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A316713
Unique representation of nonnegative numbers by iterated tribonacci A, B and C sequences.
5
1, 21, 121, 31, 1121, 221, 131, 11121, 2121, 1221, 321, 1131, 231, 111121, 21121, 12121, 3121, 11221, 2221, 1321, 11131, 2131, 1231, 331, 1111121, 211121, 121121, 31121, 112121, 22121, 13121, 111221, 21221, 12221, 3221, 11321, 2321, 111131, 21131, 12131, 3131, 11231, 2231, 1331, 11111121, 2111121, 1211121, 311121, 1121121, 221121, 131121, 1112121, 212121, 122121, 32121, 113121, 23121, 1111221, 211221, 121221, 31221, 112221, 22221, 13221, 111321, 21321, 12321
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)), ...
An equivalent such representation is given by A317206 using different complementary sequences A, B and C, related to our B = A278039, A = A278040, and C = A278041: A(n) = A003144(n) = A278039(n-1) + 1, B(n) = A003145(n) = A278040(n-1) + 1, C(n) = A003146(n) = A278041(n-1) + 1 with n >= 1.
The length of the string a(n) is A316714(n). The number of B, A and C sequences used for the ABC-representation of n (that is the number of 1s, 2s and 3s of a(n)) is A316715, A316716 and A316717, respectively.
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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KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Sep 11 2018
STATUS
approved