On Newman's phenomenon in higher bases

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 $o\left(\frac{x}{\log x}\right)$.