OFFSET
0,2
COMMENTS
This representation is the tribonacci A000073 analog of the Wythoff representation of numbers (A189921 or A317208) for the Fibonacci case.
The complementary and disjoint sets A, B and C are given by the sequences A278040, A278039, and A278041, respectively.
The present representation uses 1 for B, 2 for A and 3 for C numbers. The brackets for sequence iteration and the final argument 0 have to be added. E.g.: a(0) = 1 for B(1), a(1) = 21 for A(B(0)), a(2) = 121 for B(A(B(0))), a(3) = 31 for C(B(0)), ...
LINKS
Wolfdieter Lang, The Tribonacci and ABC Representations of Numbers are Equivalent, arXiv preprint arXiv:1810.09787 [math.NT], 2018.
EXAMPLE
The complementary and disjoint sequences A, B, C begin, for n >= 0:
n: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 ...
A: 1 5 8 12 14 18 21 25 29 32 36 38 42 45 49 52 56 58 62 65 69 73 76 ...
B: 0 2 4 6 7 9 11 13 15 17 19 20 22 24 26 28 30 31 33 35 37 39 41 ...
C: 3 10 16 23 27 34 40 47 54 60 67 71 78 84 91 97 104 108 115 121 128 135 141 ...
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The ABC representations begin:
#(1) #(2) #(3) L(a(n))
n = 0: 1 B(0) = 0 1 0 0 1
n = 1: 21 A(B(0)) = 1 1 1 0 2
n = 2: 121 B(A(B(0))) = 2 2 1 0 3
n = 3: 31 C(B(0)) = 3 1 0 1 2
n = 4: 1121 B(B(A(B(0)))) = 4 3 1 0 4
n = 5: 221 A(A(B(0))) = 5 1 2 0 3
n = 6: 131 B(C(B(0))) = 6 2 0 1 3
n = 7: 11121 B(B(B(A(B(0))))) = 7 4 1 0 5
n = 8: 2121 A(B(A(B(0)))) = 8 2 2 0 4
n = 9: 1221 B(A(A(B(0)))) = 9 2 2 0 4
n = 10: 321 C(A(B(0))) = 10 1 1 1 3
n = 11: 1131 B(B(C(B(0)))) = 11 3 0 1 4
n = 12: 231 A(C(B(0))) = 12 1 1 1 3
n = 13: 111121 B(B(B(B(A(B(0)))))) = 13 5 1 0 6
n = 14: 21121 A(B(B(A(B(0))))) = 14 3 2 0 5
n = 15: 12121 B(A(B(A(B(0))))) = 15 3 2 0 5
n = 16: 3121 C(B(A(B(0)))) = 16 2 1 1 4
n = 17: 11221 B(B(A(A(B(0))))) = 17 3 2 0 5
n = 18: 2221 A(A(A(B(0)))) = 18 1 3 0 4
n = 19: 1321 B(C(A(B(0)))) = 19 2 1 1 4
n = 20: 11131 B(B(B(C(B(0))))) = 20 4 0 1 5
...
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CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Sep 11 2018
STATUS
approved