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 $k?$
Let $x = 2^{17}$. Then the left side can be rewritten as $$\frac{x^{17}+1}{x+1}=x^{16}-x^{15}+x^{14}-x^{13}+\cdots - x^1 + x^0$$
Considering the pair $x^{16}-x^{15}$ now: $$x^{16}-x^{15}=x^15(x-1)=2^{17\times 15}(2^{17}-1)=2^{17\times 15}(2^{16}+2^{15}+\cdots +1)$$
Similarly, we can write $$x^{14}-x^{13}=2^{17\times 13}(2^{16}+2^{15}+\cdots + 1)$$
till $$x^2 - x^1 = 2^{17\times 1}(2^{16}+2^{15}+\cdots + 1)$$
Clearly these terms are not duplicate, therefore there are totally $17\times 8=136$ terms. Finally, we also need to consider the left over constant term $x^0=1$ which makes the total number of terms be $\boxed{137}$.