
COMMENTS

From A086251: "A prime factor of 2^n1 is called primitive if it does not divide 2^r1 for any r<n. Equivalently, p is a primitive prime factor of 2^n1 if ord(2,p)=n."
Are there only finitely many such composite numbers?
From Charlie Neder, Jan 09 2019: (Start)
Equivalently, composite numbers n such that, for each proper divisor d of n, 2^d1 is nsmooth.
Let S represent the set of numbers such that the greatest prime factor of 2^n1 is less than n^2. S begins {2,3,4,6,8,9,10,11,12,14,15,18,20,21,24,28,30,36,48,60} (obtained from A005420), and I conjecture that there are no further terms.
For any composite number k, if k has a divisor d >= sqrt(k) that is not in this sequence, then gpf(2^d1) > d^2 >= k and k is not in this sequence.
If S is complete, there are 15 possible choices of k, the largest of which is 121, and this sequence is complete. (End)


PROG

(PARI) lista(nn) = {forcomposite (m=1, nn, f = factor(2^m1)[, 1]~; ok = 1; for (k=1, #f, p = f[k]; if ((p > m) && (znorder(Mod(2, p)) != m), ok = 0; break); ); if (ok, print1(m, ", ")); ); } \\ Michel Marcus, Nov 11 2017
