A292364 Composites m such that each prime factor p > m of 2^m - 1 is a primitive prime factor of 2^m - 1.

4, 8, 9, 12, 24, 121

Offset: 1

Author

Thomas Ordowski, Sep 15 2017

Keywords

• nonn
• more

Comments

From A086251: "A prime factor of 2^n-1 is called primitive if it does not divide 2^r-1 for any r

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^d-1 is n-smooth.

Let S represent the set of numbers such that the greatest prime factor of 2^n-1 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^d-1) > 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)

Crossrefs

Cf. A002326, A060443, A086251.

Programs

PARI

lista(nn) = {forcomposite (m=1, nn, f = factor(2^m-1)[,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

Formula

A002326((p-1)/2) = m for every prime factor p > m of 2^m - 1.

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