I wanted to compute the Hecke eigenvalues of $\Theta^k$ for $k=3,5,7$ where
$\Theta(z)=\sum\limits_{n\in\mathbb{Z}}q^{n^2}.$
We know that if
$T(p^2)\Theta^k=\sum\limits_{n=0}^{\infty}b_nq^n$ then
$b_n=a_{p^2n}+\left(\frac{-1}{p}\right)^{\lambda}\left(\frac{n}{p}\right)p^{\lambda-1}a_n+p^{k-2}a_{\frac{n}{p^2}}$
where $\Theta^k(z)=\sum\limits_{n=0}^{\infty}a_nq^n$ and $\lambda=\frac{k-1}{2}$. For $\Theta^3,$ I know the eigenvalue to be $1+p$ but I don't know how to verify it.
$\Theta^k$ has an explicit Fourier expansion for $k=3,5,7$, so you could work it out directly, but it is simpler to go via the Shimura lift: up to proportionality, the respective lifts for $t=1$ and $k=3,5,7$ are $E_2(\tau)-2E_2(2\tau)$, $E_4(\tau)$, and $E_6(\tau)+8E_6(2\tau)$, so for $p\ne2$ the Hecke eigenvalues (which are the same by Shimura's theorem) are $1+p$, $1+p^3$, and $1+p^5$.
Edit: for other squarefree $t$, not the same linear combinations of $E_{k-1}(\tau)$ and $E_{k-1}(2\tau)$, but the result is the same for $p\ne2$.
