A304809 Solution (a(n)) of the complementary equation a(n) = b(2n) + b(4n) ; see Comments.

2, 10, 17, 23, 31, 38, 44, 52, 59, 65, 73, 80, 86, 94, 101, 107, 115, 122, 128, 136, 143, 149, 157, 164, 170, 178, 185, 191, 199, 206, 212, 220, 227, 233, 241, 248, 254, 262, 269, 275, 283, 290, 296, 304, 311, 317, 325, 332, 338, 346, 353, 359, 367, 374, 380

Offset

0,1

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..10000

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, A304810.

Sequence in context: A009387 A305093 A316753 * A304805 A214086 A127492

Adjacent sequences: A304806 A304807 A304808 * A304810 A304811 A304812

Keyword

nonn,easy

Author

Clark Kimberling, May 28 2018

Status

approved