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原始链接: https://news.ycombinator.com/item?id=43421934

Hacker News的一个帖子讨论了OEIS中所有数字均为偶数的2的幂的序列,特别是关于2048(2^11)是此类幂中最大的数的说法。评论者们对证明这个看似简单的性质如此困难感到惊讶。 一位评论者WithinReason引用了Michael S. Branicky的话,指出在2^(10^10)之前不存在其他项。这引发了人们对其如何检查如此大的范围的疑问。lifthrasiir解释了一种潜在的过滤技术:利用2^k取模递增的10的幂时必须具有偶数数字这一事实。通过检查2^k mod 10, 2^k mod 100, 2^k mod 1000等等,许多潜在的k值可以由于这些模运算结果的周期性而被快速排除。基于前几个步骤的查找表可以非常有效。另一位评论者发现,如果没有概述所使用的具体算法,则原始评论缺乏细节。

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  • 原文
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    Powers of 2 with all even digits (oeis.org)
    25 points by Hbruz0 1 hour ago | hide | past | favorite | 5 comments










    This is remarkable! I always find it fascinating that simple to express properties lack a proof. This is a very simple thing to evaluate and seems like it should be straightforward to establish that 2048 is the highest such power.


    No additional terms up to 2^(10^10). - Michael S. Branicky, Apr 16 2023

    How did he do this?



    As noted in 3) in the Shepherd's comment, 2^k has no odd digits when 2^k mod 10^n for all integer n have no odd digits as well. So about a half of all possible k would be filtered by checking whether 2^k mod 10 is odd, then another half of the remainder will get filtered with 2^k mod 100, 2^k mod 1000 and so on. All of them would be periodic, so first few steps can be made into a lookup table to filter almost every k.


    yeah that's weird - its kind of a pointless comment without an included algorithm or something


    There’s probably a smart way to rule out a lot of cases so you only have to check a relatively small number of candidates. It would be good to know what it is.






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