A298170 Solution (b(n)) of the system of 3 complementary equations in Comments.

2, 6, 8, 11, 14, 19, 22, 24, 26, 30, 32, 38, 41, 42, 44, 49, 51, 54, 55, 59, 66, 69, 71, 72, 77, 83, 84, 86, 90, 92, 93, 96, 99, 101, 109, 112, 113, 116, 119, 121, 122, 130, 131, 138, 140, 143, 147, 151, 152, 154, 156, 158, 161, 162, 165, 170, 174, 181, 184

Offset: 0

Author

Clark Kimberling, Apr 25 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 > = 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))

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

(The same limits occur in A298868 and A297469.)

Examples

			n:   0   1   2   3   4    5   6   7   8   9  10
a:   1   4   5   7   9   12  15  16  17  20  21
b:   2   6   8  11   14  19  22  24  26  30  32
c:   3  10  13  18   23  31  37  40  43  50  53
		

Links

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

Crossrefs

Cf. A299634, A298868, A297469, A297838, A298418.

Programs

Mathematica

z=200;
mex[list_,start_]:=(NestWhile[#+1&,start,MemberQ[list,#]&]);
a={1};b={2};c={3};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}];
Take[a,100] (* A297838 *)
Take[b,100] (* A298170 *)
Take[c,100] (* A298418 *)
(* Peter J. C. Moses, Apr 23 2018 *)