[seqfan] Re: Integers k of the form (2^m + 1) / (2^n + 1)

[Max Alekseyev]

Sequence of such numbers is already present as https://oeis.org/A079665
which however rather artificially excludes number 1.
I'd suggest relaxing the definition of https://oeis.org/A079665 to include
1, and discard the draft of A370377.

Regards,
Max

On Sat, Feb 17, 2024 at 2:09 AM Tomasz Ordowski <tomaszordowski at gmail.com>
wrote:

> Dear Sequence Fans!
>
> The numbers k > 0 for which the equation
> 2^m + k = k*2^n + 1 has a solution m,n > 0
> are integers of the form (2^m - 1) / (2^n - 1),
> i.e. if an only if n divides m. I wrote about it.
> See https://oeis.org/A064896 (my comment).
> Primes of the sequence A064896 are A245730.
> See https://oeis.org/A245730 (all these primes).
> Contains: all Mersenne primes, all Fermat primes,
> and other primes {73, 262657, 4432676798593, ...}.
>    Contrary to appearances, a sufficiently large number k of this form
> cannot be a dual Sierpinski number, but may be a dual Riesel number.
>
> The numbers k for which the equation
> 2^m - k = k*2^n - 1 has a solution m,n > 0
> are integers of the form (2^m + 1) / (2^n + 1).
> Note that if 2^n+1 divides 2^m+1, then n divides m.
> 1, 3, 11, 13, 43, 57, 171, 205, 241, 683, 993, 2731, 3277, ...
> This important sequence is not found on the OEIS websites.*
> Something related. See A281728 for primes in this sequence:
> https://oeis.org/A281728 (unfortunately, without definition there).
> Question: are these exactly primes of the form (2^m+1)/(2^n+1)?
>     Appearances are deceiving again,
> because a sufficiently large number k of this form
> cannot be a dual Riesel number, but may be a dual Sierpinski number.
>
> Best,
>
> Tom Ordo
> _________________
> (*) See my new draft: https://oeis.org/draft/A370377
> https://oeis.org/A076336 (Sierpinski numbers).
> https://oeis.org/A101036 (Riesel numbers).
>
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