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A297466

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

3, 7, 10, 13, 17, 20, 23, 26, 30, 33, 37, 40, 43, 47, 50, 53, 56, 60, 63, 67, 70, 73, 77, 80, 83, 86, 90, 93, 97, 100, 103, 107, 110, 113, 116, 120, 123, 127, 130, 133, 137, 140, 143, 146, 150, 153, 157, 160, 163, 167, 170, 173, 176, 180, 183, 187, 190, 193

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OFFSET

0,1

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."

The linear recurrence with signature (1,0,0,0,0,0,0,0,1,-1) and g.f. (3 + 4*x + 3*x^2 + 3*x^3 + 4*x^4 + 3*x^5 + 3*x^6 + 3*x^7 + 4*x^8)/(1 - x - x^9 + x^10) first differs from this sequence at a(67). - Georg Fischer, Jul 17 2025

LINKS

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

EXAMPLE

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

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]];