# Infinite sum with number system conversions

Any number $x \in \mathbb{Z}$ can be written as $\pm\sum\limits_{i=0}^n x_i2^i$ where $x_i \in \text{{0,1}}$ (signed binary representation). It can also be written as $\sum\limits_{i=0}^n x_i(-2)^i$ where $x_i \in \text{{0,1}}$ (negabinary representation).

We are given two functions $B(x)$ and $N(x)$, where $B(x)$ interprets the binary representation of the decimal input as a negabinary input, which it converts to binary and then to decimal and where $N(x)$ converts the binary representation of the decimal input to negabinary, which is then interpreted as binary and then as decimal.

For example:

$B(3_{10}) \rightarrow B(11_{-2}) = -1_{2}=-1_{10}$

and

$N(2_{10}) = N(10_{2})=110_{-2} \rightarrow 110_2=6_{10}$

Now the following is true:

$\sum\limits_{i=1}^{\infty} \frac{1}{B(2n+1)^2}=\frac{\pi^2}{4}$

because $B(2n+1)^2$ contains the square of all odd numbers twice and the sum of the recipricals of all odd numbers is $\sum\limits_{i=1}^{\infty} \frac{1}{(2n+1)^2}= \frac{\pi^2}{8}$

The question now is, what is $\sum\limits_{n=1}^{\infty} \frac{1}{N(n)^2}$ equal to?