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## Re: [PrimeNumbers] prime +/- 2^n = primes: A tight sequence and a potential c...

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• ... On further reflection, the correct statement (for the bit-flipping variant) is:- the chance of each of them being prime is approximately 2*C_2/log(N),
Message 1 of 4 , Jun 28, 2003
> So the chance of each of them being prime is approximately 2/log(N).

On further reflection, the correct statement (for the bit-flipping variant)
is:-
the chance of each of them being prime is approximately 2*C_2/log(N), where
C_2 = 0.66016... is the usual "twin-primes constant".
See e.g. Chris Caldwell's
http://www.utm.edu/~caldwell/preprints/Heuristics.pdf

So, as N -> oo, the expected no. of primes is 2*C_2*log2(N)/log(N) =
2*C_2/log(2) =
1.9048... And for the +/-2^n procedure, it is just twice this, or 3.8096...,
which is in rather good agreement with Jens' and my results.

Mike Oakes

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