A writeup by his doctoral advisor about his memories of Yitang Zhang: http://www.math.purdue.edu/~ttm/ZhangYt.pdf
"The most natural conjecture is that the prime numbers appear randomly..
If the twin prime conjecture is correct, then we may conclude that the
[primes] are not constructed randomly"
My intuition is the opposite: a random distribution of primes implies the twin prime conjecture is true. Because suppose there are finite twin primes, set them aside; then test pairs of numbers above the largest of them. Since we assume the twin prime conjecture is false, the chances that random pairs above that number are both prime is exacty 0; but that would mean the primes are not distributed randomly...
The prime number theorem [1] states that the probability that a natural number N is prime is approximately 1/log(N). So, the probability that N & N+2 are both prime is about 1/log(N)^2. If they were distributed randomly, there would be about
integral(1/log(x)^2 dx) from x=2 to x=N
twin primes between 1 and N. This integral converges to infinity as N grows large[2] (wish I remembered enough caclulus to do it by hand). So, there is some N for which you'd expect to have 10^100 twin primes less than N, etc.
So it seems that if you decided by a coin toss whether a number should be prime (where you'd make it prime with probability 1/log(N)), there would be infinitely many twin primes.
Wikipedia seems to confirm it's a reasonable estimate [3].
UPDATE: By the way, the same argument implies that if primeness was decided by a coin toss, there would be infinitely many N such that N and N+1 are prime. There is, however, only one such N (namely 2). So, primeness is not decided by a coin toss, and the above is not a proof of the twin primes conjecture :).
[1]: http://en.wikipedia.org/wiki/Prime_number_theorem
[2]: http://www.wolframalpha.com/input/?i=Integrate[1%2FLog[x]^2%...]
[3]: http://en.wikipedia.org/wiki/Twin_prime#First_Hardy.E2.80.93...
li(x) - x / log(x)
See also: https://en.wikipedia.org/wiki/Logarithmic_integral_function
Edit: It looks like you are right, the function is unbounded.
I a cousin comment, Retric proves that the integral of 1/log(x)^2 diverges. The point is that log(n)^2 < n for x > 2 (you can prove this by calculus: the only maximum of the function log(x)^2/x is at x=e). So, 1/log(x)^2 > 1/x, and the integral of 1/x diverges.
The most fun way to prove the last statement is that the integral of 1/x from 1 to infinity is greater than 1/2 + 1/3 + 1/4 + 1/5 + ... > 1/2 + 1/4 + 1/4 + 1/8 + ... = 1/2 + 2 * 1/4 + 4 * 1/8 + ... = 1/2 + 1/2 + 1/2 + ....
If the twin prime conjecture is correct, then we may conclude that the [prime] integers are not constructed randomly.
Another interesting tid-bit.. Looks to me that mathematics research is getting overwhelmingly complex...
For the next two semesters, I organized a seminar with five graduate students (including Yitang) on Prof. Hironaka’s monumental papers on 2 the theory of resolutions of singularities. I believed that we doubled the world population of those who had studied the papers after we finished two semesters. Prof. Grothendick once described those papers as among the most complicated thesises in the human history.
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.
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.
> 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).