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 2010-09-07, 23:51 #1 Unregistered   71 Posts 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?
2010-09-08, 02:38   #2
TimSorbet
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

11·389 Posts

Quote:
 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.

2010-09-08, 02:59   #3
CRGreathouse

Aug 2006

5,987 Posts

Quote:
 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
$\frac{4000}{\log^22}\sum_{n=10^6}^\infty n^{-2}\approx0.008$
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:
$\approx21503\sum_{n=10^6}^\infty n^{-2}\approx0.021$
takes the factors of the k-values into account.

Last fiddled with by CRGreathouse on 2010-09-08 at 03:04 Reason: make approximation clearer

2010-09-08, 05:04   #4
Unregistered

2×1,889 Posts

Quote:
 Originally Posted by CRGreathouse I don't think that's known. Heuristically, we expect something like $\frac{4000}{\log^22}\sum_{n=10^6}^\infty n^{-2}\approx0.008$ 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: $\approx21503\sum_{n=10^6}^\infty n^{-2}\approx0.021$ takes the factors of the k-values into account.
Where do the 21503 (second figure) and 4000 (first figure) numbers come from?

2010-09-08, 13:00   #5
CRGreathouse

Aug 2006

5,987 Posts

Quote:
 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...
Code:
sum(k=2,10^6-1,ff(k+k)^2,0.)/log(2)^2
So the expectations are
Code:
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.

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