A316713 Unique representation of nonnegative numbers by iterated tribonacci A, B and C sequences.

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

Author

Wolfdieter Lang, Sep 11 2018

Keywords

• nonn
• easy

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.

Examples

			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))
           a(n)                             A316715 A316716 A316717 A316714
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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Links

• Wolfdieter Lang, The Tribonacci and ABC Representations of Numbers are Equivalent, arXiv preprint arXiv:1810.09787 [math.NT], 2018.

Crossrefs

Cf. A000073, A003144, A003145, A003146, A189921, A317208, A278040, A278039, A278041, A316714, A316715, A316716, A316717, A317208.