Twin Prime Question

I'm interested in twins of the form k*2^n+/-1.

One question: What is the probablity that there is a twin of that form with k<1M and n>1M?

[TimSorbet]

Originally Posted by Unregistered

One question: What is the probablity that there is a twin of that form with k<1M and n>1M?

Probably: 1 (i.e. it's probably certain that at least one exists...unless I'm mistaken)
It's just quite hard to discover with current technology.

[CRGreathouse]

Originally Posted by Mini-Geek

Probably: 1 (i.e. it's probably certain that at least one exists...unless I'm mistaken)

I don't think that's known. Heuristically, we expect something like

examples, right?

Of course the constant factor needs work based on the residue classes 2^n takes on, as well as the factors in k.

Edit:

takes the factors of the k-values into account.

Originally Posted by CRGreathouse

I don't think that's known. Heuristically, we expect something like

examples, right?

Of course the constant factor needs work based on the residue classes 2^n takes on, as well as the factors in k.

Edit:

takes the factors of the k-values into account.

Where do the 21503 (second figure) and 4000 (first figure) numbers come from?

[CRGreathouse]

Originally Posted by Unregistered

Where do the 21503 (second figure) and 4000 (first figure) numbers come from?

4000 is about 1000 candidates * 2^2, where the 2 is because only odd numbers are used (which twice as likely to be prime). Oh! You want k < 1000000, not k < 1000; in that case it's 4000000.

The other one is a calculated number based on the factorizations of k < 1000. Let me try for k < 1000000. OK, I get 14756135.8...

sum(k=2,10^6-1,ff(k+k)^2,0.)/log(2)^2

So the expectations are

4e6/log(2)^2*(zeta(2)-sum(n=1,10^6,n^-2,0.))

8.32 (naive)

14.75 (including factorizations)

So with this larger range, the heuristic probability is high: with the Poisson model, it's something like 1 - e^-14.75 = 99.99996% likely that such a pair of primes exist.