Pisot conjugates

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 $$n\to\infty$$?

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.

ADDED "COUNTEREXAMPLE". Let now $$\beta_n$$ be the root of $$x^n-x^{n-1}-\dots-x^2-x+1$$ lying in $$(1,2)$$. Then $$\beta_n$$ is known to be a Salem number, which implies that $$\gamma_n:=\beta_n^{-1}$$ is one of its conjugates. (The rest are of modulus 1.) Thus, the result in question doesn't hold for this family, since $$\gamma_n\to\frac12$$ as $$n\to\infty$$. The trick from the previous example gives $$\gamma_n^{-n}=\frac{2-\gamma_n}{\gamma_n(2\gamma_n-1)},$$ so we cannot claim $$|\gamma_n^{-n}|=O(1)$$, since $$2\gamma_n-1\to0$$. This example shows that the problem is rather delicate, since the conjecture is invalid even for a sequence of "close relatives" of Pisot numbers.

• 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. Jan 10, 2019 at 0:42
• Fair point, Gerry. OK, I restrict $\beta_n$ to $(1,2)$ which is actually the range I care about. Jan 10, 2019 at 1:03
• 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. Jan 10, 2019 at 1:30
• @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 Jan 12, 2019 at 16:24
• @VesselinDimitrov We don't know much about the structure of $S\cap (2, \infty)$, so I guess we don't have a particular reason to believe this claim to be true. Having elements of $S''$ may add extra complications, for instance. Feb 24, 2019 at 12:01