六个速度下孤独跑者谱中的奇分母
Odd denominators in the Lonely Runner spectrum for six speeds
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中文总结 AI 辅助
本文针对六个速度的孤独跑者问题，证明ML<1/6的元组分母为奇数，仅出现k=1或3，还验证了相关猜想并给出精确计算代码。
中文摘要 AI 辅助
对于不同的正整数v₁,…,vₙ，令ML(v₁,…,vₙ)为满足在某个时刻t，每个t·vᵢ到最近整数的距离至少为L的最大数；孤独跑者猜想断言ML≥1/(n+1)。将ML表示为最简分数p/q。Kravitz猜想当ML<1/n时，q=np+1；Fan和Sun找到了n=4时的反例，猜想q=np+k（1≤k≤n）恒成立，并观察到他们n=6的数据中仅出现k=1和k=3。本文解释这一观察结果：对于六个速度，证明所有ML<1/6且除有限个外的元组满足ML=(P-1)/(6P)，其中整数P模6余1或5；特别地，k为1或3，分母q为奇数。该证明确定了集中此类值的三个无限双参数元组族，精确计算每个族上的ML，并明确描述k=3出现的位置。对速度不超过110的2×10⁹个六元组的穷尽搜索未发现反例。对于五个速度，相同方法结合Chen对达到ML=1/5的元组的分类，表明所有ML<1/5且除有限个外的元组满足Kravitz原猜想。计算是精确的，代码已提供。
英文摘要
For distinct positive integers v_1, ..., v_n let ML(v_1, ..., v_n) be the largest number L such that at some time t every t vi is at distance at least L from the nearest integer; the Lonely Runner Conjecture asserts that ML >= 1/(n+1). Write ML = p/q in lowest terms. Kravitz conjectured that whenever ML < 1/n one has q = np + 1; Fan and Sun found counterexamples for n = 4, conjectured that q = np + k with 1 <= k <= n always holds, and observed that in their data for n = 6 only k = 1 and k = 3 occur. We explain this observation. For six speeds we show that all but finitely many tuples with ML < 1/6 satisfy ML = (P-1)/(6P) for an integer P congruent to 1 or 5 modulo 6; in particular k is 1 or 3 and the denominator q is odd. The proof determines the three infinite two-parameter families of tuples on which such values concentrate, computes ML exactly on each family, and describes exactly where k = 3 occurs. An exhaustive search over the 2 x 10^9 sextuples with speeds at most 110 finds no exception. For five speeds the same method, together with Chen's classification of the tuples attaining ML = 1/5, shows that all but finitely many tuples with ML < 1/5 satisfy Kravitz's original conjecture. The computations are exact and the code is provided.
发表机构
- ENEA, Centro Ricerche Frascati（意大利国家新能源与可再生能源公司，弗拉斯卡蒂研究中心）
机构由 AI 辅助整理，请以论文原文为准。
