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A304801 Solution (a(n)) of the complementary equation a(n) = b(n) + b(3n); see Comments. 3
2, 8, 13, 17, 22, 27, 33, 38, 42, 47, 53, 58, 62, 68, 73, 77, 82, 88, 93, 97, 102, 108, 113, 117, 122, 127, 133, 138, 142, 147, 153, 158, 162, 168, 173, 177, 182, 188, 193, 198, 202, 207, 213, 218, 222, 227, 233, 238, 242, 248, 253, 257, 262, 267, 273, 278 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,1

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, A304802.

Sequence in context: A136738 A288227 A247520 * A184519 A156245 A247783

Adjacent sequences:  A304798 A304799 A304800 * A304802 A304803 A304804

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, May 19 2018

STATUS

approved

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Last modified May 12 07:21 EDT 2021. Contains 343821 sequences. (Running on oeis4.)