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A304802 Solution (b(n)) of the complementary equation a(n) = b(n) + b(3n); see Comments. 3
1, 3, 4, 5, 6, 7, 9, 10, 11, 12, 14, 15, 16, 18, 19, 20, 21, 23, 24, 25, 26, 28, 29, 30, 31, 32, 34, 35, 36, 37, 39, 40, 41, 43, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 59, 60, 61, 63, 64, 65, 66, 67, 69, 70, 71, 72, 74, 75, 76, 78, 79, 80, 81, 83 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

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

b(n) = least new,

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

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

MATHEMATICA

mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);

h = 1; 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]  (* A304801 *)

Take[b, 200]  (* A304802 *)

(* Peter J. C. Moses, May 14 2008 *)

CROSSREFS

Cf. A304799, A304801.

Sequence in context: A039179 A247914 A184518 * A288228 A002328 A039133

Adjacent sequences:  A304799 A304800 A304801 * A304803 A304804 A304805

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, May 19 2018

STATUS

approved

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Last modified May 12 06:54 EDT 2021. Contains 343820 sequences. (Running on oeis4.)