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First I want to prove that \begin{align} a_{1}(0)& = 1\\ a_{1}(0)& = (1+\operatorname{wt}(n))a_{1}(\left\lfloor\frac{n}{2}\right\rfloor,0) \end{align}

First I want to prove that \begin{align} a_{1}(0)& = 1\\ a_{1}(n)& = (1+\operatorname{wt}(n))a_{1}(\left\lfloor\frac{n}{2}\right\rfloor) \end{align}

Through this we add all possible ways to finish on any cell, which is what $a_1(n)$ counts. It remains only to add that operation above is the same as taking values of the $n$-th row of A295989, which is also the same as taking $j$ such that $$\binom{n}{j}\operatorname{mod} 2=1$$ QED

Through this we add all possible ways to finish on any cell, which is what $a_1(n)$ counts. It remains only to add that operation above is the same as taking values of the $n$-th row of A295989, which is also the same as taking $j$ such that $$\binom{n}{j}\operatorname{mod} 2=1$$ QED

Here it is obvious that we cannot put the $1$st cell on the right in front of the $1$st cell on the left, since by the condition biased rook on a white cell moves only to the left. As a result, we have $(k_1+k_2)!(k_2+1)^2$ open tours by a biased rook of the first kind.

$$a_1(2n)=(\sum\limits_{i=1}^{m}k_i)!\prod\limits_{p=1}^{m}(1+\sum\limits_{q=p}^{m}k_q)^{j_{p}+[p=1]}=(1+\sum\limits_{q=1}^{m}k_q)(\sum\limits_{i=1}^{m}k_i)!\prod\limits_{p=1}^{m}(1+\sum\limits_{q=p}^{m}k_q)^{j_p}$$ Here $$\sum\limits_{q=1}^{m}k_q=\operatorname{wt}(n)$$ so $$a_1(2n)=(1+\operatorname{wt}(n))a(n)=(1+\operatorname{wt}(2n))a(n)$$ Don't forget that for $a(n)$ we are working with a binary representation of $2n$, then for $a(2n+1)$ we have 2 options: $$\underbrace{1\cdots1}_{k_m}\underbrace{0\cdots0}_{j_m}\cdots\underbrace{1\cdots1}_{k_2}\underbrace{0\cdots0}_{j_2}\underbrace{1\cdots1}_{k_1}1\underbrace{0}_{j_1}$$ $$\underbrace{1\cdots1}_{k_m}\underbrace{0\cdots0}_{j_m}\cdots\underbrace{1\cdots1}_{k_2}\underbrace{0\cdots0}_{j_2}\underbrace{1\cdots1}_{k_1}\underbrace{0\cdots0}_{j_{1}-1}1\underbrace{0}_{j_{0}}$$ Then we have

$$a_1(2n+1)=(\sum\limits_{i=1}^{m}k_i+[i=1])!\prod\limits_{p=1}^{m}(1+\sum\limits_{q=p}^{m}k_q+[q=1])^{j_{p}}=(2+\sum\limits_{q=1}^{m}k_q)(1+\sum\limits_{i=1}^{m}k_i)!\prod\limits_{p=2}^{m}(1+\sum\limits_{q=p}^{m}k_q)^{j_p}=(2+\sum\limits_{q=1}^{m}k_q)(\sum\limits_{i=1}^{m}k_i)!\prod\limits_{p=1}^{m}(1+\sum\limits_{q=p}^{m}k_q)^{j_p}=(2+\operatorname{wt}(n))a(n)=(1+\operatorname{wt}(2n+1))a(n)$$ and $$a_1(2n+1)=(1+\sum\limits_{i=1}^{m}k_i)!\prod\limits_{p=0}^{m}(1+\sum\limits_{q=p}^{m}k_q+[q=0])^{j_{p}-[p=1]}=(1+\sum\limits_{q=0}^{m}k_q+[q=0])(1+\sum\limits_{i=1}^{m}k_i)(\sum\limits_{i=1}^{m}k_i)!\prod\limits_{p=1}^{m}(1+\sum\limits_{q=p}^{m}k_q)^{j_p-[p=1]}=(2+\sum\limits_{q=1}^{m}k_q)(\sum\limits_{i=1}^{m}k_i)!\prod\limits_{p=1}^{m}(1+\sum\limits_{q=p}^{m}k_q)^{j_p}=(2+\operatorname{wt}(n))a(n)=(1+\operatorname{wt}(2n+1))a(n)$$ QED

