Creating Normal Numbers Using the Prime Divisors of Consecutive Integers

For each integer n ≥ 2, let p 1 ≤ p 2 ≤ · · · ≤ p k be the complete list of the prime factors of a(n) := n(n+1). Consider the function s n : {p 1 , . . . , 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(p 1) . . . 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.