Earlier today I was fooling around with a recurrence relation that I thought behaved peculiarly, namely $$ X_k=\frac{k}{X_{k-1}} \qquad k\geq 0 \\ X_0=1 $$ After plotting its first 100 terms, I noticed that it was essentially two "interwoven" square-root curves. I then tried squaring the terms to confirm that I got straight lines, which I roughly did (not exactly straight, but approaching straight) which implies that whatever my piecewise interwoven square-root closed-form may be, it was only the limit of the recurrence relation as $k\rightarrow\infty$. I toyed around trying to find the coefficient of $k$, eventually finding this approximation: $$ X_k:\approx\begin{cases} \sqrt{1+\frac{\pi}{2}k} & k\text{ even} \\ \sqrt{\frac{2}{\pi}k} & k\text{ odd} \end{cases} $$
This truly seems (by inspection) to be the best approximation I can make for $X_k$ in the hour I've been working on this, but I was wondering how I could go about proving this kind of relationship. It's curious to me that pi gets involved at all, my only hypothesis so far is that some kind of inverse-trigonometric function is used somewhere in the proof.
We can solve that recurrence relation, let $ n\in\mathbb{N} $, and let $ k $ be a positive integer : \begin{aligned} X_{2k}=\frac{2k}{X_{2k-1}}&=\frac{2k}{2k-1}X_{2k-2}\\ \Longrightarrow\prod_{k=1}^{n}{\frac{X_{2k}}{X_{2k-2}}}&=\prod_{k=1}^{n}{\frac{2k}{2k-1}}\\ \iff \ \ \ \ \ \ \ \frac{X_{2n}}{X_{0}}&=\frac{2^{2n}}{\binom{2n}{n}}\end{aligned}
\begin{aligned} X_{2k+1}=\frac{2k+1}{X_{2k}}&=\frac{2k+1}{2k}X_{2k-1}\\ \Longrightarrow\prod_{k=1}^{n}{\frac{X_{2k+1}}{X_{2k-1}}}&=\prod_{k=1}^{n}{\frac{2k+1}{2k}}\\ \iff \ \ \ \ \frac{X_{2n+1}}{X_{1}}&=\frac{2n+1}{2^{2n}}\binom{2n}{n}\end{aligned}
Thus, $$ \left(\forall n\in\mathbb{N}\right),\ X_{n}=\left\lbrace\begin{aligned}\frac{2^{n}}{\binom{n}{\frac{n}{2}}} \ \ \ \ \ \ \ \ \ \ \ \ &\textrm{If }n\textrm{ is even} \\ \frac{n}{2^{n-1}}\binom{n-1}{\frac{n-1}{2}} \ \ \ \ &\textrm{If }n\textrm{ is odd}\end{aligned}\right. $$