MathDB
Sum [kn^(1/3)]

Source: Romanian 2018 TST Problem 2 Day 3

May 25, 2020
number theoryfloor functionseries summation

Problem Statement

Given a square-free integer n>2n>2, evaluate the sum k=1(n2)(n1)(kn)1/3\sum_{k=1}^{(n-2)(n-1)} \lfloor ({kn})^{1/3} \rfloor.