A304802 Solution (b(n)) of the complementary equation a(n) = b(n) + b(3n); see Comments.

1, 3, 4, 5, 6, 7, 9, 10, 11, 12, 14, 15, 16, 18, 19, 20, 21, 23, 24, 25, 26, 28, 29, 30, 31, 32, 34, 35, 36, 37, 39, 40, 41, 43, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 59, 60, 61, 63, 64, 65, 66, 67, 69, 70, 71, 72, 74, 75, 76, 78, 79, 80, 81, 83

Offset

0,2

Comments

Define complementary sequences a(n) and b(n) recursively:

b(n) = least new,

a(n) = b(n) + b(3n),

where "least new" means the least positive integer not yet placed. Empirically, {a(n) - 4*n: n >= 0} = {2,3} and {3*b(n) - 4*n: n >= 0} = {3,4,5,6,7}. See A304799 for a guide to related sequences.

Links

Clark Kimberling, Table of n, a(n) for n = 0..10000

Example

b(0) = 1, so that a(0) = 2.  Since a(1) = b(1) + b(3), we must have a(1) >= 8, so that b(1) = 3, b(2) = 4, b(3) = 5, b(4) = 6, b(5) = 7, and a(1) = 8.

Mathematica

mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
h = 1; k = 3; a = {}; b = {1};
AppendTo[a, mex[Flatten[{a, b}], 1]];
Do[Do[AppendTo[b, mex[Flatten[{a, b}], Last[b]]], {k}];
  AppendTo[a, Last[b] + b[[1 + (Length[b] - 1)/k h]]], {500}];
Take[a, 200]  (* A304801 *)
Take[b, 200]  (* A304802 *)
(* Peter J. C. Moses, May 14 2008 *)

Crossrefs

Cf. A304799, A304801.

Sequence in context: A039179 A247914 A184518 * A288228 A002328 A039133

Adjacent sequences: A304799 A304800 A304801 * A304803 A304804 A304805

Keyword

nonn,easy

Author

Clark Kimberling, May 19 2018

Status

approved