This is the story for the maximum size of a subset of [1,N] avoiding arithmetical progressions of size, 4, with gaps less-than-or-equal to, 5, . ---------------------------- Let F(N) be the largest size of a subset of [1,N] avoiding arithmetical progressions of size , 4, with spacings <=, 1, . Then: F(3 m + 1) = 4 m + 1 F(3 m + 2) = 4 m + 2 F(3 m + 3) = 4 m + 3 F(3 m + 4) = 4 m + 3 The asymptotic density is, 3/4, . ---------------------------- Let F(N) be the largest size of a subset of [1,N] avoiding arithmetical progressions of size , 4, with spacings <=, 2, . Then: F(2 m + 1) = 3 m + 2 F(2 m + 2) = 3 m + 2 F(2 m + 3) = 3 m + 3 The asymptotic density is, 2/3, . ---------------------------- Let F(N) be the largest size of a subset of [1,N] avoiding arithmetical progressions of size , 4, with spacings <=, 3, . Then: F(8 m + 1) = 12 m + 1 F(8 m + 2) = 12 m + 2 F(8 m + 3) = 12 m + 3 F(8 m + 4) = 12 m + 3 F(8 m + 5) = 12 m + 4 F(8 m + 6) = 12 m + 5 F(8 m + 7) = 12 m + 5 F(8 m + 8) = 12 m + 6 F(8 m + 9) = 12 m + 7 F(8 m + 10) = 12 m + 8 F(8 m + 11) = 12 m + 8 F(8 m + 12) = 12 m + 8 The asymptotic density is, 2/3, . ---------------------------- Let F(N) be the largest size of a subset of [1,N] avoiding arithmetical progressions of size , 4, with spacings <=, 4, . Then: F(3 m + 1) = 5 m + 1 F(3 m + 2) = 5 m + 2 F(3 m + 3) = 5 m + 3 F(3 m + 4) = 5 m + 3 F(3 m + 5) = 5 m + 4 The asymptotic density is, 3/5, . ---------------------------- Let F(N) be the largest size of a subset of [1,N] avoiding arithmetical progressions of size , 4, with spacings <=, 5, . Then: F(4 m + 1) = 7 m + 1 F(4 m + 2) = 7 m + 2 F(4 m + 3) = 7 m + 3 F(4 m + 4) = 7 m + 3 F(4 m + 5) = 7 m + 4 F(4 m + 6) = 7 m + 4 F(4 m + 7) = 7 m + 5 The asymptotic density is, 4/7, . This took, 1563.989, seconds