[darkmighty]

> Is this a correct statement?

Of course it is. By 'randomly' he refers to a maximum entropy ("uniform") distribution conforming Gauss' Prime Number Theorem. Since by that theorem the gap rises, the probability of twin primes would go to 0 and, crucially, the expected number of twin primes would be finite (or any finite gap really).

That's an interesting reason for Yitang's proof being so significant without actually reaching the small gap -- it shows primes violate eloquently the random distribution in some ways.

[Retric]

If the probably of a number being prime is ~1 / ln (N)

The probably of x and x + 2 being prime is ~ (1 / ln (N)) ^2. For large N, (ln (N)) ^2 < N so there are infinitely many twin primes. Because the Sum of 1/(f(x)) as x -> inifinity = infinity if f(x) < (x).

So, the real issue is if the number of twin primes significantly larger than (1/ln(x)) ^ 2.

[darkmighty]

Ah yes you're right I just naively assumed the gap going to infinity implied the number of small gaps has to be finite, thanks. I believe your argument argument sightly miscounts the twin primes, since you're giving a new "chance" after you had already drawn a prime/not-prime. A lower bound assuming the probabilistic model would be E[Twin Primes] > Sum ( 1/( ln(4k) ln(4k+2) ), which is still divergent so your point stands.

> so there are infinitely many twin primes

http://en.wikipedia.org/wiki/Twin_prime

Note the twin prime conjecture is not yet proven (your argument does not follow directly from PNT, but from the approximate probabilistic model that (almost surely?) displays PNT-like count).

> Because the Sum of 1/(f(x)) as x -> inifinity = infinity if f(x) < (x).

Here's a cute counterexample: x/(2*sign(sin(x))) < x, but the sum < infinity (you're missing an absolute value |f(x)| in your statement).

[Retric]

Yea, I was playing fast a loose. Though, for positive infinity you need 0 < f(x) < (x) not just |f(x)| < x. EX: f(x) = -0.5 * x

[ilyagr]

To expound on Retric's comment (I expounded much more in a cousin thread): I also first assumed that what you said must be true. When I tried to check with a calculation, it turned out that the opposite is the case.

[darkmighty]

Yea, my bad. I implicitly made the rookie mistake of assuming that f(x)->0 implies sum(f(x))->0, as shown by Retric :)