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The quadratic irrational $\frac{2221564096+283748\sqrt{462}}{491993569}$ is known as Freiman's constant and arises in the theory of continued fractions. I'm curious as to its simple continued fraction expansion, which, of course, is eventually periodic. Specifically, I'm wondering what the length $k$ of the initial part is, as well as the length $n$ of its period. With Mathematica, I compute the first 15000 terms and don't see a repetition yet. Here is the Mathematica code:

ContinuedFraction[(2221564096+283748*Sqrt[462])/491993569,15000]

The continued fraction expansion is $[4;1,1,8,2,14,2,2,1,12,117,1,1,7,16,\ldots]$, but that sequence does not appear in the OEIS. I have a suspicion that $k$ and $n$ are very large. Is computing them feasible? If not, can one provide estimates?

One can ask, more naturally, for the length of the period of an $\alpha \in \mathbb{Q}(\sqrt{462})$ with Markov constant equal to Freiman's constant. What is such an $\alpha$, and what is the period of its continued fraction?

I can't find Freiman's paper (in Russian) anywhere, and I'm unable to find any information about these questions in the literature.

EDIT: Given @JackHuizenga's Mathematica computations noted in the comments, all that is left to answer is to find an $\alpha$ with Markov constant equal to Freiman's constant, and to calculate its continued fraction expansion.

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    $\begingroup$ Taking one of the terms in particular, the integer 24249 appears at positions 4510, 71264, 138018, etc., so presumably $n = 66754$ and $k<4510$. $\endgroup$
    – Jack Huizenga
    Commented Oct 25, 2023 at 2:31
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    $\begingroup$ By printing the kth and (k+66754)th terms for the first 100 k, I am convinced that the sequence starts repeating at term 6, the 14. So the fraction is $[4;1;1,8,2,\overline{14,2,2,1,12,\ldots,1,72,1,1}]$ with the repeating part having length 66754. $\endgroup$
    – Jack Huizenga
    Commented Oct 25, 2023 at 2:37
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    $\begingroup$ That could be checked using FromContinuedFraction[{4,1,1,8,2,{14,2,2,1,12,...}}] $\endgroup$
    – Jesse Elliott
    Commented Oct 25, 2023 at 2:39
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    $\begingroup$ OK yes, running L = Drop[Take[ ContinuedFraction[(2221564096 + 283748*Sqrt[462])/491993569, 1000000], 66759], 5]; FromContinuedFraction[{4, 1, 1, 8, 2, L}] reproduces the constant! $\endgroup$
    – Jack Huizenga
    Commented Oct 25, 2023 at 2:45
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    $\begingroup$ The periodic part is $(5687655 + 283748 \cdot462^{1/2})/816983$. I was going to do it that way! $\endgroup$
    – Jesse Elliott
    Commented Oct 25, 2023 at 2:47

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