Maximum entropy probability distribution among Solomonoff priors
All versions of Solomonoff's prior have infinite entropy.
We supply a proof.
Let $m: \mathbb{Z}_+ \to [0,1]$ be Solomonoff's prior, where inputs are
encoded as integers.
Then there is $c&gt;0$ such that
$$m(n) \geq c\frac{1}{n (\log n)^2},$$
Let $S$ be the set of all $n$ such that
$$m(n) \geq \frac{1}{n}.$$
Since $\sum m(n)$ is convergent,
$$\sum_{s \in S} \frac{1}{s} \leq \sum_n m(n) &lt; \infty.$$

The entropy of $m$ is
$$
\begin{align*}
\sum_n m(n) \log(1/m(n)) &amp;\geq \sum_n c\frac{1}{n (\log n)^2} \log(1/m(n)) \\
&amp;\geq \sum_{n \not\in S} c\frac{1}{n (\log n)^2} \log(1/m(n)) \\
&amp;\geq \sum_{n \not\in S} c\frac{1}{n \log n}.
\end{align*}
$$

Now
$$
\infty = \sum_{n} c\frac{1}{n \log n} =
\sum_{n \in S} c\frac{1}{n \log n} +
\sum_{n \not\in S} c\frac{1}{n \log n},
$$
and
$$\sum_{s \in S} \frac{1}{s \log s} \leq \sum_{s \in S} \frac{1}{s} &lt; \infty,$$
so
$$
\sum_{n \not\in S} c\frac{1}{n (\log n)} = \infty.
$$