Given a theory $T$ and a formula $\phi(x)$ we say that they admit a $(\kappa, \lambda)$ model if there is a model $M$ such that $|M| = \kappa$ and $|\phi(M)| = \lambda$.
In all examples that I know of that do not admit $(\kappa, \lambda)$ models for some pairs (such as $(\mathbb R, \mathbb Q, <)$) there is a formula expressing that different elements realise different types over $\phi$ (such as $\forall x, y (x < y \to \exists z \in Q (x < z < y))$ for $(\mathbb R, \mathbb Q, <)$). Or one can iterate it finitely many times (such as in $(\cal P(P(\omega)), P(\omega), \omega, \in)$).
I am wondering if this is the only obstruction to admitting all pairs of cardinals. More concretely a question can be posed as follows. Assume that there is a model $M$ such that for every chain $\phi(x) = \psi_0(x), ..., \psi_n(x) = ``x=x\text{"}$ there is some $i < n$ with $\psi_{i+1}(M)$ containing infinitely many elements that realise the same type over $\psi_i(M)$. Does it follow that $T$ and $\phi$ admit a $(\kappa, \lambda)$ model for every $\kappa \ge\lambda$?