An Unlucky Number?

Consider $\sum_{n=0}^{m}\ln{(x^{2^n}+y^{2^n})}$

$\sum_{n=0}^{m}\ln{(x^{2^n}+y^{2^n})} = \ln{\prod_{n=0}^{m}(x^{2^n}+y^{2^n})} \\ \frac{d}{dy}(\sum_{n=0}^{m}\ln{(x^{2^n}+y^{2^n})}) = \frac{d}{dy}(\ln{\prod_{n=0}^{m}(x^{2^n}+y^{2^n})})\\ \sum_{n=0}^{m}\frac{2^ny^{2^n-1}}{x^{2^n}+y^{2^n}} = \frac{d}{dy}(\ln{(x^1+y^1)(x^2+y^2)(x^4+y^4)...(x^{2^m}+y^{2^m})}) \\ = \frac{d}{dy}(\ln{\Big(\frac{(x-y)(x+y)(x^2+y^2)(x^4+y^4)...(x^{2^m}+y^{2^m})}{(x-y)}\Big)}) \\ = \frac{d}{dy}(\ln{\Big(\frac{x^{2^{m+1}}-y^{2^{m+1}}}{(x-y)}\Big)}) \\ = \frac{d}{dy}(\ln{(x^{2^{m+1}}-y^{2^{m+1}})}-\ln{(x-y)}) \\ \sum_{n=0}^{m}\frac{2^ny^{2^n-1}}{x^{2^n}+y^{2^n}} = \frac{1}{x-y}-\frac{2^{m+1}y^{2^{m+1}-1}}{x^{2^{m+1}}-y^{2^{m+1}}}$

Now, we take limit,

$\lim_{m\rightarrow\infty}\sum_{n=0}^{m}\frac{2^n}{k^{2^n}+1} = \frac{1}{k-1}-\lim_{m\rightarrow\infty}\frac{2^{m+1}}{k^{2^{m+1}}-1} = \frac{1}{k-1}-\lim_{m\rightarrow\infty}\frac{2^{m+1}\ln{2}}{2^{m+1}k^{2^{m+1}-1}} \\ \frac{1}{k-1}-\lim_{m\rightarrow\infty}\frac{\ln{2}}{k^{2^{m+1}-1}} \\ = \frac{1}{k-1}$

With $k=13$ , we have the summation equal to $\frac{1}{12}$

I claim that $\sum_{n=0}^{k-1} \frac{2^n}{13^{2^n}+1} = \frac1{12} - \frac{2^k}{13^{2^k}-1}.$ To see this, we induct on $k$ . It's clear for $k = 1$ : $\frac1{14} = \frac1{12} - \frac2{168}$ . Now suppose this is true for $k$ . Then we get $\begin{aligned} \sum_{n=0}^k \frac{2^n}{13^{2^n}+1} &= \frac1{12} - \frac{2^k}{13^{2^k}-1} + \frac{2^k}{13^{2^k}+1} \\ &= \frac1{12} - \frac{2 \cdot 2^k}{(13^{2^k}-1)(13^{2^k}+1)} \\ &= \frac1{12} - \frac{2^{k+1}}{13^{2^{k+1}}-1} \end{aligned}$

as desired. So the claim is true for $k+1$ , hence always true by induction.

Now taking the limit as $k \to \infty$ gives a sum of $1/12$ , so the answer is $\fbox{13}$ .