[seqfan] Re: A260119; Least positive integer k such that 2^n-1 and k^n-1 are relatively prime.

[israel at math.ubc.ca]

Note that (2k)^n - 1 = k^n (2^n - 1) + k^n - 1, so gcd((2k)^n - 1, 2^n - 1) 
= gcd(k^n - 1, 2^n - 1). That takes care of your Conjecture 1. I don't 
understand your Conjecture 2. What is m?

Cheers,
Robert

On Sep 1 2015, David Corneth wrote:

>So I've put a conjecture that might help finding such values of k, but I
>don't see a proof and I guess a faster program could be found. There are
>some examples to it, but I don't know what characteristic these n have. The
>conjecture is in the history of the sequence, see
>https://oeis.org/history/view?seq=A260119&v=36
>
> Any ideas on how to proceed?
>
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