Let $\pi$ be a finitely presented group and let $\phi:\pi\to \Bbb{Z}$ be an epimorphism. Given $n\in \Bbb{N}$ we denote by $\pi_n$ the kernel of $\pi\xrightarrow{\phi} \Bbb{Z}\to \Bbb{Z}/n$ and given a group $\Gamma$ we denote by $\operatorname{rank}(\Gamma)$ its rank, i.e. the size of a minimal generating set.

The rank gradient of $(\pi,\phi)$ is then defined as $ \liminf \frac{\operatorname{rank}(\pi_n)}{n}.$ This definition, I think, basically goes back to Marc Lackenby. If $\ker(\phi)$ is finitely generated, then $\mbox{rank}(\pi_n)\leq \mbox{rank}(\ker(\phi))+1,$ i.e. the rank gradient is zero. Now suppose that $(\pi,\phi)$ is represented by an ascending HNN-extension, i.e. there exists an isomorphism $\pi\cong \langle A,t|tAt^{-1}=\varphi(A)\rangle$ where $\varphi$ is a monomorphism and such that $\phi$ is given by the obvious surjection on the HNN-extension onto $\Bbb{Z}$. In this case we also have $\operatorname{rank}(\pi_n)\leq \mbox{rank}(A)+1,$ i.e. the rank gradient is zero. The same evidently also holds for descending HNN-extensions.

My question is now, whether there exist such pairs $(\pi,\phi)$ where the ranks of the groups $\pi_n$ are not bounded but for which the rank gradient is zero.