Given a real number $x$, let $\Vert x\Vert=[x]-x$, where $[x]$ is the nearest integer to $x$.

Suppose $\lambda>1$ is a Pisot number. Let $f(x)=x^k+a_{k-1}x^{k-1}+\ldots+a_0$ be the irreducible polynomial of $\lambda$ over $\mathbb{Q}$. Let $z_1,\ldots,z_{k-1}$ be the distinct roots of $f(x)$ other than $\lambda$ (so $|z_i|<1$ for all $i$).

It is known that, for various real numbers $\alpha$, the sequence $(\Vert\alpha\lambda^n\Vert)_{n=0}^{\infty}$ is square summable. Here are two examples:

1) If $\alpha=1$ then, for sufficiently large $n$, $$ \Vert\alpha\lambda^n\Vert=z_1^n+\ldots+z_{k-1}^n. $$

2) Fix integers $c_0,\ldots,c_{k-1}$ and, for $n\geq k$, define $$ c_n=-(a_{k-1}c_{n-1}+\ldots+a_0c_{n-k}). $$ Then there are $\alpha,\beta_1,\ldots,\beta_{k-1}\in\mathbb{C}$ such that $$ c_n=\alpha\lambda^n+\beta_1z_1^n+\ldots+\beta_{k-1}z_{k-1}^n $$ Then $\Im(\alpha)\lambda^n+\Im(\beta_1z^n_1+\ldots+\beta_{k-1}z^n_{k-1})=\Im(c_n)=0$. So $\alpha$ must be real since $|z_i|<1$ and $\lambda>1$. For sufficiently large $n$, $$ \Vert\alpha\lambda^n\Vert=\beta_1z_1^n+\ldots+\beta_{k-1}z_{k-1}^n. $$

My understanding is that the original work of Pisot provides some information about what real numbers $\alpha$ will give rise to square summable sequences. But I am wondering what is known explicitly about the sequence itself. Specifically, consider some $\alpha$ such that $(\Vert\alpha\lambda^n\Vert)_{n=0}^{\infty}$ is square summable. Is the sequence always expressible using powers of $z_1,\ldots,z_{k-1}$ as in the two examples above (or perhaps powers of some other fixed "eigenvalues")? If this isn't always the case, are there conditions on $\alpha$ which will ensure this?