OFFSET

0,2

COMMENTS

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

b(n) = least new,

a(n) = b(n) + b(4n),

where "least new" means the least positive integer not yet placed. Empirically, {a(n) - 5*n: n >= 0} = {2,3} and {4*b(n) - 5*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(1) + b(4), 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) = 9.

MATHEMATICA

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

h = 1; k = 4; 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] (* A304803 *)

Take[b, 200] (* A304804 *)

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

CROSSREFS

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, May 28 2018

STATUS

approved