
COMMENTS

No more terms found up to p = 1277, 1277 being the first prime for which the complete factorization of 2^p1 is not currently known (see GIMPS link).  Michel Marcus, Jan 20 2014
Conjecture: gpf(gpf(2^p1)1) = p for finitely many p.
Conjecture: gpf(lpf(2^p1)1) = p for infinitely many p.
Namely, for p = 2, 3, 5, 7, 11, 13, 23, 29, 37, 43, 47, 53, ...  Michael B. Porter, Jan 26 2014
Note that gpf(lpf(2^p1)1) = gpf(gpf(2^p1)1) = p for p = 2, 3, 5, 7, 11, 13, 29, 53. See DATA.


EXAMPLE

For prime p=2, 2^p1=3, gpf(3)=3, gpf(31)=2, so 2 is in the sequence.
For prime p=3, 2^p1=7, gpf(7)=7, gpf(71)=3, so 3 is in the sequence.
