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Sep 3, 2015 at 9:37 comment added joro Your answer isn't rocket science, but for me this is positive result, which might be counterexample to the other answer.
Sep 3, 2015 at 7:57 comment added S. Carnahan @TerryTao I had been a bit hesitant to post this answer because it seemed rather trivial, but your observation makes it look much more interesting.
Sep 3, 2015 at 5:18 comment added joro @TerryTao Actually there is conjecture that starting from $2$ your map is genuine prime: primes.utm.edu/mersenne see "Let C0 = 2, then let C1 = 2^C0-1, C2 = 2^C1-1...Are these all prime?"
Sep 2, 2015 at 21:59 comment added Terry Tao One consequence of this is that one can rapidly and deterministically generate arbitrarily large (albeit quite sparse) probable primes via iterating the map $p \mapsto 2^{p-1}$. This is in contrast to genuine primes, for which a fast deterministic algorithm to generate large primes is still not known, see michaelnielsen.org/polymath1/index.php?title=Finding_primes
Sep 2, 2015 at 5:48 comment added joro Indeed :-) Probably more complicated proof for p prime is from the factorization of $2^p-1$.
Sep 2, 2015 at 2:28 history answered S. Carnahan CC BY-SA 3.0