A005420 -id:A005420 - cp's OEIS frontend

A359088 Odd integers k such that the multiplicative order of 2 modulo the largest prime factor of 2^k - 1 is different from k.

51, 111, 327, 1281, 1563

Offset: 1

Michel Marcus, Dec 16 2022

In other words, odd integers k that are not equal to A002326((A005420(k)-1)/2).

Inspired by former comment from Thomas Ordowski in A005420.

zn(n) = znorder(Mod(2, 2*n+1)); \\ A002326
f(n) = vecmax(factor(2^n-1)[,1]); \\ A005420
isok(k) = (k%2) && (zn((f(k)-1)/2) != k);

Edited and a(4)-a(5) added by Max Alekseyev, Feb 16 2025

A287945 a(n) = largest prime q such that q | 2^p - 2 and p - 1 | q - 1, where p = prime(n).

Original entry on oeis.org

2, 3, 5, 7, 31, 13, 257, 73, 683, 113, 331, 109, 61681, 5419, 2796203, 1613, 3033169, 1321, 599479, 122921, 38737, 22366891, 8831418697, 2931542417, 22253377, 268501, 131071, 28059810762433, 279073, 54410972897, 77158673929, 145295143558111, 2879347902817, 10052678938039, 616318177, 1133836730401, 121369

Offset: 1

Thomas Ordowski, Sep 01 2017

nonn

First conjecture: a(n) > prime(n) for all n > 6. Robert Israel tested the author's conjecture up to prime(95) = 499. The prime factorizations of the numbers 2^(p-1)-1 for larger p can be checked in available tables, see A005420.
Second conjecture: a(n) = gpf(2^prime(n) - 2) for almost all n, in the sense that the set of exceptions {10, 16, 37, 40, ...} has zero natural density.
Primes p for which p - 1 does not divide gpf(2^p - 2) - 1 are 29, 53, 157, 173, ...

			For prime(5) = 11, 2^11-2 = 2*3*11*31 and 11-1 | 31-1, so a(5) = 31.

Cf. A000040, A005420.

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

Original entry on oeis.org

4, 8, 9, 12, 24, 121

Offset: 1

Thomas Ordowski, Sep 15 2017

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)

Cf. A002326, A060443, A086251.

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

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

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A359088 Odd integers k such that the multiplicative order of 2 modulo the largest prime factor of 2^k - 1 is different from k.

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A287945 a(n) = largest prime q such that q | 2^p - 2 and p - 1 | q - 1, where p = prime(n).

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

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