[seqfan] Does A292849 contain all odd numbers ?

[Thomas Scheuerle]

Hello,

Actually I am preparing a new OEIS submission.
Therefore it would be nice if I could proof that A292849 contains all odd numbers.
>From calculation ( at least from what I could do in reasonable effort and time ) it apears true.

Now I am thinking how could that be proofed ...

Two conditions are to be fulfilled that A292849 contains all odd numbers.
First: For all odd numbers k there needs to be an odd m > 1 that fullfills A000120(k)=A000120(k*m).
A000120 shows a repeating pattern starting with each new 2^j:
0,
1,
1,2                               A000120(2^1 + 0...1)
1,2,2,3                           A000120(2^2 + 0...3)
1,2,2,3,2,3,3,4                   A000120(2^3 + 0...7)
1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5   A000120(2^3 + 0...15)
This means that we can choose for any odd number k > 1, an factor m so that A000120(k*m) becomes any Hamming weight > 1, if we could chose k*m minus biggest possible power of two to become an arbitary number r > 1. (Equation: k*m = 2^j+r = w = A063787(r).)
It has only solutions if neither k or m are divisors of r.
But it apears there are still unlimited m's as solutions if you want a r to get a certain Hamming weight w = A063787(r) > 1, because there is always a place in A063787 where you get the same value for r and r+1, and if r could be divided by k or m this will not be true for r+1.
So it apears to me the first condition is fulfilled and for a given odd m > 1,
we find an odd k such that A000120(k)=A000120(k*m) and better, there are unlimited k doing this.   Is this right ?

Second: For all k there needs to be an m where k is the smallest solution to A000120(k)=A000120(k*m).
It apears to me:
If m = (2^j+r)/k and r the smallest r which fulfills A063787(r)=A000120(k), then k is the smallest solution to A000120(k)=A000120(k*m)
if k shares no common prime factor with 2^j+r.
Here I am stuck, has any one an idea ?

Thank you very much

Thomas Scheuerle

-- 
T.S. <ts181 at mail.ru>