A304811 Solution (a(n)) of the complementary equation a(n) = b(2n) + b(5n) ; see Comments.

2, 11, 19, 26, 34, 43, 51, 58, 67, 75, 83, 91, 99, 107, 114, 123, 131, 138, 146, 155, 163, 170, 179, 187, 195, 203, 211, 219, 226, 235, 243, 250, 258, 267, 275, 282, 291, 299, 306, 314, 323, 331, 338, 347, 355, 363, 370, 379, 387, 394, 403, 411, 418, 426

Offset

0,1

Comments

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

b(n) = least new,

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

Mathematica

mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
h = 2; 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]  (* A304811 *)
Take[b, 200]  (* A304812 *)
(* Peter J. C. Moses, May 14 2008 *)

Crossrefs

Cf. A304799, A304812.

Sequence in context: A019375 A078784 A117196 * A064739 A112860 A152312

Adjacent sequences: A304808 A304809 A304810 * A304812 A304813 A304814

Keyword

nonn,easy

Author

Clark Kimberling, May 30 2018

Status

approved