Saye 2022 two conjectures concerning ternary digits powers

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lemma_1: States that u_k = 2·3^(k-1) is the least positive u with 2^u congruent to 1 modulo 3^k, that exponents whose powers of two agree modulo 3^k differ by a multiple of u_k, and that adding i·u_k to an exponent j shifts the (k+1)st ternary digit of 2^j by i times its last digit, modulo 3.

main_theorem: Reports a computer search showing that for every n at most 2 times 3^45, about 5.9 times 10^21, the ternary expansion of 2^n contains a 2 unless n is 0, 2 or 8, and contains a 0 unless n is 0, 1, 2, 3, 4 or 15.

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Saye, Robert I., On two conjectures concerning the ternary digits of powers of two. J. Integer Seq. 25 (2022), Art. 22.3.4, 9 pp.

Saye tests by computer two conjectures on the ternary digits of powers of two: Erdős's, that only 2^0, 2^2 and 2^8 have no digit 2, and Sloane's, that every power of two other than 2^0, 2^1, 2^2, 2^3, 2^4 and 2^15 has a digit 0 (p. 1). The main result (unnumbered, p. 5) is that no counterexample to either exists with exponent n <= 2·3^45, about 5.9 × 10^21, extending the earlier range n <= 2·3^20 that the paper attributes to Vardi; the paper restates this as 2^n containing every ternary digit for 16 <= n <= 2·3^45 (p. 2). The search rests on Lemma 1 (p. 3): u_k = 2·3^(k-1) is the order of 2 modulo 3^k, and adding 0, u_k or 2u_k to an exponent keeps the last k ternary digits of the power while running its (k+1)st digit through all three values, so the exponents whose trailing digits avoid a given digit can be generated in increasing order (Algorithm 1, p. 4). The paper also computes and plots the least n for which the last k digits of 2^n avoid a given digit (pp. 5-7), entered in the OEIS as A351927 and A351928 (p. 8). Nothing in the paper is a proof of either conjecture.

Read status. Claims checked: the main result and Lemma 1 were read clause by clause on the print. The proof of Lemma 1 (pp. 7-8) was read but not checked step by step; the computation has not been rerun.

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Bears on. #406: the main result finds no power 2^n with 8 < n <= 2·3^45 whose ternary expansion uses only the digits 0 and 1; a finite check, it does not settle whether there are only finitely many.

Results. Main result (p. 5, unnumbered; also pp. 1, 2, 8); Lemma 1 (p. 3, proved pp. 7-8).

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