A Quadratic Recurrence Relation that Computes $\pi$

The Recurrence Relation

Define the sequence $ A(x,y) = \{ a_0, a_1, a_2, … \} $ such that

\[a_n = \begin{cases} x & n = 0 \\ y & n = 1 \\ 6a_{n-1} + (2n-3)^2a_{n-2} &\text{Otherwise} \end{cases}\]

Now define $ P = \{p_0, p_1, p_2, …\} = A(1,3) $ and $ Q = \{q_0, q_1, q_2, …\} = A(0,1) $. In other words, the sequence $P$ and $Q$ are generated from the same recurrence relation, but have different starting values.

Now, I posit that

\[\lim_{n \rightarrow \infty} \cfrac{p_n}{q_n} = \pi\]

Here's a code version:

Why it works

$P$ and $Q$ are the coefficients of the n-th term of the continued fraction:

\[\pi = \cfrac{1^2}{6+\cfrac{3^2}{6+\cfrac{5^2}{6+\cfrac{7^2}{...}}}}\]

Which is a well known approximation of $\pi$.

Detailed proof

Lemma:

\[\cfrac{xp_{n-1}+\left(2n-3\right)^{2}p_{n-2}}{xq_{n-1}+\left(2n-3\right)^{2}q_{n-2}}=\cfrac{\left[6+\cfrac{(2n-3)^{2}}{x}\right]p_{n-2}+(2n-5)^{2}p_{n-3}}{\left[6+\cfrac{(2n-3)^{2}}{x}\right]q_{n-2}+\left(2n-5\right)^{2}q_{n-3}}\]

Proof:

\[\begin{aligned} \cfrac{xp_{n-1}+\left(2n-3\right)^{2}p_{n-2}}{xq_{n-1}+\left(2n-3\right)^{2}q_{n-2}} &= \cfrac{x\left[6p_{n-2}+\left(2n-5\right)^{2}p_{n-3}\right]+\left(2n-3\right)^{2}p_{n-2}}{x\left[6q_{n-2}+\left(2n-5\right)^{2}q_{n-3}\right]+\left(2n-3\right)^{2}q_{n-2}}& \\ &= \cfrac{\left[6x+\left(2n-3\right)^{2}\right]p_{n-2}+x\left(2n-5\right)^{2}p_{n-3}}{\left[6x+\left(2n-3\right)^{2}\right]q_{n-2}+x\left(2n-5\right)^{2}q_{n-3}}& \\ &= \cfrac{\left[6+\cfrac{(2n-3)^{2}}{x}\right]p_{n-2}+(2n-5)^{2}p_{n-3}}{\left[6+\cfrac{(2n-3)^{2}}{x}\right]q_{n-2}+\left(2n-5\right)^{2}q_{n-3}} &\quad \blacksquare \end{aligned}\]

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Using the lemma:

\[\begin{aligned} \cfrac{p_n}{q_n} &= \cfrac{6p_{n-1}+\left(2n-3\right)^{2}p_{n-2}}{6q_{n-1}+\left(2n-3\right)^{2}q_{n-2}} \\ &=\cfrac{\left[6+\cfrac{(2n-3)^{2}}{6}\right]p_{n-2}+\left(2n-5\right)^{2}p_{n-3}}{\left[6+\cfrac{(2n-3)^{2}}{6}\right]q_{n-2}+\left(2n-5\right)^{2}q_{n-3}} \\ &= \cfrac{\left[6+\cfrac{(2n-5)^{2}}{6+\cfrac{(2n-3)^{2}}{6}}\right]p_{n-3}+\left(2n-7\right)^{2}p_{n-4}}{\left[6+\cfrac{(2n-5)^{2}}{6+\cfrac{(2n-3)^{2}}{6}}\right]q_{n-3}+\left(2n-7\right)^{2}q_{n-4}}\\ & ... \\ &=\cfrac{F_{n}p_{1}+1^{2}p_{0}}{F_{n}q_{1}+1^{2}q_{0}}=\cfrac{3F_{n}+1}{F_{n}}=3+\cfrac{1}{F_{n}} \end{aligned}\]

Where I've applied the lemma repeatedly and

\[\displaystyle F_n=6+\cfrac{3^2}{6+\cfrac{5^2}{6+\cfrac{7^2}{...+\cfrac{...}{6+\cfrac{\left(2n-3\right)^2}{6}}}}}\]

Hence

\[\lim_{n \rightarrow \infty} \frac{p_n}{q_n} = \lim_{n \rightarrow \infty} 3+\frac{1}{F_n}=3+\cfrac{1^2}{6+\cfrac{3^2}{6+\cfrac{5^2}{6+\cfrac{7^2}{...}}}} = \pi \quad \blacksquare\]