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## 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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Recognitions:      Homework  Help     Science  Advisor  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

Recognitions:      Homework  Help     Science  Advisor  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.

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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

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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.

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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 $$X\in\mathbb{N}$$ with

$$\lim_{n\rightarrow\infty}\frac1n\sum_{x\in X|x\le n}x=0$$

$$\forall n\in\mathbb{N}\;\;\exists m\in\mathbb{N}$$ such that there is no $$x\in X$$ with $$m\le x\le m+n$$. (The set of primes is of course such a set by the PNT.) I'm sorry if I was ambiguous.

Recognitions:      Homework  Help     Science  Advisor  You want:  $$\lim_{n\rightarrow\infty}\frac1n\sum_{x\in X|x\le n}1=0$$  or equivalently here pi(n)/n->0 as n->infinity. In otherwords, the asymptotic density of the primes is zero.

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.

Recognitions:      Homework  Help     Science  Advisor  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].

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Quote by CRGreathouse 					 				 				What I mean is that for a set $$X\in\mathbb{N}$$ with  $$\lim_{n\rightarrow\infty}\frac1n\sum_{x\in X|x\le n}x=0$$  $$\forall n\in\mathbb{N}\;\;\exists m\in\mathbb{N}$$ such that there is no $$x\in X$$ with $$m\le x\le m+n$$.  (The set of primes is of course such a set by the PNT.)  I'm sorry if I was ambiguous.
density, not measure.

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Quote by shmoe 					 				 				You want:  $$\lim_{n\rightarrow\infty}\frac1n\sum_{x\in X|x\le n}1=0$$  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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