How to find the general formula of the recurrence sequence? Let two sequences $\{a_n\},\{b_n\}$ such that
$a_{2n-1}=a_{2n}=a_{2n-2}+\frac{1}{b_{2n-2}}$,
$b_{2n}=b_{2n+1}=b_{2n-1}+\frac{1}{a_{2n-1}}$,
$a_1=b_1=1$.
With the help of OEIS, I find a solution:
$a_{2n-1}=a_{2n}=\frac{(2n-1)!!}{(2n-2)!!}$,
$b_{2n}=b_{2n+1}=\frac{(2n)!!}{(2n-1)!!}$,
where $(2n-1)!!=1\times3\times\cdots\times(2n-1)$ and $(2n)!!=2\times4\times\cdots\times(2n)$.
How can I get it without mathematical induction?
 A: I only manage to decouple the recurrence relations: We have
\begin{array}{rlcrl}
a_{2n} &= a_{2n-2} + 1/b_{2n-2} \quad (*) & \quad &
b_{2n+1} &= b_{2n-1} + 1/a_{2n-1} \quad (**)\\
a_{2n} &= a_{2n-1} & \quad &
b_{2n+1} &= b_{2n} \\
a_1 &= 1 & \quad &
b_1 &= 1
\end{array}
then solving equation $(**)$ for $a_{2n-1}$ in terms of the $b$ (and similar solving $(*)$ for $b_{2n-2}$ in terms of the $a$), and some index translation, we get
$$
a_{2n} = a_{2n-1} = \frac{1}{b_{2n+1}-b_{2n-1}} \quad\quad
a_{2n-2} = a_{2n-3} = \frac{1}{b_{2n-1}-b_{2n-3}} \\
b_{2n-1} = b_{2n-2} = \frac{1}{a_{2n} - a_{2n-2}} \quad\quad
b_{2n+1} = b_{2n} = \frac{1}{a_{2n+2} - a_{2n}} \\
$$
and inserting back, plus changing $b_{2n-2}\to b_{2n-1}$ and $a_{2n-1} \to a_{2n}$, we get
$$
\frac{1}{b_{2n+1}-b_{2n-1}} 
= \frac{1}{b_{2n-1}-b_{2n-3}} + \frac{1}{b_{2n-1}} \\
\frac{1}{a_{2n+2}-a_{2n}} 
= \frac{1}{a_{2n}-a_{2n-2}} + \frac{1}{a_{2n}}
$$
A recurrence relation
$$
\frac{1}{c_{k+1} - c_k} 
= \frac{1}{c_k-c_{k-1}} + \frac{1}{c_k} 
$$
is non-linear. We can try to transform it:
$$
\frac{1}{c_k} = -\frac{\Delta(c_k) - \Delta(c_{k-1})}{\Delta(c_k) \Delta(c_{k-1})} \iff \\
\frac{\Delta(c_k)}{c_k} 
= -\frac{\Delta\Delta(c_{k-1})}{\Delta(c_{k-1})}
$$
or
$$
\frac{1}{c_{k+1} - c_k} 
= \frac{2c_k-c_{k-1}}{c_k(c_k-c_{k-1})} \iff \\
$$
\begin{align}
c_{k+1} 
&= c_k + \frac{c_k(c_k-c_{k-1})}{2c_k-c_{k-1}} \\
&= c_k \left( 1 + \frac{c_k-c_{k-1}}{2c_k-c_{k-1}} \right) \\
&= c_k \frac{3c_k-2c_{k-1}}{2c_k-c_{k-1}} \\
\end{align}
