OFFSET
0,2
COMMENTS
Define sequences a(n) and b(n) recursively, starting with b(0) = 1:
b(n) = least new;
a(n) = b(n) + b(floor(n/2)),
where "least new k" means the least positive integer not yet placed.
***
Conjectures: a(n)/n -> 5/2 and -1 <= 5/2 - a(n)/n <= 2 for n >= 1;
b(n)/n -> 5/3 and -1 <= 5/3 - b(n)/n <= 2 for n >= 1.
EXAMPLE
a(0) = b(0) + b(0) = 2;
a(1) = b(1) + b(2) >= 3 + 4, so that b(2) = 3, b(2) = 4, b(3) = 5, b(4) = 6, and a(1) = 7.
MATHEMATICA
mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
z = 1000; a = {}; b = {1};
Do[AppendTo[a, Last[b] + b[[Floor[(1 + Length[b])/2]]]];
AppendTo[b, mex[Flatten[{a, b}], 1]], {z}];
Take[a, 100] (* A304451 *)
Take[b, 100] (* A304452 *)
(* Peter J. C. Moses, May 10 2018 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, May 16 2018
STATUS
approved