Wallis concluded that
$$\frac{\pi}{2} = \frac{4}{3}\cdot\frac{16}{15}\cdot\frac{36}{35}\cdot\frac{64}{63}\cdot....$$
How can one define the right hand side recursively? After a few hours all I could muster was this:
$$\begin{cases} a_1=3. \\ a_n=\frac{(a_{n-1}+2(n-1)+1)^2}{(a_{n-1}+2(n-1)+1)^2-1}.\end{cases}$$
But this only seems to hold true for the first two factors.
It seems like: $$ \frac \pi 2 = \frac 43 \frac{16}{15} \frac{36}{35}\frac{64}{63} ... = \frac{2^2}{2^2-1} \frac{4^2}{4^2-1}\frac{6^2}{6^2 - 1} ... $$
The key point here is to note that $a_n = \frac{4n^2}{4n^2 - 1}$, and $a_1 = \frac 43$.
So, all we need to do, is: $$ \frac{a_n}{a_{n-1}} = \frac{\frac{4n^2}{(4n^2 - 1)}}{\frac{4(n-1)^2}{4(n-1)^2 - 1}} = \frac{4n^2 (4(n-1)^2 - 1)}{4(n-1)^2 (4n^2 - 1)} = \frac{n^2(2n-3)}{1 + n^2 (2n-3)} $$
(The last formula came after simplification)
Hence, we can write $a_1 = \frac 43$ and $a_n = a_{n-1} \frac{n^2(2n-3)}{1+n^2(2n-3)}$.
To check: If $n=2$, then $a_2 = \frac 43 \frac{4(1)}{1+4} = \frac{16}{15}$.
$n=3 \implies a_3 = \frac{16}{15} \frac{9(3)}{1+9(3)} = \frac{432}{420} = \frac{36}{35}$.
$n=4 \implies a_4 = \frac{36}{35} \frac{16(5)}{1+16(5)} = \frac{2880}{2835} = \frac{64}{63}$.
You can prove that this works by induction. I leave you to do that.