A304805 Solution (a(n)) of the complementary equation a(n) = b(n) + b(5n) ; see Comments.

2, 10, 17, 24, 31, 37, 44, 51, 59, 66, 73, 79, 86, 93, 101, 108, 115, 121, 128, 135, 143, 150, 157, 164, 170, 177, 185, 192, 199, 206, 212, 220, 227, 234, 241, 247, 254, 262, 269, 276, 283, 289, 296, 304, 311, 318, 325, 331, 338, 345, 353, 360, 367, 373, 380

Offset

0,1

Comments

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

b(n) = least new,

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

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

Mathematica

mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
h = 1; 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]  (* A304805 *)
Take[b, 200]  (* A304806 *)
(* Peter J. C. Moses, May 14 2008 *)

Crossrefs

Cf. A304799, A304806.

Sequence in context: A305093 A316753 A304809 * A214086 A127492 A258974

Adjacent sequences: A304802 A304803 A304804 * A304806 A304807 A304808

Keyword

nonn,easy

Author

Clark Kimberling, May 28 2018

Status

approved