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Statement

For a $3$-adic integer $\lambda=\sum_{j\ge0}d_j3^j$ with each $d_j\in\{0,1,2\}$, its $3$-adic expansion is $(\lambda)_3=(\cdots d_2d_1d_0)_3$ (p. 4). Put $\alpha_0=\log_32\approx0.63092$.

Theorem 1.4 (p. 4). For each nonzero $\lambda\in\mathbb Z_3$ and each $X\ge2$,

$$\tilde N_\lambda(X)=\#\{n\le X:(\lambda2^n)_3\text{ omits the digit }2\}\le2X^{\alpha_0}. \qquad(1.9)$$

The statement prints the range as $n\le X$; the proof (p. 20, display (3.2)) counts $1\le n\le X$. For $\lambda=1$ the expansions are the ternary expansions of the integers $2^n$, and the paper presents the theorem (p. 4) as an extension, by essentially the same proof, of Narkiewicz's bound $N_1(X)\le1.62X^{\alpha_0}$ (p. 1) to all nonzero $\lambda$.

Source. Theorem 1.4, p. 4, 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 (p. 20) was followed. Nothing here is independently reviewed.

Proof pointer

P. 20. Dividing out the power of $3$ in $\lambda$ shifts digits and does not change the count, so $\lambda$ may be taken prime to $3$. Since $2$ is a primitive root modulo $3^k$, the residues $\lambda2^n$ for $1\le n\le2\cdot3^{k-1}$ run once through the $2\cdot3^{k-1}$ unit classes modulo $3^k$, of which exactly $2^{k-1}$ have no digit $2$ among their lowest $k$ digits. Choosing $k$ with $2\cdot3^{k-2}<X\le2\cdot3^{k-1}$ gives $\tilde N_\lambda(X)\le2^{k-1}\le2X^{\alpha_0}$.

Dependencies

None beyond the fact that $2$ is a primitive root modulo every power of $3$.

Bears on

• Problem 406: with $\lambda=1$ the theorem says that for every $X\ge2$ at most $2X^{\log_32}$ exponents $n\le X$ give a power $2^n$ with only the digits $0$ and $1$ in base $3$; the argument uses only the lowest digits of $2^n$. It is a density bound, slightly weaker in its constant than Narkiewicz's bound for that case, and does not decide whether there are finitely many such powers.