Formula for designing Lynn's low pass filter For the second order Lynn's low pass filter, the general form of the transfer function is:
$$
H(z)=\frac{(1−z^{-m})^2}{ (1−z^{-1})^2 }
$$
where $m$ is a positive integer. The gain for this is $m^2$, time delay is $m-1$ samples and the nominal frequency corresponding to the location of the lowest frequency zero is $f_l = f_s /m$. The frequency response is given by:
$$
|H(e^{j\omega T})|=\frac{\sin^2(m\omega /2f_s)}{\sin^2(\omega /2f_s)}
$$
At the 3 dB cutoff frequency, for unit gain we have:
$$
|H(e^{j\omega_c T})|=\frac{\sin^2(m\omega_c /2f_s)}{m^2\sin^2(\omega_c /2f_s)} \\ \therefore  \frac{1}{\sqrt2}=\frac{\sin^2(m\pi f_c /f_s)}{m^2\sin^2(\pi f_c /f_s)}
$$
Now, to design the filter for our required cutoff frequency $f_c$ and sampling frequency $f_s$, we need to find the value of $m$ corresponding to these inputs.
So how can we get this equation to the form $m = f(f_c, f_s)$, so that we can avoid getting the $m$ value by trial-and-error or numerical methods? Or is there some other quick approximation formula that we can use?
Just analyzing your equation, let $a = f_c / f_s$
Also we know that for $m = k/(2a)$ it reduces to $1/(m^2 sin(\pi a))$ for odd $k$ and $0$ for even $k$.
$$\lim_{m \to 0} \frac{sin^2( \pi m a)}{m^2 sin(\pi a)} = \frac{a^2 \pi^2}{sin(\pi a)} = \frac{1}{\sqrt{2}}$$
Even for this case, determining $a$ is not an easy task. I computed numerically the curve defined by this equation.