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d(n -1) + d(n) + d(n +1) <= 8 , d(n) is no. of all positive divisors of n>=2

Source: 49th Austrian Mathematical Olympiad Regional Competition (Qualifying Round) 5th April 2018 p4

May 25, 2019
number of divisorsDivisorsnumber theory

Problem Statement

Let d(n)d(n) be the number of all positive divisors of a natural number n≥2n \ge 2. Determine all natural numbers n≥3n \ge 3 such that d(n−1)+d(n)+d(n+1)≤8d(n -1) + d(n) + d(n + 1) \le 8.
Proposed by Richard Henner
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