Erdős三元数字问题的Repunit坐标
Repunit Coordinates for the Erdős Ternary-Digit Problem
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中文总结 AI 辅助
本文用广义基4 repunit重新表述Erdős关于2的幂三元展开不含数字2的猜想，通过模3^m置换和提升恒等式构造3-adic同胚，但未证明原猜想。
中文摘要 AI 辅助
Paul Erdős猜想，2的幂中，其三元展开不含数字2的仅有1、4和256。尽管进行了广泛的计算验证，该猜想仍未解决。我们使用广义的基4 repunit重新表述该问题。任何允许的指数都是偶数，设n=2s，且2^{2s}=4^s=3R_s(4)+1，其中R_s(4)=(4^s-1)/3。因此，2^{2s}省略三元数字2当且仅当R_s(4)省略该数字。对于每个深度m，映射Φ_m:s mod 3^m ↦ R_s(4) mod 3^m是模3^m的剩余类的一个置换。因此，每个指定的m位三元后缀都有一个唯一的规范指数种子模3^m，并且产生该后缀的所有指数构成一个等差数列。提升恒等式R_{r+q3^m}(4)≡R_r(4)+q3^m (mod 3^{m+1})在每个深度给出三个提升，其中恰好两个保持允许性。这些有限映射组装成3-adic整数上的等距同胚，将x↦x+1与y↦4y+1共轭。未解决的全局整数问题仍然存在。我们不声称证明了Erdős猜想。
英文摘要
Paul Erdős conjectured that the only powers of two whose ternary expansions contain no digit 2 are 1, 4, and 256. Despite extensive computational verification, the conjecture remains open. We reformulate the problem using generalised base-4 repunits. Any admissible exponent is even, say \(n=2s\), and \[ 2^{2s}=4^s=3R_s(4)+1,\qquad R_s(4)=\frac{4^s-1}{3}. \] Thus \(2^{2s}\) omits the ternary digit 2 if and only if \(R_s(4)\) does. For each depth \(m\), the map \(Φ_m:s\bmod 3^m\mapsto R_s(4)\bmod 3^m\) is a permutation of the residue classes modulo \(3^m\). Hence every prescribed \(m\)-digit ternary suffix has a unique canonical exponent seed modulo \(3^m\), and all exponents producing that suffix form an arithmetic progression. The lifting identity \[ R_{r+q3^m}(4)\equiv R_r(4)+q3^m\pmod{3^{m+1}} \] gives three lifts at each depth, exactly two of which preserve admissibility. The finite maps assemble into an isometric homeomorphism of the 3-adic integers conjugating \(x \mapsto x+1\) with \(y \mapsto 4y+1\). The unresolved global integer problem remains. No proof of the Erdős conjecture is claimed.
