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al-mahed
al-mahed is offline
#42
Dec27-07, 12:45 PM
P: 258
This is the point: seems to me that you already are considering that there are such primes of the form [tex]p + 2^n[/tex], and we don't know yet. Or did I not understand some particular point?

My idea on measuring the primes between [itex]2^n[/itex] and [itex]2^{n+1}[/itex]:

Lets discriminate the numbers in classes according to the highest exponent, such that the nth-class has [itex]2^n[/itex] numbers, etc. Considering some prime number in nth-class like [itex]2^0 + 2^a + 2^b + 2^n[/itex], we should have [itex]2^0 + 2^b + 2^n[/itex], [itex]2^0 + 2^a + 2^n[/itex] or [itex]2^0 + 2^a + 2^b[/itex] = prime, or if for some reason this particular prime couldn't be written of the form [itex]p + 2^n[/itex], we should have [itex]2^0 + 2^a + 2^b + 2^n + 2^k[/itex] a prime number for some [itex]2^k[/itex] exponent, writting this prime in the [itex]p - 2^n[/itex] form.

My idea: [tex]\frac{2^n}{ln(2)}*(\frac{n-1}{n^2+n})[/tex] should give us the number of primes expected in the nth-class, and then we'll be able do find the probability that some particular prime number have a permutation of the exponents that represents another prime number. By doing these calculations I hope that the probability increases when n increases.

Honestly I don't feel that these statistical procedures could give us an absolute proof, only heuristics evidences.

Quote Quote by CRGreathouse View Post
I'm not sure why you're measuring the primes between [itex]2^n[/itex] and [itex]2^{n+1}[/itex]. The relevant probabilities are

[tex]1/\ln(p+2)[/tex]
[tex]1/\ln(p+4)[/tex]
[tex]1/\ln(p+8)[/tex]
. . .