Parseval's theorem states that for two sequences $a,b \in \ell^2(\mathbb{Z}) $, we have
$$\sum_{n \in \mathbb{Z} }a(n)\overline{b(n)} = \int_0^{2\pi} A(\omega)\overline{B(\omega)} \ d\omega$$
where $A(\omega), B(\omega)$ are the Fourier series of $a$ and $b$ respectively, that is $A(\omega) = \sum_{n \in \mathbb{Z}} a(n)e^{in\omega} $. (And yes the equality I wrote may not be exactly true and may require normalization but this is irrelevant). Is there a generalization of this if we restrict the sum on the left to be over $c+d\mathbb{Z} $ for some non-zero integers $c$ and $d$. I am also interested in the sum being as simple as possible. I am actually working on simplifying the case where $c = 0$ and $d = 2^j\mathbb{Z}$ for some particular $a,b$ but I'm wondering if there's a generalization. In that case, what I got was that \begin{align} \sum_{n \in \mathbb{Z}}a(2^jn)b(2^jn) & = \int_0^{2\pi} \bigg(\frac{1}{2^j}\sum_{m=0}^{2^j -1} A(\omega + \frac{2\pi m}{2^j}) \bigg)\bigg(\frac{1}{2^j} \sum_{n=0}^{2^j -1} B(\omega + \frac{2\pi n}{2^j}) \bigg) \\ \text{By periodicity} \ \ & = \frac{1}{2^j}\int_0^{2\pi/2^j } \bigg(\sum_{m=0}^{2^j -1} A(\omega + \frac{2\pi m}{2^j}) \bigg)\bigg( \sum_{n=0}^{2^j -1} B(\omega + \frac{2\pi n}{2^j}) \bigg) \end{align}
I personally think this formula is a bit ugly and would appreciate a simpler looking generalization of Parsevals. So again, I am wondering if there is a simple formula for
$$\sum_{n \in \mathbb{Z}}a(c+dn)b(c+dn) $$ on the Fourier domain for non-zero integers $c,d$.