A304815 Solution (a(n)) of the complementary equation a(n) = b(4n) + b(5n); see Comments.

2, 13, 22, 33, 43, 53, 63, 72, 83, 92, 103, 112, 123, 133, 143, 153, 163, 173, 182, 193, 203, 213, 223, 233, 243, 253, 263, 272, 283, 292, 303, 313, 323, 333, 342, 353, 362, 373, 382, 393, 403, 413, 423, 432, 443, 452, 463, 472, 483, 493, 503, 513, 522, 533

Offset

0,1

Comments

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

b(n) = least new,

a(n) = b(4n) + 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} = {8,9,10,11,12,13,14,15,16,17}. 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(4) + b(5), we must have a(1) >= 11, so that b(1) = 3, b(2) = 4, b(3) = 5, b(4) = 6, b(5) = 7, and a(1) = 13.

Mathematica

mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
h = 4; 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]  (* A304815 *)
Take[b, 200]  (* A304816 *)
(* Peter J. C. Moses, May 14 2008 *)

Crossrefs

Cf. A304799, A304816.

Sequence in context: A127485 A061385 A366208 * A156179 A090519 A018540

Adjacent sequences: A304812 A304813 A304814 * A304816 A304817 A304818

Keyword

nonn,easy

Author

Clark Kimberling, May 30 2018

Status

approved