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There is a sequence $$ F_n= \begin{cases} aF_{n-1}+q^{n-2}F_{n-2},& n \text{ is even}\\ bF_{n-1}+q^{n-2}F_{n-2},& n \text{ is odd} \end{cases} $$ with the initial conditions $F_0 = 0 $, $F_1 = 1$.Here $q$ is just a smybol. The first terms of this sequence are $\{0,1,a,ab+q,\ldots\}$. The following function satisfies $F_n$ as $$ F[n] = a^{1-\xi(n)}\sum_{k=0}^{\lfloor \frac{n-1}{2} \rfloor} {n-k-1\brack k}(ab)^{\lfloor \frac{n-1}{2}\rfloor -k}q^{k^2}, $$ where ${n\brack k}$ $=$ ${\frac{[n]_q !}{[k]_q ! [n-k]_q !}}$, $[n]_q != [1]_q[2]_q[3]_q \ldots[n-1]_q [n]_q$ and $[n]_q = 1+q+q^2+\ldots q^{n-1}$. Also $ \xi(n)= \begin{cases} 0,& n \text{ is even}\\ 1,& n \text{ is odd} \end{cases} $ is a parity function and $\lfloor ,\rfloor$ is floor function. Now, I have a different sequence as

$$ C_n= \begin{cases} bC_{n-1}+q^{n-2}C_{n-2},& n \text{ is even}\\ aC_{n-1}+q^{n-2}C_{n-2},& n \text{ is odd} \end{cases} $$ with the initial conditions $C_0=2$ and $C_1=a$. The first terms of this sequence are $\{2,a,ab+2,a^2 b+a q+2a,\ldots\}$. I try the following functions, to obtain the same results with the new sequence, which are

$$ C[n] = a^{\xi(n)}\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \frac{[n]_q}{[n-k]_q}{n-k\brack k}(ab)^{\lfloor \frac{n}{2}\rfloor -k}q^{k^2}, $$

and $$ C[n] = a^{\xi(n)}\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \frac{[n]_q}{[n-k]_q}{n-k\brack k}(ab)^{\lfloor \frac{n}{2}\rfloor -k}q^{\binom{k}{2}}. $$

Unfortunately both $C[n]$ doesn't give the same results with $C_n$. How can I modify $C[n]$ to obtain elements of $C_n$?

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