Evil number differences

[Max A.]

On 8/28/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
> This is equivalent to showing that A001969(n) = 2*n + A001285(n) - 1,
> where A001285 is the Thue-Morse sequence.

I'd better say that  A001969(n) = 2*n + A010060(n). And the proof is
rather simple:
It is enough to notice that [A001969(n)/2] = n (i.e., all bits except
the last/lowest one of elements of A001969 form the sequence of all
positive integers) while the last bits serve for the parity
correction: if n is odd the last bit=1, if n is even the last bit=0,
forming the sequence A010060. Therefore, A001969(n) = 2*n +
A010060(n).

Max