A297465 Solution (b(n)) of the system of 4 complementary equations in Comments.

2, 5, 9, 12, 15, 19, 22, 25, 29, 32, 35, 39, 42, 45, 49, 52, 55, 59, 62, 65, 69, 72, 76, 79, 82, 85, 89, 92, 95, 99, 102, 105, 109, 112, 115, 119, 122, 125, 129, 132, 135, 139, 142, 145, 149, 152, 155, 159, 162, 166, 169, 172, 175, 179, 182, 185, 189, 192

Offset: 0

Author

Clark Kimberling, Apr 22 2018

Keywords

• nonn
• easy

Comments

Define sequences a(n), b(n), c(n), d(n) recursively, starting with a(0) = 1, b(0) = 2, c(0) = 3;:

a(n) = least new;

b(n) = least new;

c(n) = least new;

d(n) = a(n) + b(n) + c(n);

where "least new k" means the least positive integer not yet placed.

***

Conjecture: for all n >= 0,

0 <= 10n - 6 - 3 a(n) <= 2

0 <= 10n - 2 - 3 b(n) <= 3

0 <= 10n +1 - 3 c(n) <= 3

0 <= 10n - 3 - d(n) <= 2

***

The sequences a,b,c,d partition the positive integers. The sequence d can be called the "anti-tribonacci sequence"; viz., if sequences a and b are defined as above, and c(n) is defined by c(n) = a(n) + b(n), then the resulting system of 3 complementary sequences gives c = A036554, the "anti-Fibonacci sequence."

Examples

			n:   0    1    2    3    4    5    6    7    8    9
a:   1    4    8   11   14   18   21   24   28   31
b:   2    5    9   12   15   19   22   25   29   32
c:   3    7   10   13   17   20   23   26   30   33
d:   6   16   27   36   46   57   66   75   87   96
		

Links

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

Crossrefs

Cf. A036554, A299634, A297464, A297466, A265389.

Programs

Mathematica

z = 400;
mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
a = {1}; b = {2}; c = {3}; d = {}; AppendTo[d, Last[a] + Last[b] + Last[c]];
Do[{AppendTo[a, mex[Flatten[{a, b, c, d}], 1]],
   AppendTo[b, mex[Flatten[{a, b, c, d}], 1]],
   AppendTo[c, mex[Flatten[{a, b, c, d}], 1]],
   AppendTo[d, Last[a] + Last[b] + Last[c]]}, {z}];
Take[a, 100]  (* A297464 *)
Take[b, 100]  (* A297465 *)
Take[c, 100]  (* A297466 *)
Take[d, 100]  (* A265389 *)