a)
We prove by induction on $n$ that $a_n = s_2(n)$ for all $n \ge 1$ where $s_2(n)$ is the sum of the digits of $n$ in binary representation.

The base case $n = 1$ is trivial. Now, we assume that $a_n = s_2(n)$ is proved for $n = 1, 2, \dots, k$, and prove it for $n = k + 1$.

We shall prove the statement for $k$ with some cases as follows.

• $2 \mid k+1$, let $k+1 = \overline{x_1x_2...x_{m-1}x_m}(2)$ with $x_1 = 1, x_m = 0$.
It follows $\frac{k+1}{2} = \overline{x_1x_2...x_{m-1}}(2)$, so
$$a_{k+1} = a_{\frac{k+1}{2}} = s_2\left(\frac{k+1}{2}\right) = s_2(k+1).$$
• $2 \mid k$, let $k+1 = \overline{x_1x_2...x_{m-1}x_m}(2)$ with $x_1 = 1, x_m = 1$.
It follows $\frac{k}{2} = \overline{x_1x_2...x_{m-1}}(2)$, so
$$a_{k+1} = a_{\frac{k}{2}} + 1 = s_2\left(\frac{k}{2}\right) + 1 = s_2(k+1).$$
Therefore in all cases, we always have $a_{k+1} = s_2(k+1)$. So that $a_n = s_2(n)$ for all positive integers $n$. We distinguish two cases regarding the value of $n$.

Case 1. $n < 3$, it is easy to see that $n \in \{1, 2\}$ satisfied.

Case 2. $n \ge 3$, we investigate three subcases

Subcase 1. if $n = \overline{100...0}(2)$ then choose $k=3$, so $kn = \overline{1100...0}(2)$, which is not satisfying $a_n = a_{kn}$.

Subcase 2. if $n = \overline{111...1}(2)$ then we will prove $n$ in this case satisfied. Assume that $n = 2^t - 1$, with $k < n$ and $k$ is odd, expand $k = \overline{x_1x_2...x_t}(2)$. We have
$$kn = \overline{x_1x_2...x_t\overline{00...0}(2)} - \overline{x_1x_2...x_t\overline{0}(2)} = \overline{x_1x_2...x_{t-1}\overline{0y_1y_2...y_{t-1}}\overline{1}}$$
with number $\overline{x_1x_2...x_t\overline{00...0}(2)}$ have $t$ digits 0 after $x_t$ and $y_i = 1+x_i \pmod 2$ for all $i < t$ because
$$\overline{y_1y_2...y_{t-1}}\overline{1}(2) + \overline{x_1x_2...x_{t-1}}\overline{1} = \overline{100...0}(2) \pmod{2} \text{ (t numbers 0).}$$
Thus, $s_2(kn) = 1+0+\sum_{i=1}^{t-1}(x_i+y_i) = t$ for all $k$ odd.

On the other hand, if $k$ is even then $s_2(kn) = s_2\left(\frac{k}{2^{v_2(k)}} \cdot n\right) = t$.

Thus, $s_2(kn) = t$ for all $k < n+1$.

Subcase 3. $n$ does not have the form of $2^t$ or $2^t-1$. Let $n = 2^p \cdot q$ where $q$ is odd and greater than 1, we have

* $p > 0$, let $n = \overline{x_1x_2\dots x_m00\dots0}_{(2)}$ with $p$ digits 0, so $x_m = 1$. Choose $k = \overline{100\dots01}$ with $k-1$ digits 0, so
$$nk = 2^p (\overline{x_1\dots x_m 0\dots 0}_{(2)} + \overline{x_1\dots x_m}_{(2)}) = 2^p \overline{x_1\dots x_m x_1\dots x_m}.$$
Therefore, $s_2(kn) = 2s_2(n)$, which is not satisfied.

* $p = 0$, let $n = \overline{x_1x_2\dots x_kx_{k+1}x_{k+2}\dots x_t}$, with $x_1 = 1 = x_{k+1} = 1$ and $x_{k+2} = \dots = x_t = 0$ because $t$ does not have the form of $2^t - 1$.
Choose $k = \overline{100\dots01}_{(2)}$ with $t-2$ digits 0. It follows that
$$\begin{aligned} kn &= 2^{t-1}n + n = \overline{x_1x_2\dots x_t00\dots0}_{(2)} + \overline{x_1x_2\dots x_t}_{(2)} \\ &= \overline{x_1x_2\dots x_k100\dots0x_2x_3\dots x_t}_{(2)} \end{aligned}$$
with $\overline{x_1x_2\dots x_k100\dots0x_2x_3\dots x_t}_{(2)}$ have $t-k+1$ digits 0 from $t^{th}$ in the left to $k^{th}$ in the right and $\overline{x_1x_2\dots x_t00\dots0}_{(2)}$ have $t-1$ digits 0 at the end.
Thus, $s_2(kn) = s_2(n) + x_1 + x_2 + \dots + x_k > s_2(n)$ since $x_1 = 1$, which do not satisfied $a_{kn} = a_n$ for all $k \le n$.

Hence, there are only $n=2$ and $n=2^t-1$ satisfied.

b)
Choose $m = 2^p$, so $a_m = 1 \le a_{km}$ for all positive integers $k$.

$\square$