MathDB
1/2^n \sum_{k = 1}^{2^n} \frac{u(k)}{k}> 2/3, greatest odd divisor

Source: Germany 1997 p2

February 22, 2020
inequalitiesgreatest odd divisordivisornumber theory

Problem Statement

For a positive integer kk, let us denote by u(k)u(k) the greatest odd divisor of kk. Prove that, for each nNn \in N, 12nk=12nu(k)k>23\frac{1}{2^n} \sum_{k = 1}^{2^n} \frac{u(k)}{k}> \frac{2}{3}.