three cases. We put for a positive real number $A$, $$\label{HAs}
H^A_s(z_1,z_2) = \sum_{\theta\in{\mathcal{O}}}  Q_{s-1}\left(1+\frac{ |(z_1+\theta)(z_2+\theta')+A|^2  }{2y_1 y_2
A}\right),$$ and denote (for $y_1y_2>A$) the Fourier expansion by $$H^A_s(z_1,z_2) = \sum_{\nu\in{\mathfrak d}^{-1}} b_s^A(\nu,y_1,y_2) e(\nu x_1+\nu'x_2).$$ Moreover, we let $R(mp^2)$ be a set of representatives for $$\begin{aligned}
&\{\lambda\in \partial_F^{-1}/a{\mathcal{O}}_F;\quad {\operatorname{N}}(\lambda\sqrt{p})\equiv mp^2 \pmod{ap}\}\\
&=\{\lambda\in {\mathcal{O}}_F/a{\mathcal{O}}_F;\quad {\operatorname{N}}(\lambda)\equiv -mp \pmod{a}\}.\end{aligned}$$ We start with the case that $a$ is positive and coprime to $p$. Here we can argue as in [@Br1]. We may write $$\begin{aligned}
\tilde\Phi^a_{mp^2} (z_1,z_2,s)&= \epsilon_p(a)\sum_{\lambda\in R(mp^2)}\sum_{\theta\in{\mathcal{O}}} Q_{s-1}\left(1+\frac{ |(z_1+\theta+\tfrac{\lambda'}{a})(z_2+\theta'+\tfrac{\lambda}{a})+\tfrac{mp}{a^2}|^2 }{ 2y_1 y_2 mp/a^2}\right)\\
&=\sum_{\nu\in\partial_F^{-1}}\epsilon_p(a)\sum_{\lambda\in R(mp^2)} e\left(\frac{{\operatorname{tr}}(\nu\lambda)}{
a}\right)b_s^{mp/a^2}(\nu,y_1,y_2) e(\nu z_1+\nu'z_2).\end{aligned}$$ When $a$ is positive and divisible by $p$, one finds in a similar way that $$\begin{aligned}
\tilde\Phi^a_{mp^2} (z_1,z_2,s)&= \sum_{\lambda\in R(mp^2)}\sum_{\theta\in{\mathcal{O}}} \epsilon_p(\tfrac{\lambda\lambda'+mp}{a}) Q_{s-1}\left(1+\frac{ |(z_1+\theta+\tfrac{\lambda'}{a})(z_2+\theta'+\tfrac{\lambda}{a})+\tfrac{mp}{a^2}|^2 }{ 2y_1 y_2 mp/a^2}\right)\\
&=\sum_{\nu\in\partial_F^{-1}}
\sum_{\lambda\in R(mp^2)} \epsilon_p(\tfrac{\lambda\lambda'+mp}{a})  e\left(\frac{{\operatorname{tr}}(\nu\lambda)}{
a}\right)b_s^{mp/a^2}(\nu,y_1,y_2) e(\nu z_1+\nu'z_2).\end{aligned}$$ If we define $$\label{tildeGa}
\tilde G_a(mp^2,\nu)=\begin{cases} \epsilon_p(a)\sum_{\lambda\in R(mp^2)}
e\left(\frac{{\operatorname{tr}}(\nu\lambda)}{ a}\right),&\text{if $p\nmid a$,}\\[1ex]
\sum_{\lambda\in R(mp^2)} \epsilon_p(\tfrac{\lambda\lambda'+mp}{a}) e\left(\frac{{\operatorname{tr}}(\nu\lambda)}{
a}\right),&\text{if $p\mid a$,}
             \end{cases}$$ we may finally write $$\label{Phim2}
\tilde\Phi_{mp^2} (z_1,z_2,s) = \tilde\Phi^0_{mp^2}(z_1,z_2,s) + 2
\sum_{\nu\in\partial_F^{-1}}\left[\sum_{a=1}^{\infty} \tilde G_a(mp^2,\nu)b_s^{mp/a^2}(\nu,y_1,y_2) \right]e(\nu
x_1+\nu'x_2).$$

For $r_1,r_2\in {\mathbb{R}}$ we briefly write $$\begin{aligned}
\alpha(r_1,r_2)&:= \max(|r_1|,|r_2|),\\
\beta(r_1,r_2)&:= \min(|r_1|,|r_2|).\end{aligned}$$

\[lphi0\] The function $\tilde\Phi^0_{mp^2}(z_1,z_2,s)$ has the Fourier expansion $$\begin{gathered}
\label{lphi0a}
\tilde\Phi_{mp^2}^0(z_1,z_2,s) =  2 \sqrt{p} \sum_{\substack{ \lambda\in\partial_F^{-1}\\ {\operatorname{N}}(\lambda)=-m/p}}
\sum_{n\geq 1} \frac{\epsilon_p(n)}{n}
 {\mathcal{I}}_{s}(2\pi n \beta(\lambda 