Context free grammar for $bin(n)bin(n+1)^R$
Here is one solution:
$$
\begin{align*}
&amp;S \to 01 \mid 1A1 \mid 1B1 \\
&amp;A \to 1A0 \mid 0 \\
&amp;B \to 1B1 \mid 0B0 \mid 0C1 \\
&amp;C \to 1C0 \mid \epsilon
\end{align*}
$$

Explanation:


$A$ generates $1^n 0^{n+1}$, and so $1A1$ generates $1^{n+1} 0^{n+1}1$, which handles numbers whose binary expansion is $1^{n+1}$.
$C$ generates $1^n 0^n$, and so $0C1$ generates $01^n 0^n1$.
$B$ generates $w01^n0^n1w^R$, and so $1B1$ generates $1w01^n0^n1w^R1$, which handles numbers whose binary expansion is $1w01^n$.
Finally, $01$ handles $0$.