cp's OEIS Frontend

Let a(0) = 0 and construct a(n) from a(n-1) by (i) incrementing the rightmost digit and (ii) if any digit is 2, replace the rightmost 2 with a 0 and increment the digit immediately to its left. (Note that changing "if" to "while" in this recipe gives the standard binary representation of n, A007088(n)).

Equivalently, a(2n+1) = a(n):1 and a(2n+2) = b(n):0, where b(n) is obtained from a(n) by incrementing the least significant digit and : denotes string concatenation.

If the digits of a(n) are d_k, d_{k-1}, ..., d_2, d_1, d_0, then n = Sum_{i=0..k} d_i*2^i, just as in standard binary notation. The difference is that here we are a bit lazy, and allow a few digits to be 2's. The number of 2's in a(n) appears to be A037800(n+1). - N. J. A. Sloane, Jun 03 2023

Every pair of 2's is separated by a 0 and every pair of significant 0's is separated by a 2.

a(n) has exactly floor(log_2((n+2)/3))+1 digits [cf. A033484] and their sum is exactly floor(log_2(n+1)) [A000523].

The i-th digit of a(n) is ceiling( floor( ((n+1-2^i) mod 2^(i+1))/2^(i-1) ) / 2).

A137951 gives values of terms interpreted as ternary numbers, a(n)=A007089(A137951(n)). - Reinhard Zumkeller, Feb 25 2008