Since the $\sf GCH$ cannot fail everywhere everyway (see here), the question here is if it can fail everywhere in every finite manner, that if we have a strictly increasing function $f$ on the ordinals at finite intervals $\geq 2 $, then we can have: $$\forall\alpha: 2^{\aleph_\alpha}= \aleph_{f(\alpha)}$$
Formally:
$$ \exists M: M \equiv \operatorname {CTM}(\mathsf {ZF}) \land \forall f \subseteq M \ \big{(}\\f: \operatorname {Ord}^M \to \operatorname {Ord}^M \land \forall \alpha \forall \beta \, ( \beta > \alpha \to f(\beta) > f(\alpha) ) \\ \land \forall \alpha \exists n \in \mathbb N_2 \, (f(\alpha)=\alpha+n) \\ \implies \\ \exists N : N \equiv \operatorname {CTM}(\mathsf {ZF}) \land \operatorname {Ord}^N =\operatorname {Ord}^M \\ \land (N \models \forall \alpha: 2^{\aleph_\alpha}= \aleph_{f(\alpha)})\big{)} $$
where "$\equiv\operatorname {CTM}(\mathsf {ZF})$" means "is a countable transitive model of ZF"; $\mathbb N_2= \mathbb N \setminus \{0,1\}$.
Is this possible in $\sf ZFC$?
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