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To 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, $r_i(m)=m-(\lfloor{\frac{m}{2^k}}\rfloor -(-1)^{\lfloor{\frac{m}{2^k}}\rfloor})2^k$.

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, $r_i(m)=m-(\lfloor{\frac{m}{2^k}}\rfloor -(-1)^{\lfloor{\frac{m}{2^k}}\rfloor})2^k$.

To 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 $O(m)$ ($O(0)=1,O(1)=2...$) is the $g(m)$-th number of in the sequence of $g(2k+1), k=0,1,2,...,$

Hence, $O(m)=g(2g(m)+1)$.

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, $r_i(m)=m-(\lfloor{\frac{m}{2^k}}\rfloor -(-1)^{\lfloor{\frac{m}{2^k}}\rfloor})2^k$.

To 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, $r_i(m)=m-(\lfloor{\frac{m}{2^k}}\rfloor -(-1)^{\lfloor{\frac{m}{2^k}}\rfloor})2^k$.

To 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 $O(m)$ ($O(0)=1,O(1)=2...$) is the $g(m)$-th number of in the sequence of $g(2k+1), k=0,1,2,...,$

Hence, $O(m)=g(2g(m)+1)$.

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, $r_i(m)=m-(\lfloor{\frac{m}{2^k}}\rfloor -(-1)^{\lfloor{\frac{m}{2^k}}\rfloor})2^k$.

To 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 $O(m)$ ($O(0)=1,O(1)=2...$) is the $g(m)$-th number of in the sequence of $g(2k+1), k=0,1,2,...,$

Hence, $O(m)=g(2g(m)+1)$.

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, $r_i(m)=m-(\lfloor{\frac{m}{2^k}}\rfloor -(-1)^{\lfloor{\frac{m}{2^k}}\rfloor})2^k$.