A304801 Solution (a(n)) of the complementary equation a(n) = b(n) + b(3n); see Comments.

2, 8, 13, 17, 22, 27, 33, 38, 42, 47, 53, 58, 62, 68, 73, 77, 82, 88, 93, 97, 102, 108, 113, 117, 122, 127, 133, 138, 142, 147, 153, 158, 162, 168, 173, 177, 182, 188, 193, 198, 202, 207, 213, 218, 222, 227, 233, 238, 242, 248, 253, 257, 262, 267, 273, 278

Offset

0,1

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, A304802.

Sequence in context: A136738 A288227 A247520 * A184519 A156245 A247783

Adjacent sequences: A304798 A304799 A304800 * A304802 A304803 A304804

Keyword

nonn,easy

Author

Clark Kimberling, May 19 2018

Status

approved