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Old 2018-01-28, 19:21   #12
CRGreathouse
 
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Quote:
Originally Posted by a1call View Post
Can you please provide links to the references that you mentioned.
My googling did not return any relevant hits.
Strange, Googleing the authors gave me references on the first page of the search results.

R. K. Guy, C. B. Lacampagne and J. L. Selfridge, Primes at a glance, Math. Comp. 48 (1987), 183-202.

Agoh, Erdos, and Granville, Primes at a (somewhat lengthy) glance, The American Mathematical Monthly Vol. 104, No. 10, Dec., 1997, pages 943 to 945
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Old 2018-01-28, 20:03   #13
a1call
 
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Originally Posted by science_man_88 View Post
Depends on n if odd neither will n!-(n+1) as it will be even. If n is 2 mod 3 then n+1 is 0 mod 3 and the result of subtracting it if n is more than 3 will be 0 mod 3.
Can you please provide an example for n, m where m is a prime number such that
n!-n <= m < n! -1

Thanks in advance.

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Old 2018-01-28, 20:08   #14
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Quote:
Originally Posted by a1call View Post
Can you please provide an example for n, m where m is a prime number such that
n!-n <= m < n! -1

Thanks in advance.
Not what was claimed. I claimed your lower bound, is not the lowest without being prime it could go in some cases.
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Old 2018-01-28, 20:11   #15
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What is the correct lower bound in your opinion?

Noting that the difference in your quoted post is only in the upper bounds.

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Old 2018-01-28, 20:16   #16
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Quote:
Originally Posted by a1call View Post
What is the correct lower bound in your opinion?
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.

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Old 2018-01-28, 20:31   #17
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Quote:
Originally Posted by science_man_88 View Post
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.
So this is sum of the series all over again.

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Old 2018-01-29, 00:50   #18
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Quote:
Originally Posted by a1call View Post
Actually it seems to me that the correct optimised range is neither

n!-n^2 < P < n!

Nor

n!-n^2 < P < n!-1

But rather

n!-n^2 < P < n!-n
Since none of the integers m such that
n!-n <= m < n! -1
Can be prime.
Right, you can take the upper bound as n!-n or n!-1.

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
\[\prod_{p\le n}\frac{p}{p-1} \approx e^{-\gamma}\log n\]
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
\[\int_{500.5}^{\infty}\exp\left(-\frac{e^{-\gamma}n}{2\log n}\right) \approx 4\cdot10^{-9}.\]
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