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

[Tomasz Ordowski]

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).