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?
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}