Creating Normal Numbers Using the Prime Divisors of Consecutive Integers

For each integer n ≥ 2, let p₁ ≤ p₂ ≤ ··· ≤ p_k be the complete list of the prime factors of a(n):= n(n+1). Consider the function s_n : {p₁,..., p_k}→ {0, 1} defined by s_n(p_j)= 0 if p_j | n and 1 if p_j | n + 1. Then consider the binary number h(n) := s_n(p₁) ...s_n(p_k). In an earlier paper, we proved that the number 0.h(2) h(3) h(4) ... is a binary normal number and in fact we proved the more general statement when, for a fixed integer t ≥ 2, we set a(n) := n(n +1) ··· (n + t − 1), thus allowing for the construction of a normal number in base t. Here, we give a much shorter and simpler proof of this result and then we consider a more general result when a(n) is the product of linear functions.