Prime Number Gaps - What's the Largest Integer Difference?

[rad0786]

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

 

[shmoe]

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

 

[CRGreathouse]

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.

 

[mathwonk]

i refer you to the work of helmut maier.

Helmut Maier
Primes in short intervals.

Source: Michigan Math. J. 32, iss. 2 (1985), 221

 

[matt grime]

Since the primes have measure 0

the integers have measure zero, but the gaps between consecutive integers is not unbounded.

 

[CRGreathouse]

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.

 

[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.

 

[erszega]

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.

 

[shmoe]

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].

 

[matt grime]

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.

 

[CRGreathouse]

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.