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Prime Number Gaps

 
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Jul10-06, 10:40 PM   #1
 

Prime Number Gaps


Hello everyone,

I'd first like to say that I am uninformed on this subject and that I have a question to the mathematicians on these forums who know about the subject.

In the set of all prime numbers, has the integer gaps between two prime numbers been studied? I mean, do mathematicans know what the largest difference is between two prime numbers?

Im interested in this subject and I would like to know..

Thanks in advance.

--rad
 
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Jul10-06, 11:28 PM   #2
 
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The gap can be arbitrarily large. Just consider n!+2, n!+3,...n!+n.

Lots of work has been done, you might want to take a look at Caldwell's prime pages: http://primes.utm.edu/notes/gaps.html
 
Aug5-06, 05:41 PM   #3
 
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Since the primes have measure 0, prime gaps must be unbounded in length. Think about it -- if every k integers had a prime number for some fixed k, then some primes would have common (nontrivial) factors.
 
Aug5-06, 10:35 PM   #4
 
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Prime Number Gaps


i refer you to the work of helmut maier.

Helmut Maier
Primes in short intervals.

Source: Michigan Math. J. 32, iss. 2 (1985), 221
 
Aug7-06, 06:41 PM   #5
 
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Quote by CRGreathouse
Since the primes have measure 0
the integers have measure zero, but the gaps between consecutive integers is not unbounded.
 
Aug8-06, 12:31 AM   #6
 
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Quote by matt grime
the integers have measure zero, but the gaps between consecutive integers is not unbounded.
What I mean is that for a set [tex]X\in\mathbb{N}[/tex] with

[tex]\lim_{n\rightarrow\infty}\frac1n\sum_{x\in X|x\le n}x=0[/tex]

[tex]\forall n\in\mathbb{N}\;\;\exists m\in\mathbb{N}[/tex] such that there is no [tex]x\in X[/tex] with [tex]m\le x\le m+n[/tex]. (The set of primes is of course such a set by the PNT.) I'm sorry if I was ambiguous.
 
Aug8-06, 01:48 AM   #7
 
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You want:

[tex]\lim_{n\rightarrow\infty}\frac1n\sum_{x\in X|x\le n}1=0[/tex]

or equivalently here pi(n)/n->0 as n->infinity. In otherwords, the asymptotic density of the primes is zero.
 
Aug8-06, 02:20 PM   #8
 
According to Bertrand's postulate, there is at least one prime between n and 2n-2, for any n>3. I wonder if there is a theorem about the number of primes between n and 2n exclusive (see http://www.research.att.com/~njas/sequences/A060715 ), because that number seems to be steadily increasing over a sufficiently large period of the sequence (sorry if this is not precise enough); what I mean is that, for n=5, for instance, the number of primes between n and 2n is 1, but, it seems, for any n > 5 the number of primes between n and 2n is greater than 1; similarly, for n= 8, the number of primes between n and 2n is 2, but for any n>8, the number of primes between n and 2n is greater than 2 (?), and so on.

If it is true that for any m >= 1 there is an n for which the number of primes between k and 2k, k>=n, is greater than m (is it?) then there is a limit on prime gaps as well, depending on m (or n), I think (although Bertrand's postulate itself puts a limit on prime gaps, depending on n).

What I mean by Bertrand's postulate putting a limit on prime gaps is that for any prime p, there is another prime between p+1 and 2p.
 
Aug8-06, 03:04 PM   #9
 
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pi(2n)-pi(n)~n/log(n) by the prime number theorem, so the number of primes in [n,2n] tends to infinity as n does.

The bound Bertrands puts on the gap to the next prime is pretty far from what's known to be true (though correspondingly simpler to prove!). For example, if n is large enough, we can guarantee a prime in [n,n+n^0.525].
 
Aug8-06, 07:52 PM   #10
 
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Quote by CRGreathouse
What I mean is that for a set [tex]X\in\mathbb{N}[/tex] with

[tex]\lim_{n\rightarrow\infty}\frac1n\sum_{x\in X|x\le n}x=0[/tex]

[tex]\forall n\in\mathbb{N}\;\;\exists m\in\mathbb{N}[/tex] such that there is no [tex]x\in X[/tex] with [tex]m\le x\le m+n[/tex]. (The set of primes is of course such a set by the PNT.) I'm sorry if I was ambiguous.
density, not measure.
 
Aug9-06, 10:44 AM   #11
 
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Quote by shmoe
You want:

[tex]\lim_{n\rightarrow\infty}\frac1n\sum_{x\in X|x\le n}1=0[/tex]

or equivalently here pi(n)/n->0 as n->infinity. In otherwords, the asymptotic density of the primes is zero.
Oops, you're absolutely right. That's what I meant.
 
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