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P: 258
This is the point: seems to me that you already are considering that there are such primes of the form $$p + 2^n$$, and we don't know yet. Or did I not understand some particular point?

My idea on measuring the primes between $2^n$ and $2^{n+1}$:

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

My idea: $$\frac{2^n}{ln(2)}*(\frac{n-1}{n^2+n})$$ 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 by CRGreathouse I'm not sure why you're measuring the primes between $2^n$ and $2^{n+1}$. The relevant probabilities are $$1/\ln(p+2)$$ $$1/\ln(p+4)$$ $$1/\ln(p+8)$$ . . .