What is known about the consistency of $2^{\aleph_\alpha} = \aleph_{\alpha+\gamma}$ for all $\alpha$?

Question

For $\gamma$ an ordinal, let “$H_\gamma$” be the statement:

For all ordinals $\alpha$, we have $2^{\aleph_\alpha} = \aleph_{\alpha+\gamma}$.

So clearly $H_0$ is false, and so is $H_\omega$; in fact, $H_\gamma$ implies that $\gamma$ is successor (because otherwise $\operatorname{cf}\gamma = \omega_\alpha$, say, and $\aleph_{\alpha+\gamma} = 2^{\aleph_\alpha} = (2^{\aleph_\alpha})^{\aleph_\alpha} = (\aleph_{\alpha+\gamma})^{\aleph_\alpha} = (\aleph_{\alpha+\gamma})^{\operatorname{cf}\aleph_{\alpha+\gamma}}$ gives a contradiction).

On the other hand, $H_1$ is precisely the generalized continuum hypothesis (GCH) and it is consistent relative to ZFC.

I understand from this question and this one that $H_2$ is known to be consistent relative to certain large cardinal assumptions, and perhaps even $H_k$ for any concrete $k < \omega$.

What else, if anything, is known about the consistency of the various $H_\gamma$? Might we perhaps construct¹ a successor ordinal $\gamma$ for which $H_\gamma$ is demonstrably false?

1. Admittedly, I don't know how to phrase this question properly, because clearly “the smallest successor ordinal for which $H_\gamma$ is false” is a definable successor ordinal for which $H_\gamma$ is provably false, which is clearly not what I'm asking about. But a proof in ZFC that $H_{\omega+1}$ is false (say) would be a good answer to my question.

Answer 1

1. By a result of Patai, $\gamma$ should be finite (this is exercise 5.15 in Jech's book).

2. For any finite $n>0, H_n$ is consistent, see Merimovich's paper A power function with a fixed finite gap everywhere.

---

For completeness and since not everyone has Jech's book at hand, here's how the proof of (1) goes (note that Jech uses "$\beta$" in place of "$\gamma$"). Suppose $\gamma\ge \omega$ is such that $2^{\aleph_\eta}=\aleph_{\eta+\gamma}$ for all $\eta$. Letting $\alpha$ be minimal such that $\alpha+\gamma>\gamma$ we have $0<\alpha\le\gamma$ and $\alpha$ is a limit. Consider $\kappa=\aleph_{\alpha\cdot 2}$. This $\kappa$ is obviously singular and by choice of $\alpha$ we have $$2^{\aleph_{\alpha+\xi}}=\aleph_{\alpha+\xi+\gamma}=\aleph_{\alpha+\gamma}$$ for each $\xi<\alpha$. From this we get $$2^\kappa=\aleph_{\alpha+\gamma}$$ (more generally, if $\lambda$ is singular and $\mu=\max\{2^\theta:\theta<\lambda\}$ exists then $2^\lambda=\mu$). On the other hand, we have $$2^\kappa=\aleph_{\alpha\cdot 2+\gamma}=\aleph_{\alpha+(\alpha+\gamma)}>\aleph_{\alpha+\gamma}$$ again by choice of $\alpha$. So we have a contradiction.