21. Given $\left\{a_{n}\right\}$ is an increasing sequence of non-negative integers, i.e., $a_{1}<a_{2}<\cdots<a_{k}$, satisfying:
$$
\frac{2^{289}+1}{2^{17}+1}=2^{a_{1}}+2^{a_{2}}+\cdots+2^{a_{k}} \text {. }
$$

Then the value of $k$ is ( ).
(A) 117
(B) 136
(C) 137
(D) 273
(E) 306
21. C.
Let $x=2^{17}$. Then
$$
\begin{array}{l} 
\frac{2^{289}+1}{2^{17}+1}=\frac{x^{17}+1}{x+1} \\
= x^{16}-x^{15}+x^{14}-x^{13}+\cdots+x^{2}-x+1 \\
=(x-1)\left(x^{15}+x^{13}+x^{11}+\cdots+x^{3}+x\right)+1 \\
=\left(2^{17}-1\right)\left(2^{15 \times 17}+2^{13 \times 17}+2^{11 \times 17}+\cdots+\right. \\
\left.2^{3 \times 17}+2^{17}\right)+1 \\
=\left(2^{16}+2^{15}+\cdots+2+1\right)\left(2^{15 \times 17}+2^{13 \times 17}+\right. \\
\left.2^{11 \times 17}+\cdots+2^{3 \times 17}+2^{17}\right)+1 .
\end{array}
$$

From the above expansion, there are a total of $17 \times 8+1=137$ terms, hence $k=137$.