A304810 Solution (b(n)) of the complementary equation a(n) = b(2n) + b(4n) ; see Comments.

1, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 18, 19, 20, 21, 22, 24, 25, 26, 27, 28, 29, 30, 32, 33, 34, 35, 36, 37, 39, 40, 41, 42, 43, 45, 46, 47, 48, 49, 50, 51, 53, 54, 55, 56, 57, 58, 60, 61, 62, 63, 64, 66, 67, 68, 69, 70, 71, 72, 74, 75, 76, 77, 78

Offset

0,2

Comments

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

b(n) = least new,

a(n) = b(2n) + b(4n),

where "least new" means the least positive integer not yet placed. Empirically, {a(n) - 7*n: n >= 0} = {2,3} and {6*b(n) - 7*n: n >= 0} = {5,6,7,8,9,10,11}.. See A304799 for a guide to related sequences.

Links

Clark Kimberling, Table of n, a(n) for n = 0..9999

Example

b(0) = 1, so that a(0) = 2.  Since a(1) = b(2) + b(4), 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) = 10.

Mathematica

mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
h = 2; 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]  (* A304809 *)
Take[b, 200]  (* A304810 *)
(* Peter J. C. Moses, May 14 2008 *)

Crossrefs

Cf. A304799, A304809.

Sequence in context: A047565 A026466 A304806 * A026469 A258187 A039237

Adjacent sequences: A304807 A304808 A304809 * A304811 A304812 A304813

Keyword

nonn,easy

Author

Clark Kimberling, May 30 2018

Status

approved