Series connection of digital PT1 filters
You could have mentioned where you got that formula from. But anyway, I'll try to reconstruct it.

A first order discrete-time low pass filter has the following transfer function:

$$H(z)=\frac{1+c}{1+cz^{-1}},\qquad -1&lt;c&lt;0\tag{1}$$

where the constant $c$ determines the cut-off frequency. If you cascade $N$ of these filters, the total transfer function is

$$H_N(z)=\prod_{k=1}^NH(z)=\frac{(1+c)^N}{(1+cz^{-1})^N}\tag{2}$$

Using the Binomial Theorem this can be written as

$$H_N(z)=\frac{(1+c)^N}{\displaystyle\sum_{k=0}^N\dbinom{N}{k}c^kz^{-k}}=\frac{(1+c)^N}{\displaystyle\sum_{k=0}^Na_kz^{-k}}\tag{3}$$

with denominator coefficients

$$a_k=\dbinom{N}{k}c^k\tag{4}$$

Note that the numerator $(1+c)^N$ is just there to make sure that the scaling is correct, i.e., that the transfer function equals $1$ at $z=1$ (DC).

Following MBaz's suggestion, I add the relationship between the constant $c$ in $(1)$ and the filter's $3\,\text{dB}$ cut-off frequency $\omega_c$. For this we have to solve

$$|H(e^{j\omega_c})|^2=\frac12|H(1)|^2=\frac12\tag{5}$$

From $(1)$ we get

$$\frac{(1+c)^2}{1+2c\cdot \cos(\omega_c)+c^2}=\frac12\tag{6}$$

from which we obtain after some elementary algebra

$$c=\cos(\omega_c)-2+\sqrt{\cos^2(\omega_c)-4\cos(\omega_c)+3}\tag{7}$$