An Euler totient sum inequality

•Define χ(1)=1χ(1)=1, χ(n)=ϕ(n)+χ(n/q)χ(n)=ϕ(n)+χ(n/q) (n>1n>1), where q is the least prime factor of n.•When n=dℓn=dℓ and the prime factors of d exceed those of ℓ  , χ(n)=ϕ(d)(χ(ℓ)−1)+χ(d)χ(n)=ϕ(d)(χ(ℓ)−1)+χ(d).•When (m,n)=1(m,n)=1, χ(mn)≤χ(m)χ(n)χ(mn)≤χ(m)χ(n), with equality if and only if one of m or n is 1.•GL(2,p)GL(2,p) (p>11p>11) has clique number greater than that of its largest cyclic subgroup.

TextDefine χ(n)χ(n) recursively by χ(1)=1χ(1)=1 and χ(n)=ϕ(n)+χ(n/q)χ(n)=ϕ(n)+χ(n/q) for all integers n>1n>1, where q is the least prime factor of n, and where ϕ   is the Euler totient function. We show that χ(n)=ϕ(d)(χ(ℓ)−1)+χ(d)χ(n)=ϕ(d)(χ(ℓ)−1)+χ(d), where n=dℓn=dℓ and the prime factors of d are greater than the prime factors of ℓ  . We also show χ(nm)≤χ(n)χ(m)χ(nm)≤χ(n)χ(m) when n and m   are coprime numbers. As an application, we show that for all primes p≥11p≥11, χ(p2−p)>χ(p2−1)χ(p2−p)>χ(p2−1). We discuss the interpretation of χ as the clique number of the power graph of a finite cyclic group and the significance of the inequality in this context.VideoFor a video summary of this paper, please visit https://youtu.be/p8finzAEJps.