A284884 Positions of 1's in A284881.

1, 3, 7, 9, 13, 17, 19, 21, 25, 27, 31, 35, 37, 39, 43, 47, 49, 51, 55, 57, 61, 63, 67, 71, 73, 75, 79, 81, 85, 89, 91, 93, 97, 101, 103, 105, 109, 111, 115, 117, 121, 125, 127, 129, 133, 137, 139, 141, 145, 147, 151, 153, 157, 161, 163, 165, 169, 171, 175

Offset: 1

Author

Clark Kimberling, Apr 16 2017

Keywords

• nonn
• easy

Comments

This sequence and A284882 and A284883 form a partition of the positive integers. Conjecture: for n>=1, we have a(n)-3n-3 in {0,1,2}, 3*n+1-A284883(n) in {0,1,2,3}, and 3*n-1-A284884(n) in {0,1,2}.

A284881 = (1,-1,1,0,-1,0,1,-1,1,0,-1,0,1,0,...); thus

A284882 = (2,5,8,11,15,18,...)

A284883 = (4,6,10,12,14,16,...)

A284884 = (1,3,7,9,13,17,...).

Links

• Clark Kimberling, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A284793, A284881, A284882, A284883.

Programs

Mathematica

s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0, 1, 1, 0}}] &, {0}, 6] (* A284878 *)
d = Differences[s]  (* A284881 *)
Flatten[Position[d, -1]] (* A284882 *)
d2 = Flatten[Position[d, 0]]  (* A284883 *)
Flatten[Position[d, 1]]  (* A284884 *)
d2/2  (* A284885 *)