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Source. Robert I. Saye, On two conjectures concerning the ternary digits of powers of two, J. Integer Seq. 25 (2022), Article 22.3.4, 9 pp., as identified on the source card: Lemma 1, stated on p. 3, proved in Section 5, pp. 7--8.

Read depth. Claims checked: the statement and the notation it uses (Section 2, p. 2) were read clause by clause on the print. The proof was read but not checked step by step; nothing here is independently reviewed.

Statement

Notation (p. 2). For integers $a,b$ and a positive integer $k$, $a\equiv_k b$ means $a\equiv b\pmod{3^k}$. For $a$ with ternary expansion $a=\sum_{i=0}^{n}a_i3^i$, the $k$-th ternary digit is $d_k(a)=a_{k-1}$, so $d_1(a)$ is the least significant digit.

Lemma 1 (p. 3). Let $k$ be a positive integer and put $u_k=2\cdot3^{k-1}$. Then:

(i) $u_k$ is the least positive integer $u$ with $2^{u}\equiv_k1$;

(ii) for $i,j\in\mathbb N$, if $2^i\equiv_k2^j$ then $i$ and $j$ differ by a multiple of $u_k$;

(iii) for $i,j\in\mathbb N$,

$$d_{k+1}\bigl(2^{iu_k+j}\bigr)\equiv d_{k+1}\bigl(2^j\bigr)+i\,d_1\bigl(2^j\bigr) \pmod 3.$$

The paper notes (p. 3, footnote 1) that $u_k=\varphi(3^k)$, so part (i) says that $2$ has the full order $\varphi(3^k)$ modulo $3^k$. In part (iii) $d_1(2^j)$ is $1$ or $2$, so as $i$ runs over $0,1,2$ the $(k+1)$st digit takes all three values while, by part (i), the last $k$ digits stay fixed; this is the step the paper's search uses (p. 3).

Proof pointer

Section 5 (pp. 7--8). The key fact, proved by induction on $k$ by cubing, is that $2^{u_k}\equiv1+3^k\pmod{3^{k+1}}$. Part (i) follows by induction on $k$, ruling out $u_{k-1}$ and $2u_{k-1}$ as the order modulo $3^k$; part (ii) reduces to part (i) by cancelling a power of two, a unit modulo $3^k$; part (iii) follows by expanding $(1+3^k)^i$ modulo $3^{k+1}$ and multiplying by $2^j$.

Dependencies

None outside the paper.

Bears on

• Problem 406: only as the tool behind the computer search on the main result's page. The lemma itself says nothing about which powers of two avoid the digit $2$.