On Newman's phenomenon in higher bases

TL;DR

This paper extends Newman's phenomenon, showing that the dominance of numbers with even number of 1's in their base-b expansion persists in higher bases and analyzing its strength and distribution among primes.

Contribution

It generalizes Newman's phenomenon to higher bases and establishes conditions for its occurrence, also comparing the strength of the phenomenon across bases.

Findings

Newman's phenomenon exists for bases greater than 2.

The phenomenon is stronger in higher bases than in binary.

Number of primes with the phenomenon grows slower than x/log x.

Abstract

A well known result of Newman says that upto a limit, multiples of $3$ with even number of 1's in binary representation always exceed multiples of $3$ with odd number of 1's. The phenomenon of preponderance of even number of 1's is now known as Newman's phenomenon. We show that this phenomenon exists for higher bases. Let $b$ be a positive integer($\geq 2$). Let $A_{b}$ be the set of all natural numbers which contain only 0's and 1's in b-ary expansion and $S^{(b)}_{q,i}(n)$ be the difference between the corresponding number of $k_e<n$, $k_e\equiv i \mod q$, $k_e\in A_{b}$ and $k_e$ has even number of 1's in b-ary expansion and the number of $k_o$ $k_o<n$, $k_o\equiv i \mod q$, $k_o\in A_{b}$ and $k_o$ has odd number of 1's in b-ary expansion. Let $q$ be a multiple or divisor of $b+1$ which is relatively prime to $b$ then we show that $S^{(b)}_{q,0}(n)>0$ for sufficiently large $n$. We show that there is a stronger Newman's phenomenon in $A_b$ in the following sense. If $b>2$ and $n=\sum_{i=0}^{k-1}b_i2^i$ with $b_i\in \{0,1\}$, let $b(n)=\sum_{i=0}^{k-1}b_ib^i$ then $\lim_{n\rightarrow \infty} \frac{S^{(2)}_{3,0}(n)}{S^{(b)}_{b+1,0}(b(n))}=0$. That is, for the same number of terms there is stronger preponderance in $A_b$ than in $A_2=\mathbb{N}$. In the last section we show that number of primes $p\leq x$ for which $S_{p,0}^{(b)}(n)>0$ for sufficiently large $n$ is $o\left(\frac{x}{\log x}\right)$.