This is a result I accidentally derived while reviewing my notes from The Feynman Lectures on Physics, Vol I, Chapter 48 https://www.feynmanlectures.caltech.edu/I_48.html
Begin with Feynman's eq. 48.22
$$\left(\frac{\hbar\omega}{\mathrm{c}}\right)^{2}-\left(\hbar k\right)^{2}=\left(m\mathrm{c}\right)^{2}.$$
Solve for $\omega$
$$\omega=\mathrm{c}\sqrt{k^{2}+\left(\frac{m\mathrm{c}}{\hbar}\right)^{2}}.$$
The phase velocity is
$$v_{p}=\frac{\omega}{k}=\mathrm{c}\sqrt{1+\left(\frac{m\mathrm{c}}{\hbar k}\right)^{2}}=\mathrm{c}\sqrt{1+\left(\frac{m\mathrm{c}}{p}\right)^{2}},$$
where I have used $\hbar k=p$ in the last expression. Setting $\mathrm{c}=1$ for simplicity, and using the standard hyperbolic trigonometric notation gives
$$\begin{aligned}\frac{v_{p}}{\mathrm{c}}=\beta_{p}=&\sqrt{1+\left(\frac{m}{p}\right)^{2}}\\ =&\sqrt{1+\left(\frac{1}{\sinh\theta}\right)^{2}}\\ =&\sqrt{\frac{\sinh^{2}\theta+1}{\sinh^{2}\theta}}\\ =&\frac{1}{\tanh\theta}=\beta^{-1}\\ =&\left(\frac{v}{\mathrm{c}}\right)^{-1}. \end{aligned}$$
So the phase velocity is $\mathrm{c}^2$ over the relativistic velocity
$$v_{p}=\frac{\mathrm{c}^2}{v}.$$
Is this correct? And if it is, what does it mean?
I'm happy to follow the links suggested in the comments (time permitting), but perhaps there is a "short answer" to my question?
