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Statement

Notation as on Theorem 1.6: $\Sigma_{3,\bar2}$ is the $3$-adic Cantor set of digits $0$ and $1$, and $\mathcal C(M_1,\ldots,M_k)$ is the set of $\lambda\in\mathbb Z_3$ with every $M_j\lambda$ in $\Sigma_{3,\bar2}$.

Theorem 1.7 (p. 6). Let $1\le M_1<M_2<\cdots<M_k$ be positive integers, and suppose there is a positive integer $N$ in $\Sigma_{3,\bar2}$ such that $NM_i\in\Sigma_{3,\bar2}\cap\mathbb Z$ for every $i$, display (1.19). Then

$$\dim_H(\mathcal C(M_1,M_2,\ldots,M_k))\ge\frac{\log_3(2)}{\lceil\log_3(NM_k)\rceil}. \qquad(1.20)$$

The print places $N$ in "$\Sigma_{3,\bar{2}}\cup\mathbb{Z}$" [sic] and indexes (1.19) by "$1\le j\le k$" [sic] while writing $NM_i$; read as above, $N$ is a positive integer whose ternary expansion omits the digit $2$, and the condition is on every $NM_i$. The paper notes (p. 6) that the condition is sufficient but not necessary: $N=1$, $M_1=1$, $M_2=52$ fails it, yet $\mathcal C(1,52)$ has positive dimension.

Source. Theorem 1.7, p. 6, of Jeffrey C. Lagarias, Ternary expansions of powers of 2, J. Lond. Math. Soc. (2) 79 (2009), no. 3, 562--588; labels and pages are those of the arXiv:math/0512006v4 edition (11 July 2008) identified on the source card.

Read depth. Claims checked: the statement was read clause by clause on the page image, and the proof (pp. 24--25) was followed. Nothing here is independently reviewed.

Proof pointer

Pp. 24--25. With $n=\lceil\log_3(NM_k)\rceil$, the $3$-adic integers built from blocks of $n$ digits, each block either all zeros or the ternary digits of $N$, form a Cantor set of dimension $\log_32/n$. Multiplying by $M_j$ replaces each copy of $N$ by $NM_j$, which still fits in $n$ digits without carries, so the set lies in $\mathcal C(M_1,\ldots,M_k)$.

Dependencies

None beyond the paper's definitions.

Bears on

• Problem 406: the paper offers the theorem (p. 6) as a possible route to a positive lower bound for $\dim_H(\mathcal E^{(k)}(\mathbb Z_3))$ with $k\ge4$, if suitable $M_i=2^{n_i}$ can be found. A lower bound on these sets does not bear on whether $1$ lies in the $3$-adic exceptional set, and the theorem does not decide the problem.