cp's OEIS Frontend

Sequence A005186 tells how many of these numbers are in [2^n, 2^(n+1)-1].

From Rémy Sigrist, Aug 19 2017: (Start)

a(2^n) = 2^n for any n >= 0.

A000120(a(n)) - 1 = A006667(n) for any n > 0.

A070939(a(n)) - 1 = A006577(n) for any n > 0.

All terms are Fibbinary numbers (A003714).

(End)

Sequence A005186 tells how many of these numbers are in [2^n, 2^(n+1)-1].

Working from right to left, the sequence of 0's and 1's in a(n) encode, respectively, the sequence of 3x+1 and x/2 steps in the Collatz iteration of n. This is reverse one's complement of Garner's parity vector. Criswell mentions this encoding.

The length of a(n) is A006577(n). The number of 1's in a(n) is A006666(n). The number of 0's in a(n) is A006667(n). The number of terms having length k is A005186(k).