A304807 Solution (a(n)) of the complementary equation a(n) = b(2n) + b(3n) ; see Comments.

2, 9, 14, 20, 27, 32, 39, 44, 51, 57, 62, 69, 75, 81, 87, 92, 99, 105, 111, 117, 122, 129, 134, 140, 147, 152, 159, 164, 170, 177, 182, 189, 194, 200, 207, 212, 219, 225, 231, 237, 242, 249, 255, 260, 267, 272, 279, 285, 290, 297, 302, 309, 314, 320, 327

Offset

0,1

Comments

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

b(n) = least new,

a(n) = b(2n) + b(3n),

where "least new" means the least positive integer not yet placed. Empirically, {a(n) - 6*n: n >= 0} = {2,3} and {5*b(n) - 6*n: n >= 0} = {4,5,6,7,8,9}. 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(2) + b(3), we must have a(1) >= 7, so that b(1) = 3, b(2) = 4, b(3) = 5, b(4) = 6, and a(1) = 9.

Mathematica

mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
h = 2; 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]  (* A304807 *)
Take[b, 200]  (* A304808 *)
(* Peter J. C. Moses, May 14 2008 *)

Crossrefs

Cf. A304799, A304808.

Sequence in context: A274133 A288483 A353308 * A045920 A242466 A071344

Adjacent sequences: A304804 A304805 A304806 * A304808 A304809 A304810

Keyword

nonn,easy

Author

Clark Kimberling, May 28 2018

Status

approved