Odd b(n) = 2n/2^m-2^m with m > 0

[Tomasz Ordowski]

Hello everyone! 

 

Let us define a sequence of odd numbers by the formula 

   a(n) = 2n/2^m+2^m = n/2^(m-1)+2^m with m > 0. 

Data a(n) : 3, 5, 5, 9, 7, 7, 9, 17, 11, 9, ... for n > 0.  

We have a(n) = Odd(2n)+2n/Odd(2n) = Odd(n)+2n/Odd(n), 

where Odd(n) = A000265(n) is the odd part of n. 

Cf. https://oeis.org/history/view?seq=A386220&v=40

   Every positive odd number k appears exactly 

floor(log_2(k)) times in this sequence.*  

 

I also considered the sequence of odd integers given by  

   b(n) = 2n/2^m-2^m = n/2^(m-1)-2^m with m > 0. 

Data b(n) : -1, -3, 1, -7, 3, -1, 5, -15, 7, 1, ... for n > 0.  

We have b(n) = Odd(2n)-2n/Odd(2n) = Odd(n)-2n/Odd(n), 

where Odd(n) as above: https://oeis.org/A000265 

   How many times does each odd number k 

(positive and negative) appear in this sequence? 

   Perhaps infinitely many times [sic]. 

Note that b(n) < 0 iff 2n+1 is a Proth number: 

https://oeis.org/A080075 (in ascending order). 

https://oeis.org/A369901 (as a permutation). 

 

Best, 

 

Tom Ordo 

__________________

(*) Cf. https://oeis.org/A387016 and 

https://oeis.org/history/view?seq=A386386&v=19 

The formula of (a) is an inverse of the formula of this permutation.  

 

[Tomasz Ordowski]

PS. See my current drafts: 

https://oeis.org/draft?user=Thomas%20Ordowski 

   And https://oeis.org/A387016 (recently published).

I intend to propose the sequence (b), after discussion.

 

Thomas Ordowski (2/3)