sum of odious numbers to the power of k In number theory, an odious number is a positive integer that has an odd number of $1$s in its binary expansion.
The first odious numbers are:
$1, 2, 4, 7, 8, 11, 13, 14, 16, 19, 21, 22, 25, 26, 28, 31, 32, 35, 37, 38...$
Let $a$ denote the increasing sequence of odious numbers where $a_0=1$, $a_1=2$, etc.
Whats the summatory function of the sequence $b$ where $b_i=a_i^k$?
Also: this is useful; but it didn't give out the actual function. Please help.
 A: I think there is no simple formula here, although we can get some recurrence relations and related identities for generating functions as explained below.

Similarly to odious numbers, we have evil numbers that contain even number of 1s in their binary representation.
Let $S^o_k(n)$ and $S^e_k(n)$ denote the sum of $k$-th powers of all odious and evil numbers $\leq n$, respectively.
It can be easily seen that each odious number has one of the two forms: $2m$ where $m$ is an odious number, or $2m+1$ where $m$ is an evil number. It follows that for $n\geq 1$
$$S^o_k(n) = S^o_k(\lfloor\tfrac{n}2\rfloor)2^k + \sum_{i=0}^k \binom{k}{i} S^e_i(\lfloor\tfrac{n-1}2\rfloor) 2^i$$
and similarly
$$S^e_k(n) = S^e_k(\lfloor\tfrac{n}2\rfloor)2^k + \sum_{i=0}^k \binom{k}{i} S^o_i(\lfloor\tfrac{n-1}2\rfloor) 2^i.$$
We also have $S^o_k(0)=S^e_k(0)=0$ for all $k\geq 1$, and $S^o_0(0)=0$ and $S^e_0(0)=1$.

In terms of generating functions $F^o(x,y):=\sum_{n,k\geq 0} S^o_k(n) x^n \frac{y^k}{k!}$ and $F^e(x,y):=\sum_{n,k\geq 0} S^e_k(n) x^n \frac{y^k}{k!}$, we have
$$F^o(x,y) = (1+x)F^o(x^2,2y) + (x+x^2) F^e(x^2,2y) e^y,$$
$$F^e(x,y) = (1+x)F^e(x^2,2y) + (x+x^2) F^o(x^2,2y) e^y.$$
Defining $G(x,y):=F^e(x,y) - F^o(x,y)$, we have
$$G(x,y) = (1+x)(1 - x e^y)G(x^2,2y).$$
Also, it is easy to get
$$F^e(x,y) + F^o(x,y) = \frac{1}{(1-x)(1-xe^y)}.$$
Since $G(x,y) =  \frac{1}{(1-x)(1-xe^y)} - 2F^o(x,y)$, we a standalong functional equation for $F^o$:
$$F^o(x,y) = (1+x)(1 - x e^y)F^o(x^2,2y) + \frac{xe^y}{(1-x)(1-x^2e^{2y})}.$$
A: Too big to comment:
If we take the sequence of Gray code numbers to be $\{ g(m) \}_{m=0}^{\infty}$ then the $m$-th Odious number $Q(m)$ ($Q(0)=1,Q(1)=2...$) is the $g(m)$-th number of in the sequence of $g(2k+1), k=0,1,2,...,$
Hence, $Q(m)=g(2g(m)+1)$.
A formula for $m$-th Gray number is $$g(m)=\sum_{k=0} \sigma_i(m)2^{k-1}$$ where $\sigma_i(m)=\lfloor{\frac{|r_i(m)|}{2^k}}\rfloor$.
And, $r_i(m)=m-(\lfloor{\frac{m}{2^k}}\rfloor -(-1)^{\lfloor{\frac{m}{2^k}}\rfloor})2^k$.
