OFFSET
1,1
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.
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
FORMULA
Composites of A014741(n) + 1. - Thomas Ordowski, Jul 01 2024
MATHEMATICA
Select[Range[75000], CompositeQ[#] && Divisible[PowerMod[2, #, # - 1] - 2, # - 1] &]
PROG
(PARI) forcomposite(k=1, 75000, if(!((2^k-2)%(k-1)), print1(k, ", "))) \\ Hugo Pfoertner, Dec 12 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Amiram Eldar and Thomas Ordowski, Dec 12 2019
STATUS
approved