <= m < n! 1 Thanks in advance. Last fiddled with by a1call on 20180128 at 20:04 

20180128, 20:08  #14 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 

20180128, 20:11  #15 
"Rashid Naimi"
Oct 2015
Remote to Here/There
87E_{16} Posts 
What is the correct lower bound in your opinion?
Noting that the difference in your quoted post is only in the upper bounds. Last fiddled with by a1call on 20180128 at 20:14 
20180128, 20:16  #16 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Like I said it depends on n, you can go as low as n!nextprime(n) at least. That follows from the fact that all numbers under nextprime(n) have a factor under n.
Last fiddled with by science_man_88 on 20180128 at 20:21 
20180128, 20:31  #17 
"Rashid Naimi"
Oct 2015
Remote to Here/There
4176_{8} Posts 
So this is sum of the series all over again.
Last fiddled with by a1call on 20180128 at 20:31 
20180129, 00:50  #18  
Aug 2006
13533_{8} Posts 
Quote:
Either way you have ~ n^2 numbers of size roughly n! which are divisible by none of the primes up to n. Heuristically this makes them \[\prod_{p\le n}\frac{p}{p1} \approx e^{\gamma}\log n\] times more likely to be prime than the average prime of its size, for an overall probability of \[\frac{e^{\gamma}\log n}{\log n!} \approx \frac{e^{\gamma}\log n}{n\log n} = \frac{e^{\gamma}}{n}\] and an expected \[\frac{n^2}{\log(n^2)}\cdot\frac{e^{\gamma}}{n} = \frac{e^{\gamma}n}{2\log n}\] primes in the interval. The chance of having none is then \[\exp\left(\frac{e^{\gamma}n}{2\log n}\right)\] and since \[\int\exp\left(\frac{e^{\gamma}n}{2\log n}\right)\] converges there should be only finitely many you'd expect only finitely many intervals without primes. A quick check shows that the first 500 have primes, making the odds of any being empty around \[\int_{500.5}^{\infty}\exp\left(\frac{e^{\gamma}n}{2\log n}\right) \approx 4\cdot10^{9}.\] 

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Goldbach Conjecture  MattcAnderson  MattcAnderson  4  20210404 19:21 
Goldbach's Conjecture  Patrick123  Miscellaneous Math  242  20110315 14:28 
Proof of Goldbach Conjecture  vector  Miscellaneous Math  5  20071201 14:43 
Goldbach's conjecture  Citrix  Puzzles  3  20050909 13:58 