Different Euler Bases

The number $N_0(n) + N_1(n)$ is just the number of digits in the binary expansion of $n$ . Thus $N_0(n) + N_1(n) = m$ for $2^{m-1} \le n < 2^m$ for any $m \ge 1$ . Thus the sum is $\begin{array}{rcl} \mathcal{S} & = & \displaystyle \sum_{m=1}^\infty \sum_{n=2^{m-1}}^{2^m-1} \frac{m}{n(2n+1)} \; = \; \sum_{m=1}^\infty 2m \sum_{n=2^{m-1}}^{2^m-1}\left(\frac{1}{2n} - \frac{1}{2n+1}\right) \\ & = & \displaystyle \sum_{m=1}^\infty 2m \sum_{n=2^m}^{2^{m+1}-1} \frac{(-1)^n}{n} \; = \; \sum_{m=1}^\infty 2m \left\{ 2\sum_{n=2^{m-1}}^{2^m-1} \frac{1}{2n} -\sum_{n=2^m}^{2^{m+1}-1} \frac{1}{n}\right\} \\ & = & \displaystyle \sum_{m=1}^\infty 2m\left\{ 2\sum_{n=1}^{2^m-1} n^{-1} - \sum_{n=1}^{2^{m-1}-1} n^{-1} - \sum_{n=1}^{2^{m+1}-1}n^{-1}\right\} \\ & = & \displaystyle \sum_{m=1}^\infty 2m\left\{ 2\sum_{n=1}^{2^m} n^{-1} - \sum_{n=1}^{2^{m-1}} n^{-1} - \sum_{n=1}^{2^{m+1}}n^{-1} + 2^{-m-1}\right\} \\ & = & \displaystyle 2 + \sum_{m=1}^\infty 2m\left\{ 2\big[ S_{2^m} + \ln 2^m\big] - \big[S^{2^{m-1}} + \ln 2^{m-1}\big] - \big[S_{2^{m+1}} + \ln2^{m+1}\big]\right\} \\ & = & \displaystyle 2 + \sum_{m=1}^\infty 2m\big[ 2S_{2^m} - S_{2^{m-1}} - S_{2^{m+1}}\big] \end{array}$ where $S_n \; = \; \sum_{m=1}^n m^{-1} - \ln n \;,$ noting that $S_n \,=\, \gamma + o(n^{-1})$ as $n \to \infty$ . But then $\begin{array}{rcl} \mathcal{S} & = & \displaystyle 2 + \lim_{M\to\infty} \sum_{m=1}^M 2m\big[ 2S_{2^m} - S_{2^{m-1}} - S_{2^{m+1}}\big] \\ & = & \displaystyle 2 + \lim_{M \to \infty} \left\{ \sum_{m=1}^M 4m S_{2^m} - \sum_{m=0}^{M-1}2(m+1)S_{2^m} - \sum_{m=2}^{M+1}2(m-1)S_{2^m}\right\} \\ & = & \displaystyle \lim_{M \to \infty} \big[(2M+2)S_{2^M} - 2MS_{2^{M+1}}\big] \\ & = & \displaystyle 2\gamma + \lim_{M\to\infty}\big[(2M+2)(S_{2^M}-\gamma) - 2M(S_{2^{M+1}}-\gamma)\big] \\ & = & 2\gamma \end{array}$ making the answer $\boxed{2}$ ..