[seqfan] Integers k > 0 such that (k-2^m)2^n+1 = k has a solution m, n > 0

[Tomasz Ordowski]

Dear number lovers!

Let us define these positive integers a bit differently, namely:
Numbers k > 0 such that (k-1)/(k-2^m) = 2^n for some m,n > 0.
In parametric approach (with parameter t) assuming (in advance)
k = t+2^m = t 2^n+1, we get t = (2^m-1)/(2^n-1) where n|m, QED.
Is this complete proof that the values of t are exactly A064896?
If "yes", we have (after eliminating t) two equivalent formulas:
k = (2^m-1)/(2^n-1)+2^m = (2^n)(2^m-1)/(2^n-1)+1 with n|m.
They seem to be the same numbers A064896 except 1.
If so, we already know their new interesting properties!
Note that a prime p is a term of A064896 if and only if
p is a Mersenne prime or p is a Fermat prime.

Best,

Tom Ordo
 _______________
See https://oeis.org/A064896
This is a subset of sturdy numbers,
by Stolarsky's (1980) Theorem 2.1 in the work
http://matwbn.icm.edu.pl/ksiazki/aa/aa38/aa3825.pdf
(as T. D. Noe rightly notes in his comment to A064896).
See also https://oeis.org/A125121 (the sturdy numbers).