There exists a unique strictly increasing sequence of nonnegative integers $a_1 < a_2 < … < a_k$ such that\[\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.\]What is $k?$
$\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306$
First, substitute $2^{17}$ with $x$. 
Then, the given equation becomes $\frac{x^{17}+1}{x+1}=x^{16}-x^{15}+x^{14}...-x^1+x^0$ by sum of powers factorization.
Now consider only $x^{16}-x^{15}$. This equals $x^{15}(x-1)=x^{15} \cdot (2^{17}-1)$.
Note that $2^{17}-1$ equals $2^{16}+2^{15}+...+1$, by difference of powers factorization (or by considering the expansion of $2^{17}=2^{16}+2^{15}+...+2+2$).
Thus, we can see that $x^{16}-x^{15}$ forms the sum of 17 different powers of 2. 
Applying the same method to each of $x^{14}-x^{13}$, $x^{12}-x^{11}$, ... , $x^{2}-x^{1}$, we can see that each of the pairs forms the sum of 17 different powers of 2. This gives us $17 \cdot 8=136$.
But we must count also the $x^0$ term. 
Thus, Our answer is $136+1=\boxed{\textbf{(C) } 137}$.
~seanyoon777