OFFSET
0,1
COMMENTS
Define sequences a(n), b(n), c(n) recursively, starting with a(0) = 1, b(0) = 2:
a(n) = least new k >= 2*b(n-1);
b(n) = least new k;
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 = 11/6. Conjectures:
a(n) - 2*n*x = 0 for infinitely many n;
b(n) - n*x = 0 for infinitely many n;
c(n) - 3*n*x = 0 for infinitely many n;
(a(n) - 2*n*x) is unbounded below and above;
(b(n) - n*x) is unbounded below and above;
(c(n) - 3*n*x) is unbounded below and above;
***
Let d(a), d(b), d(c) denote the respective difference sequences. Conjectures:
12 occurs infinitely many times in d(a); 6 occurs infinitely many times in d(b);
2 occurs infinitely many times in d(c).
LINKS
Clark Kimberling, Table of n, a(n) for n = 0..1000
EXAMPLE
n: 0 1 2 3 4 5 6 7 8 9
a: 1 4 10 12 14 17 23 26 30 37
b: 2 5 6 7 8 11 13 15 18 20
c: 3 9 16 19 22 28 36 41 48 57
MATHEMATICA
z = 1000;
mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
a = {1}; b = {2}; c = {}; AppendTo[c, Last[a] + Last[b]];
Do[{
AppendTo[a, mex[Flatten[{a, b, c}], 2 Last[b]]],
AppendTo[b, mex[Flatten[{a, b, c}], 1]],
AppendTo[c, Last[a] + Last[b]]}, {z}];
Take[a, 100] (* A299634 *)
Take[b, 100] (* A299635 *)
Take[c, 100] (* A299636 *)
(* Peter J. C. Moses, Apr 08 2018 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Apr 17 2018
STATUS
approved