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A304808 Solution (b(n)) of the complementary equation a(n) = b(2n) + b(3n) ; see Comments. 3
1, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 15, 16, 17, 18, 19, 21, 22, 23, 24, 25, 26, 28, 29, 30, 31, 33, 34, 35, 36, 37, 38, 40, 41, 42, 43, 45, 46, 47, 48, 49, 50, 52, 53, 54, 55, 56, 58, 59, 60, 61, 63, 64, 65, 66, 67, 68, 70, 71, 72, 73, 74, 76, 77, 78, 79 (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(2n) + b(3n),

where "least new" means the least positive integer not yet placed.  Empirically, {a(n) - 6*n: n >= 0} = {2,3} and {5*b(n) - 6*n: n >= 0} = {4,5,6,7,8,9}.  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(3), we must have a(1) >= 7, so that b(1) = 3, b(2) = 4, b(3) = 5, b(4) = 6, and a(1) = 9.

MATHEMATICA

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

h = 2; 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]  (* A304807 *)

Take[b, 200]  (* A304808 *)

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

CROSSREFS

Cf. A304799, A304807.

Sequence in context: A304804 A077656 A039236 * A039180 A073071 A039134

Adjacent sequences:  A304805 A304806 A304807 * A304809 A304810 A304811

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, May 28 2018

STATUS

approved

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