A299636 Solution (c(n)) of the system of 3 complementary equations in Comments.

3, 9, 16, 19, 22, 28, 36, 41, 48, 57, 61, 66, 74, 77, 83, 89, 94, 97, 101, 103, 108, 115, 121, 130, 133, 136, 139, 146, 154, 157, 161, 166, 171, 178, 183, 191, 200, 209, 214, 217, 222, 229, 238, 241, 244, 248, 253, 257, 265, 275, 282, 290, 295, 298, 306, 317

Offset: 0

Author

Clark Kimberling, Apr 17 2018

Keywords

• nonn
• easy

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).

Examples

			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
		

Links

• Clark Kimberling, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A299634, A299635.

Programs

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 *)