arXivDaily arXiv每日学术速递 周一至周五更新
arXiv周末暂无论文更新,休息一下吧,周末愉快~~

乘法持久性猜想:解决非零偶目标的2进障碍

The Multiplicative Persistence Conjecture: Resolving the \(2\)-Adic Obstruction for Nonzero Even Targets

Patrick Nyadjo Fonga

arXiv 2608.27802首次发表:更新:

AI 中文总结

本文证明了乘法持久性猜想中关于非零偶目标的2进指数一致有界的部分,得到显式界并推广至任意基,结合已有方法得到分析非零偶终端数字的有限算法。

AI 中文摘要

乘法持久性问题研究将正整数反复替换为其各位数字的乘积,直至得到单个数字(称为终端数字)的过程。经典猜想断言,没有十进制整数需要超过11次迭代。Brier、Clavier、Gutsche和Naccache证明了所有奇终端数字的猜想成立。在处理非零偶终端数字时,他们得到了无限族十进制整数,其中数字2到9的个数固定,可插入任意多个数字1。他们猜想,尽管该族无限,其元素中2的指数是一致有界的。我们证明了该猜想,并得到仅依赖于规定数字重数的显式界。更一般地,我们的论证适用于每个基b≥3及每个素数p|b。在基10下,将我们的界与Brier等人的方法结合,得到了用于进一步分析每个非零偶终端数字的有限算法。

英文摘要

The multiplicative persistence problem studies the process of repeatedly replacing a positive integer by the product of its digits until a single digit, called the terminal digit, is reached. The classical conjecture asserts that no decimal integer requires more than \(11\) iterations. Brier, Clavier, Gutsche, and Naccache proved the conjecture for all odd terminal digits. In their approach to nonzero even terminal digits, they were led to infinite families of decimal integers in which the numbers of digits \(2,\ldots,9\) are fixed, while arbitrarily many digits \(1\) may be inserted. They conjectured that, despite the infinitude of such a family, the exponent of \(2\) dividing its elements is uniformly bounded. We prove this conjecture and obtain an explicit bound depending only on the prescribed digit multiplicities. More generally, our argument applies in every base \(b\geq3\) and to every prime \(p\mid b\). In base \(10\), combining our bound with the method of Brier, Clavier, Gutsche, and Naccache yields a finite algorithm for a further analysis of each nonzero even terminal digit.

Comments14 pages

论文原文

arXiv 摘要页 · PDF 原文 · HTML 原文

↑