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
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, May 28 2018
STATUS
approved