A304803 Solution (a(n)) of the complementary equation a(n) = b(n) + b(4n); see Comments.

2, 9, 15, 21, 26, 32, 38, 45, 51, 56, 62, 68, 75, 81, 87, 92, 98, 105, 111, 117, 122, 129, 135, 141, 146, 152, 159, 165, 171, 176, 182, 189, 195, 201, 206, 212, 218, 225, 231, 236, 242, 248, 255, 261, 267, 272, 279, 285, 291, 297, 302, 309, 315, 321, 326

Offset

0,1

Comments

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

b(n) = least new,

a(n) = b(n) + b(4n),

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

Mathematica

mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
h = 1; k = 4; 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]  (* A304803 *)
Take[b, 200]  (* A304804 *)
(* Peter J. C. Moses, May 14 2008 *)

Crossrefs

Cf. A304799, A304804.

Sequence in context: A083288 A272044 A354972 * A184531 A063105 A215035

Adjacent sequences: A304800 A304801 A304802 * A304804 A304805 A304806

Keyword

nonn,easy

Author

Clark Kimberling, May 19 2018

Status

approved