Balmer and Paschen Series
$$E_n = Z^2\mathcal{R}\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

You can easily verify that if you replace $Z$ with $2Z$, then the energy of the corresponding transition will be four times the original energy. Now if you replace both $n_1$ and $n_2$ with $2n_1$ and $2n_2$, then you can take out a factor of $1/4$; so the transition $2n_1 \leftarrow 2n_2$ in an ion with nuclear charge $2Z$ occurs at the same energy/wavelength/wavenumber as the transition $n_1 \leftarrow n_2$ in an ion with nuclear charge $Z$.

Same applies for any other factor $k$: the $1 \leftarrow 2$ transition in hydrogen occurs at the same wavelength as the $4 \leftarrow 8$ transition in $\ce{Be^3+}$. Obviously this discussion ignores the variation in the Rydberg constants for different nuclei.