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Source. J. Geneson, Deletion thresholds and exponential examples for complete sequences, arXiv:2609.25107v1 (20 September 2026): Section 5 "Incomplete exponential sequences" with Theorem 9, Lemma 10 and Proposition 11 on physical p. 11 and the proof of Theorem 9 on p. 12. Read in the canonical conversion beside the held PDF; the artifact is identified on the library source card, Geneson (2026), and the theorem on its result page.

Standing. This is an author-recorded reconstruction. It is not an independent review, changes no status and assigns no tier. Two inputs are imported from Dubickas's paper as the source quotes them and are not reproved here: Lemma 10, and the identification of the displayed polynomial as the minimal polynomial of a Salem number. The two exact polynomial evaluations in the proof are rechecked by the folder's evidence.

Definitions

A Salem number is a real algebraic integer $\gamma>1$ whose other conjugates lie in the closed unit disk, at least one on the unit circle; a Pisot number is one whose other conjugates lie in the open unit disk. $\{x\}=x-\lfloor x\rfloor$. A sequence of integers is complete, in the source's convention (p. 1), if every sufficiently large integer is a sum of terms with distinct indices. $\varphi=(1+\sqrt5)/2$.

Imported (Lemma 10, Dubickas, his Theorem 6). Let $\gamma$ be a Pisot or Salem number with minimal polynomial $P$ and $P(1)=-q$ for an integer $q\ge2$. For every $\epsilon>0$ there is a real $\xi\in\mathbb Q(\gamma)$ with

$$\frac1q-\epsilon<\{\xi\gamma^n\}<\frac1q+\epsilon\qquad(n\ge1).$$

The sign of $\xi$ is not specified.

Imported (Dubickas, p. 332 as cited). The polynomial

$$P(x)=x^{18}-x^{12}-x^{11}-x^{10}-x^9-x^8-x^7-x^6+1$$

is the minimal polynomial of a Salem number $\gamma$.

Statement

Proposition 11. Let $\gamma$ be a Pisot or Salem number with minimal polynomial $P$ and $P(1)=-q$ for an integer $q\ge3$. There is a real $\eta>0$ with

$$\frac3{4q}<\{\eta\gamma^n\}<\frac5{4q}\qquad(n\ge1).$$

Consequently there are arbitrarily large real $t>0$ such that $\lfloor t\gamma^n\rfloor$ is even for every integer $n\ge0$.

Theorem 9. Let $\gamma>1$ be the Salem number with minimal polynomial $P$ above. Then $6/5<\gamma<13/10<\varphi$, and there are arbitrarily large real $t>0$ such that $\lfloor t\gamma^n\rfloor$ is even for every integer $n\ge0$. In particular these sequences are not complete.

Proof

Proposition 11

Put $\epsilon=1/(4q(q-1))$. Then $\epsilon\le1/(4q)$, $(q-1)\epsilon=1/(4q)$ and $\epsilon<1/q$. Take $\xi$ from Lemma 10 for this $\epsilon$. The lower bound $1/q-\epsilon$ is positive, so no $\xi\gamma^n$ is an integer and $\xi\ne0$. For $n\ge1$ let $a_n=\lfloor\xi\gamma^n\rfloor$ and $e_n=\{\xi\gamma^n\}-1/q$, so that

$$\xi\gamma^n=a_n+\frac1q+e_n,\qquad a_n\in\mathbb Z,\qquad|e_n|<\epsilon .$$

Since $q\ge3$, $5/(4q)<1/2$, so the interval $(3/(4q),5/(4q))$ lies in $(0,1/2)$.

If $\xi>0$, set $\eta=\xi$: the residual $1/q+e_n$ after the integer $a_n$ differs from $1/q$ by less than $\epsilon\le1/(4q)$.

If $\xi<0$, set $\eta=-(q-1)\xi>0$. Multiplying the display by $-(q-1)$,

$$\eta\gamma^n=-(q-1)a_n-\frac{q-1}q-(q-1)e_n =\bigl(-(q-1)a_n-1\bigr)+\frac1q-(q-1)e_n ,$$

where $-(q-1)a_n-1$ is an integer and the residual $1/q-(q-1)e_n$ differs from $1/q$ by $(q-1)|e_n|<(q-1)\epsilon=1/(4q)$.

In both cases $\eta\gamma^n$ is an integer plus a residual in $(3/(4q),5/(4q))\subset(0,1)$, so the residual is $\{\eta\gamma^n\}$, and the display of the proposition holds. Since $\{\eta\gamma^n\}<1/2$,

$$\lfloor2\eta\gamma^n\rfloor =2\lfloor\eta\gamma^n\rfloor+\lfloor2\{\eta\gamma^n\}\rfloor =2\lfloor\eta\gamma^n\rfloor$$

is even for $n\ge1$. For an integer $m\ge1$ put $t_m=2\eta\gamma^m$; then $\lfloor t_m\gamma^n\rfloor=\lfloor2\eta\gamma^{m+n}\rfloor$ is even for every $n\ge0$, and $t_m\to\infty$ as $m\to\infty$ because $\gamma>1$ and $\eta>0$.

Theorem 9

$P(1)=1-7+1=-5$, so Proposition 11 applies with $q=5$ and gives the coefficients $t$. For the location of $\gamma$, exact evaluation gives

$$5^{18}P(6/5)=-41745565065959<0,\qquad 10^{18}P(13/10)=28586401421206393129>0$$

(both integers rechecked by the folder's evidence), so by the intermediate value theorem $P$ has a real root in $(6/5,13/10)$. That root has modulus greater than $1$, and $\gamma$ is the only root of $P$ outside the closed unit disk, so it is $\gamma$. Thus $6/5<\gamma<13/10<3/2<\varphi$, the last because $\sqrt5>2$. The terms $\lfloor t\gamma^n\rfloor$ are nonnegative and all even, so every finite sum of terms with distinct indices is even, no odd integer is represented, and the sequence is not complete.

Scope. The construction is existential: it gives no explicit $t$ and no numerical upper bound on one. The source adds (p. 12) that by van Doorn's computer-assisted Proposition 8 the sequence is complete for $1.2<\gamma\le1.3$ and $0<t\le5$, so every counterexample coefficient at this base exceeds $5$; that comparison is not reconstructed here. The theorem concerns one base and one sequence; its bearing on Problem 354 is through Corollary 12.