bibkey: hasler2016a260407
authors: M. F. Hasler; Jaroslav Krizek
year: 2016
title: "OEIS A260407, Numbers n such that (n-1)^2+1 divides 2^(n-1)-1"
doi: null
url: https://oeis.org/A260407
claim: "A260407 %N: Numbers n such that (n-1)^2+1 divides 2^(n-1)-1. A260407 %C: Conjecture: also numbers n such that ((2^k)^(n-1)-1) == 0 mod ((n-1)^2+1) for all k >= 1. - _Jaroslav Krizek_, Jun 02 2016"
strata_touched:
  - D5/S3/Arith/Congruence/KrizekAllMultiplesOfTheOrder
license: citation-only
triage: anchor

OEIS A260407

The NAME of A260407 states:

Numbers n such that (n-1)^2+1 divides 2^(n-1)-1.

Jaroslav Krizek’s COMMENT of June 2, 2016 states:

Conjecture: also numbers n such that ((2^k)^(n-1)-1) == 0 mod ((n-1)^2+1) for all k >= 1. - Jaroslav Krizek, Jun 02 2016

The AUTHOR line states:

M. F. Hasler, Jul 24 2015

The sequence author and AUTHOR-line date are Hasler and July 24, 2015. The bibliographic year 2016 is the year of Krizek’s target conjecture.

Verified locator

• URL: https://oeis.org/A260407
• NAME (verbatim): Numbers n such that (n-1)^2+1 divides 2^(n-1)-1.
• COMMENT, Jun 02 2016 (verbatim): Conjecture: also numbers n such that ((2^k)^(n-1)-1) == 0 mod ((n-1)^2+1) for all k >= 1. - Jaroslav Krizek, Jun 02 2016
• AUTHOR (verbatim): M. F. Hasler, Jul 24 2015

The formal result proves that the NAME condition is equivalent to the target COMMENT condition for every natural n at least one. It does not claim any other comment in A260407.

The proof specializes a textbook divisibility fact: divisibility of 2^d-1 implies divisibility of 2^(k*d)-1; it does not claim that general fact as a new result. At n=1, the modulus is 1 and 2^0-1=0, so both sides of the equivalence are true. A bounded scan over n=2,...,59999, sampling the universal condition through k=24, found zero violations; this scan carries no proof.