A330382 Composite numbers k such that k-1 divides 2^k-2.

55, 295, 343, 1027, 1135, 1315, 1807, 2059, 2395, 3403, 4375, 5335, 6175, 6499, 7183, 7939, 9235, 10207, 12643, 13123, 14155, 16003, 16255, 19495, 21547, 23815, 27595, 27703, 30619, 35479, 37927, 43219, 45487, 48007, 48763, 50275, 55567, 58483, 64387, 64639, 74899

Offset: 1

Author

Amiram Eldar and Thomas Ordowski, Dec 12 2019

Keywords

• nonn

Comments

If k is in this sequence, then 2^k-1 is also a term, so this sequence is infinite.

Also 2^p-1 is in this sequence for such prime p in A069051 that 2^p-1 is composite.

Theorem: if k-1 | 2^k-2, then m-1 | 2^m-2, where m = 2^k-1.

Conjecture: k-1 | 2^k-2 for k = (2^n-1)^3 if and only if n(n-1) | 2^n-2 for n > 2.

It seems that A007013(n)^3 for n > 1 and A007013(n) for n > 4 are in this sequence.

These are the composites k for which M - 1 divides 2^M - 2 where M = 2^k - 1. - Thomas Ordowski, Jul 01 2024

Links

• Amiram Eldar, Table of n, a(n) for n = 1..5000

Crossrefs

Cf. A007013, A014741, A054723, A069051, A217468.

A217468 is a subsequence.

Programs

Mathematica

Select[Range[75000], CompositeQ[#] && Divisible[PowerMod[2, #, # - 1] - 2, # - 1] &]

PARI

forcomposite(k=1,75000,if(!((2^k-2)%(k-1)),print1(k,", "))) \\ Hugo Pfoertner, Dec 12 2019

Formula

Composites of A014741(n) + 1. - Thomas Ordowski, Jul 01 2024