Fourier coefficients of discrete difference of a square wave
It's quite straightforward to express the Fourier coefficients of a shifted version of a function in terms of the Fourier coefficients of the original function.
If the Fourier series of a $T$-periodic function $f(t)$ is given by
$$f(t)=\sum_{k=-\infty}^{\infty}c_ke^{j2\pi kt/T}\tag{1}$$
then it follows that the Fourier series of a shifted version must be
$$\begin{align}f(t-t_0)&amp;=\sum_{k=-\infty}^{\infty}c_ke^{j2\pi k(t-t_0)/T}\\&amp;=\sum_{k=-\infty}^{\infty}c_ke^{-j2\pi kt_0/T}e^{j2\pi kt/T}\\&amp;=\sum_{k=-\infty}^{\infty}\tilde{c}_ke^{j2\pi kt/T}\tag{2}\end{align}$$
Hence, the Fourier coefficients of $f(t-t_0)$ are given by
$$\tilde{c}_k=c_ke^{-j2\pi kt_0/T}\tag{3}$$
And, by linearity, the Fourier series coefficients of $g(t)=f(t)-f(t-t_0)$ are
$$d_k=c_k-\tilde{c}_k=c_k\big(1-e^{-j2\pi kt_0/T}\big)\tag{4}$$

For periodic discrete-time sequences we can use the Discrete Fourier Tansform (DFT), which is just a discrete Fourier series representation:
$$x[n]=\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j2\pi nk/N}\tag{5}$$
with
$$X[k]=\sum_{k=0}^{N-1}x[n]e^{-j2\pi nk/N}\tag{6}$$
In a completely analogous manner to the continuous-time case, we can derive an expression for the DFT coefficients of the sequence $y[n]=x[n]-x[n-n_0]$, $n_0\in\mathbb{Z}$:
$$Y[k]=X[k]\big(1-e^{-j2\pi n_0k/N}\big)\tag{7}$$