OFFSET
0,1
COMMENTS
Define sequences a(n), b(n), c(n) recursively:
a(n) = least new;
b(n) = least new > = a(n) + n + 1;
c(n) = a(n) + b(n);
where "least new k" means the least positive integer not yet placed.
***
The sequences a,b,c partition the positive integers.
***
Let x = be the greatest solution of 1/x + 1/(x+1) + 1/(2x+1) = 1. Then
x = 1/3 + (2/3)*sqrt(7)*cos((1/3)*arctan((3*sqrt(111))/67))
LINKS
Clark Kimberling, Table of n, a(n) for n = 0..1000
EXAMPLE
n: 0 1 2 3 4 5 6 7 8 9 10
a: 1 2 6 8 9 11 12 14 17 19 22
b: 3 5 10 13 15 18 20 23 27 30 34
c: 4 7 16 21 24 29 32 37 44 49 56
MATHEMATICA
z = 200;
mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
a = {}; b = {}; c = {}; n = 0;
Do[{n++;
AppendTo[a,
mex[Flatten[{a, b, c}], If[Length[a] == 0, 1, Last[a]]]],
AppendTo[b, mex[Flatten[{a, b, c}], Last[a] + n + 1]],
AppendTo[c, Last[a] + Last[b]]}, {z}];
(* Peter J. C. Moses, Apr 23 2018 *)
Take[a, 100] (* A297469 *)
Take[b, 100] (* A299533 *)
Take[c, 100] (* A299423 *)
(* Peter J. C. Moses, Apr 23 2018 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, May 01 2018
STATUS
approved