Is it really true that, for any n > 0 , there is a prime between 2 ^ n

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This morning, while thinking over my first conjecture, I came up with a second conjecture, which is closely related to my first conjecture and which could lend more substance to my first conjecture. My second conjecture is that, for any k > 2, there is always a composite number in the range (2^k , 2^{k + 1}) whose range size is less than that of 2^k. The range size of a number n is how many prime factors n has \times the sum of n's prime factors. The range size of 2^k is therefore 2k².

So far, I've tested my second conjecture up to k = 30. I would need more than unsigned int to handle powers of 2 over 2³¹.

Proof: For k>2, 2^k is divisible by 8 and the "range size" (never heard of that before, but I have no idea about number theory) of 9/8 * 2^k is 2k(k-1) < 2k². The number is in your range, as 1<9/8<2.
Between 2^k and 2^k + 2 k², we expect 2 k/ln(2) primes. Neglecting correlations, the probability of finding a prime is $\displaystyle p(k) \approx 1-\exp\left(-\frac{2k}{ln(2)}\right)$. The product over all k should converge to some finite value, so the non-existence of a small counterexample is a good argument that could be no such gap. It is not a proof, of course.