A298870 Solution (c(n)) of the system of 3 complementary equations in Comments.

3, 9, 13, 18, 23, 30, 35, 39, 44, 49, 55, 62, 65, 69, 75, 80, 84, 88, 97, 102, 108, 112, 116, 123, 129, 132, 138, 143, 145, 150, 155, 162, 169, 175, 179, 183, 187, 193, 199, 204, 211, 218, 225, 228, 231, 235, 240, 246, 249, 255, 259, 263, 270, 277, 282, 288

Offset: 0

Author

Clark Kimberling, Apr 18 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;

b(n) = least new k >= a(n) + n;

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))

x = 2.07816258732933084676..., and a(n)/n - > x, b(n)/n -> x+1, and c(n)/n - > 2x+1.

Examples

			n:   0    1    2    3    4    5    6    7    8    9
a:   1    4    6    8   11   14   15   17   19   21
b:   2    5    7   10   12   16   20   22   25   28
c:   3    9   13   18   23   30   35   39   44   49
		

Links

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

Crossrefs

Cf. A299634, A298868, A298869.

Programs

Mathematica

z = 400;
mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
a = {1}; b = {2}; c = {}; AppendTo[c, Last[a] + Last[b]]; n = 0;
Do[{n++, AppendTo[a, mex[Flatten[{a, b, c}], 1]],
   AppendTo[b, mex[Flatten[{a, b, c}], a[[n]] + n]],
   AppendTo[c, Last[a] + Last[b]]}, {z}];
Take[a, 100] (* A298868 *)
Take[b, 100] (* A298869 *)
Take[c, 100] (* A298870 *)
(* Peter J. C. Moses, Apr 08 2018 *)