A304814 Solution (b(n)) of the complementary equation a(n) = b(3n) + b(5n); see Comments.

1, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 18, 19, 20, 22, 23, 24, 25, 26, 27, 28, 30, 31, 32, 33, 34, 35, 36, 37, 38, 40, 41, 42, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 54, 55, 56, 58, 59, 60, 61, 62, 63, 64, 65, 67, 68, 69, 70, 71, 72, 73, 75, 76

Offset

0,2

Comments

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

b(n) = least new,

a(n) = b(3n) + b(5n),

where "least new" means the least positive integer not yet placed. Empirically, {a(n) - 8*n: n >= 0} = {2,3} and {7*b(n) - 8*n: n >= 0} = {7,8,9,10,11,12,13,14,15}. See A304799 for a guide to related sequences.

Links

Clark Kimberling, Table of n, a(n) for n = 0..9999

Example

b(0) = 1, so that a(0) = 2.  Since a(1) = b(3) + b(5), we must have a(1) >= 10, so that b(1) = 3, b(2) = 4, b(3) = 5, b(4) = 6, b(5) = 7, and a(1) = 12.

Mathematica

mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
h = 3; k = 5; 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]  (* A304813 *)
Take[b, 200]  (* A304814 *)
(* Peter J. C. Moses, May 14 2008 *)

Crossrefs

Cf. A304799, A304813.

Sequence in context: A105048 A026511 A321153 * A052404 A298112 A026417

Adjacent sequences: A304811 A304812 A304813 * A304815 A304816 A304817

Keyword

nonn,easy

Author

Clark Kimberling, May 30 2018

Status

approved