Evil number differences

[Max A.]

On 8/28/06, Max A. <maxale at gmail.com> wrote:

> 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.

Ops. I means if the number of bits in n is odd then the last bit of
A001969(n) is 1, and if the number of bits in n is even then the last
bit of A001969(n) is 0. Hence, the sequence of last bits is A010060.

> Therefore, A001969(n) = 2*n + A010060(n).

Max