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#1 |
Nov 2016
111012 Posts |
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Excluding a gap of one, it seems that the first prime before and after a factorial is always a prime distance away.
For instance: 5! = 120, the next prime after is 127, this is 7 away, which is also prime. This conjecture is unproven. So far it has been tested true up to 4000! for positive and 1000! for negative. This does not apply to 2^n - the first prime after 2^n is not always a prime distance away. Why the difference? Data: a(n) = p-n!, where p is the smallest prime > n!+1. https://oeis.org/A037153 |
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#2 |
May 2020
5·7 Posts |
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If it's not a prime distance away, it would have to be at least (n+1)^2.
Otherwise, n!+d will be divisible by the smallest prime factor of d, since the smallest prime factor of d will be smaller than n and, trivially, divide into n!. The merit of a prime gap this size would be at least (n+1)^2 / ln(n!), or something like n/(ln(n)-1), which gets larger as n gets bigger. The largest prime gap merit ever found is 38.07, so for n = 150 and bigger, finding an example of a non-prime distance away from a factorial being the first prime would lead to record-breaking prime gap merits. It's hyper-unlikely (and likely conjecturable) that an example of a non-prime gap away from a factorial being the first prime does not exist. |
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#3 |
Feb 2017
Nowhere
10001011000002 Posts |
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Let n > 2, and let Q be the largest prime less than n!. Let k be the least integer greater than 1 for which n! + k is prime.
Clearly k can be composed only of prime factors greater than n. If p is the least prime greater than n, the smallest composite k composed of primes greater than n is p^2. So for k to be composite, the gap between Q and the next prime is at least p^2 > n^2. Now log(n!) = n*log(n) apprixomately, so n = log(n!)/loglog(n!) approximately, so p^2 = log^2(Q)/loglog^2(Q), approximately. This would be (to say the least) unusually large for a gap between Q and the next prime, though not beyond the "infinitely often" size conjectured by Granville. |
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#4 |
Nov 2016
29 Posts |
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There are no primes in the range [n!+2, n!+n] ?
Last fiddled with by a nicol on 2021-03-02 at 17:57 |
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#5 | |
"Robert Gerbicz"
Oct 2005
Hungary
26568 Posts |
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As you noted to find composite G we need G>n^2 so the [n!+1,n!+n^2] interval contains only composites. A random number near x is composite with 1-1/log(x) probability, so using log(n!)~n*log(n) we get composite G with at most (1-1/(n*log(n)))^(n^2) probability, that is approx. exp(-1/log(n)^2) and the sum of these is convergent [elementary way: for fixed K sum of these in the exp(K)<n<exp(K+1) interval, giving that the sum of these is <sum(exp(K-K^2)) what is clearly a convergent serie.] Last fiddled with by R. Gerbicz on 2021-03-02 at 20:36 Reason: typo |
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#6 |
Nov 2016
29 Posts |
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Line graphs of A037153 in order and ascending numerical order:
https://oeis.org/A037153/a037153_1.png |
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#7 |
Romulan Interpreter
Jun 2011
Thailand
33·347 Posts |
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#8 |
"Seth"
Apr 2019
3×89 Posts |
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These properties are used by the prime gap search.
Also true of primorials, for P# if P# +-1 are not prime then there are no primes in [P#-P, P#+P] For P#/2 if P# +- 2^n are not prime then there are no primes in [P#-2P, P#+2P] Last fiddled with by Dr Sardonicus on 2021-03-24 at 14:50 Reason: Insert omitted word |
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