A297464 Solution (a(n)) of the system of 4 complementary equations in Comments.

1, 4, 8, 11, 14, 18, 21, 24, 28, 31, 34, 38, 41, 44, 48, 51, 54, 58, 61, 64, 68, 71, 74, 78, 81, 84, 88, 91, 94, 98, 101, 104, 108, 111, 114, 118, 121, 124, 128, 131, 134, 138, 141, 144, 148, 151, 154, 158, 161, 164, 168, 171, 174, 178, 181, 184, 188, 191

Offset: 0

Author

Clark Kimberling, Apr 19 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 = A075326, the "anti-Fibonacci sequence." See A299409 for the "anti-tetranacci" sequences.

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
• Wieb Bosma, Rene Bruin, Robbert Fokkink, Jonathan Grube, Anniek Reuijl, and Thian Tromp, Using Walnut to solve problems from the OEIS, arXiv:2503.04122 [math.NT], 2025. See p. 8.

Crossrefs

Cf. A036554, A299634, A297465, 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 *)

Formula

a(n) = a(n-1) + a(n-3) - a(n-4) (conjectured).

d(n) = A275389(n) for n >= 0.