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