A304800 Solution (b(n)) of the complementary equation a(n) = b(n) + b(2n); see Comments.

1, 3, 4, 5, 6, 8, 9, 11, 12, 13, 15, 16, 17, 19, 20, 21, 22, 24, 25, 27, 28, 29, 30, 32, 33, 35, 36, 37, 39, 40, 41, 42, 44, 45, 47, 48, 49, 51, 52, 53, 54, 56, 57, 58, 60, 61, 63, 64, 65, 67, 68, 69, 70, 72, 73, 75, 76, 77, 78, 80, 81, 83, 84, 85, 87, 88

Offset

0,2

Comments

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

b(n) = least new,

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

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

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

Mathematica

mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
h = 1; k = 2; 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]  (* A304799 *)
Take[b, 200]  (* A304800 *)
(* Peter J. C. Moses, May 14 2008 *)

Crossrefs

Cf. A304799.

Sequence in context: A047428 A218784 A039066 * A197911 A298007 A026363

Adjacent sequences: A304797 A304798 A304799 * A304801 A304802 A304803

Keyword

nonn,easy

Author

Clark Kimberling, May 19 2018

Status

approved