Well, there's a slight problem ... 3 = 2 + 1, and I'm assuming that we're not considering 1 to be prime ... so then a = 0 (a is alpha ... too early for me to be using latex, hehehe). So you should allow for a = 0 .

Just some other things, for odd primes, it's obvious that P = p + 2n ... i.e. the difference between any odd primes is even .

As for P = p + 2^a, we might be able to reconstruct P so that since where is the next prime before P, then where is the next prime before , so we keep doing this until we get where ... and I guess it's up to you to show that it's possible ... or not

Also, not quite the same, but you might want to look up the twin primes conjecture (and related/modified conjectures), it might give you some ideas ....