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partition of 1,2,...,2^n into 2 classes, none has arithmetic progression

Source: Romania BMO TST 1993 p3

February 17, 2020
arithmetic sequencepartitionSubsetscombinatorics

Problem Statement

Show that the set {1,2,....,2n}\{1,2,....,2^n\} can be partitioned in two classes, none of which contains an arithmetic progression of length 2n2n.