Here it is obvious that we cannot put the $1$st cell on the right after of the $1$st cell on the left, since by the condition biased rook on a white cell moves only to the left. As a result, we have $(k_1+k_2)!(k_2+1)^2$ open tours by a biased rook of the first kind.

$$a_1(2n)=(\sum\limits_{i=1}^{m}k_i)!\prod\limits_{p=1}^{m}(1+\sum\limits_{q=p}^{m}k_q)^{j_{p}+[p=1]}=(1+\sum\limits_{q=1}^{m}k_q)(\sum\limits_{i=1}^{m}k_i)!\prod\limits_{p=1}^{m}(1+\sum\limits_{q=p}^{m}k_q)^{j_p}$$ Here $$\sum\limits_{q=1}^{m}k_q=\operatorname{wt}(n)$$ so $$a_1(2n)=(1+\operatorname{wt}(n))a(n)=(1+\operatorname{wt}(2n))a(n)$$ Don't forget that for $a(n)$ we are working with a binary representation of $2n$, then for $a(2n+1)$ we have 2 options: $$\underbrace{1\cdots1}_{k_m}\underbrace{0\cdots0}_{j_m}\cdots\underbrace{1\cdots1}_{k_2}\underbrace{0\cdots0}_{j_2}\underbrace{1\cdots1}_{k_1}1\underbrace{0}_{j_1}$$ $$\underbrace{1\cdots1}_{k_m}\underbrace{0\cdots0}_{j_m}\cdots\underbrace{1\cdots1}_{k_2}\underbrace{0\cdots0}_{j_2}\underbrace{1\cdots1}_{k_1}\underbrace{0\cdots0}_{j_{1}-1}\underbrace{1}_{k_{0}}\underbrace{0}_{j_{0}}$$ Then we have

$$a_1(2n+1)=(\sum\limits_{i=1}^{m}k_i+[i=1])!\prod\limits_{p=1}^{m}(1+\sum\limits_{q=p}^{m}k_q+[q=1])^{j_{p}}=(2+\sum\limits_{q=1}^{m}k_q)(1+\sum\limits_{i=1}^{m}k_i)!\prod\limits_{p=2}^{m}(1+\sum\limits_{q=p}^{m}k_q)^{j_p}=(2+\sum\limits_{q=1}^{m}k_q)(\sum\limits_{i=1}^{m}k_i)!\prod\limits_{p=1}^{m}(1+\sum\limits_{q=p}^{m}k_q)^{j_p}=(2+\operatorname{wt}(n))a(n)=(1+\operatorname{wt}(2n+1))a(n)$$ and $$a_1(2n+1)=(1+\sum\limits_{i=1}^{m}k_i)!\prod\limits_{p=0}^{m}(1+\sum\limits_{q=p}^{m}k_q)^{j_{p}-[p=1]}=(1+\sum\limits_{q=0}^{m}k_q)(1+\sum\limits_{i=1}^{m}k_i)(\sum\limits_{i=1}^{m}k_i)!\prod\limits_{p=1}^{m}(1+\sum\limits_{q=p}^{m}k_q)^{j_p-[p=1]}=(2+\sum\limits_{q=1}^{m}k_q)(\sum\limits_{i=1}^{m}k_i)!\prod\limits_{p=1}^{m}(1+\sum\limits_{q=p}^{m}k_q)^{j_p}=(2+\operatorname{wt}(n))a(n)=(1+\operatorname{wt}(2n+1))a(n)$$ QED