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An informal version of my question is "If we have a Pisot number between 1 and 2 of a very large degree, is it true that all its other conjugates are very close to 1 in modulus?"

A more formal version is as follows: let $\beta_n\in(1,2)$ be a convergent sequence of Pisot numbers with $d_n$ being the degree of $\beta_n$ and $\gamma^{(n)}_1,\dots, \gamma^{(n)}_{d_n-1}$ be the Galois conjugates of $\beta_n$ which are different from $\beta_n$. We have then by definition, $|\gamma_k^{(n)}|<1$ for all $n\ge1$ and all $k=1,\dots,d_n-1$. Put $$ \mu_n=\min_{1\le k\le d_n-1} |\gamma_k^{(n)}|. $$ Question. Is it true that $\mu_n\to1$ as $n\to\infty$?

A colleague of mine believes he has seen this claim somewhere, so this may be a matter of a straightforward reference.

EDIT. For those familiar with the literature, this claim can be proved directly for the so-called regular Pisot numbers $\beta=\lim_{n\to\infty}\beta_n$. Unfortunately, there exist other (irregular) Pisot numbers which may or may not accumulate at 2. This class does not appear to exhibit any pattern. So I've been wondering if there is a way to prove this without addressing this issue.

ADDED EXAMPLE. I think it might be helpful to give an example. Let $\beta_n$ be the $n$th multinacci number, i.e., the root of $x^{n+1}-2x^n+1$ lying between 1 and 2. Let $\gamma=\gamma_k^{(n)}$ be one of its conjugates. We have $\gamma^{n+1}=2\gamma^n-1$, whence $$ |\gamma^{-n}|=\left|\frac2{\gamma}-1\right|\le \left|\frac2{\gamma}\right|+1\le5, $$ since $|\gamma|\ge\frac12$. Thus, $|\gamma|\ge 1/\sqrt[n]5\to1$ as $n\to\infty$. This works so well here, because $\beta_n$ is a root of a polynomial with small coefficients and a small number of nonzero coefficients. It is unclear whether this is always the case.

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    $\begingroup$ If $\beta$ is a Pisot number, then so is $\beta^m$, $m=2,3,\dots$, so high degree alone does not imply closeness to 1 in modulus. $\endgroup$ – Gerry Myerson Jan 10 at 0:42
  • $\begingroup$ Fair point, Gerry. OK, I restrict $\beta_n$ to $(1,2)$ which is actually the range I care about. $\endgroup$ – Nikita Sidorov Jan 10 at 1:03
  • $\begingroup$ I believe there are Salem numbers less than, but arbitrarily close to, 2. Such Salem numbers necessarily have a conjugate close to $1/2$. If there's a way to find Pisot numbers that, together with their conjugates, shadow Salem numbers and their conjugates.... $\endgroup$ – Gerry Myerson Jan 10 at 1:14
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    $\begingroup$ I'm not on top of that, I only know that David Boyd did some work on Salem numbers as limits of sequences of Pisot numbers, and others have continued that work. Boyd conjectured that the union of the Salems and Pisots is a closed set, which I think is still an open question. $\endgroup$ – Gerry Myerson Jan 10 at 1:30
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    $\begingroup$ @VesselinDimitrov, What you describe is pretty much the regular Pisot numbers. Unfortunately, there exist irregular Pisot numbers; it is known that for any $\delta>0$, there are only finitely many of those in $(1,2-\delta)$ and there is an algorithm due do D. Boyd which - at least, in theory - allows to list them all in any $(a,b)\subset(1,2)$ with $b<2$. It is not known if there are infinitely many irregulars in $(1,2)$. For more details see, e.g., arxiv.org/abs/1103.2147 $\endgroup$ – Nikita Sidorov Jan 12 at 16:24

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