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A lower bound for Cusick’s conjecture on the digits of n + t
Mathematical Proceedings of the Cambridge Philosophical Society ( IF 0.6 ) Pub Date : 2021-02-24 , DOI: 10.1017/s0305004121000153
LUKAS SPIEGELHOFER

Let S be the sum-of-digits function in base 2, which returns the number of 1s in the base-2 expansion of a nonnegative integer. For a nonnegative integer t, define the asymptotic density $${c_t} = \mathop {\lim }\limits_{N \to \infty } {1 \over N}|\{ 0 \le n < N:s(n + t) \ge s(n)\} |.$$ T. W. Cusick conjectured that ct > 1/2. We have the elementary bound 0 < ct < 1; however, no bound of the form 0 < αct or ctβ < 1, valid for all t, is known. In this paper, we prove that ct > 1/2 – ε as soon as t contains sufficiently many blocks of 1s in its binary expansion. In the proof, we provide estimates for the moments of an associated probability distribution; this extends the study initiated by Emme and Prikhod’ko (2017) and pursued by Emme and Hubert (2018).

中文翻译:

Cusick 猜想关于 n + t 的数字的下界

小号是基数为 2 的数字和函数,它返回非负整数的基数为 2 的扩展中 1 的个数。对于非负整数, 定义渐近密度$${c_t} = \mathop {\lim }\limits_{N \to \infty } {1 \over N}|\{ 0 \le n < N:s(n + t) \ge s(n)\ } |.$$TW Cusick 推测C> 1/2。我们有基本界限 0 <C< 1; 但是,没有 0 < 形式的界限αC要么Cβ< 1,对所有人有效, 是已知的。在本文中,我们证明C> 1/2 –ε立刻在其二进制扩展中包含足够多的 1 块。在证明中,我们提供了对相关概率分布的矩的估计;这扩展了由 Emme 和 Prikhod'ko (2017) 发起并由 Emme 和 Hubert (2018) 进行的研究。
更新日期:2021-02-24
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