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A319195 Irregular triangle with the unique representation of positive integers in the tribonacci ABC-representation.

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%I A319195 #16 Apr 06 2020 18:58:13
%S A319195 1,0,0,1,0,2,0,0,0,1,0,1,1,0,0,2,0,0,0,0,1,0,1,0,1,0,0,1,1,0,2,1,0,0,
%T A319195 0,2,0,1,2,0,0,0,0,0,1,0,1,0,0,1,0,0,1,0,1,0,2,0,1,0,0,0,1,1,0,1,1,1,
%U A319195 0,0,2,1,0,0,0,0,2,0,1,0,2,0,0,1,2,0,2,2,0,0,0,0,0,0,1,0,1,0,0,0,1,0
%N A319195 Irregular triangle with the unique representation of positive integers in the tribonacci ABC-representation.
%C A319195 The row length L(n) of this irregular triangle is A316714(n), n >= 1.
%C A319195 This representation is based on the complementary and disjoint sets A, B and C given by the sequences A278040, A278039 and A278041, respectively. In the present notation sequences A, B and C are denoted by 1, 0 and 2.
%C A319195 The numbers are represented by iterations of these sequences always starting with B(0) = 0 (in analogy to the Wythoff B sequence in the Fibonacci case). Uniqueness requires that the representations end in A(B(0)) or C(B(0)).
%C A319195 B^[k](0) (k-fold iterations) for k >= 2 are forbidden. One could represent the number 0 by B(0), but this is not done here, because it is found that the ABC-representations of positive numbers is equivalent to the tribonacci representation of positive numbers given in A278038 for n >= 1 (n = 0 is not represented by T(1) = A000073(1) = 0. This representation uses the tribonacci numbers {T(k)}_{k >= 3} = {1, 2, 4, 7, 13, ...} for uniqueness reason).
%C A319195 For this table the operation of sequences A, B and C is denoted by 1, 0 and 2, respectively, and the brackets and the final argument (0) of B(0) are not recorded. E.g., A(B(C(B(0)))) is written as 1020.
%C A319195 Another form of this table is given in A316713 where A, B and C are denoted 2, 1 and 3, respectively.
%C A319195 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.
%C A319195 The present representation is the analog to the Wythoff representation of positive numbers (A189921 or A317208) using the Wythoff A and B sequences A000201 and A001950, respectively.
%C A319195 The number length of the ABC-representation of n >= 1 is L(n) = A316714(n). The number of 0's (B's), 1's (A's) and 2's (C's) of the representation of n is A316715, A316716, A316717.
%H A319195 Wolfdieter Lang, <a href="https://arxiv.org/abs/1810.09787">The Tribonacci and ABC Representations of Numbers are Equivalent</a>, arXiv preprint arXiv:1810.09787 [math.NT], 2018.
%e A319195 The complementary and disjoint sequences A, B, C begin, for n >= 0:
%e A319195 n: 0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15  16  17  18  19  20  21  22 ...
%e A319195 A: 1  5  8 12 14 18 21 25 29 32 36 38 42 45 49 52  56  58  62  65  69  73  76 ...
%e A319195 B: 0  2  4  6  7  9 11 13 15 17 19 20 22 24 26 28  30  31  33  35  37  39  41 ...
%e A319195 C: 3 10 16 23 27 34 40 47 54 60 67 71 78 84 91 97 104 108 115 121 128 135 141 ...
%e A319195 ---------------------------------------------------------------------------------
%e A319195 The ABC-representation of the positive integers begins:
%e A319195                                               #(1)   #(2)    #(3)    L(n)
%e A319195                                            A316715 A316716 A316717 A316714
%e A319195 n = 1:      10               A(B(0)) =  1      1      1       0       2
%e A319195 n = 2:     010            B(A(B(0))) =  2      2      1       0       3
%e A319195 n = 3:      20               C(B(0)) =  3      1      0       1       2
%e A319195 n = 4:    0010         B(B(A(B(0)))) =  4      3      1       0       4
%e A319195 n = 5:     110            A(A(B(0))) =  5      1      2       0       3
%e A319195 n = 6:     020            B(C(B(0))) =  6      2      0       1       3
%e A319195 n = 7:   00010      B(B(B(A(B(0))))) =  7      4      1       0       5
%e A319195 n = 8:    1010         A(B(A(B(0)))) =  8      2      2       0       4
%e A319195 n = 9:    0110         B(A(A(B(0)))) =  9      2      2       0       4
%e A319195 n = 10:    210            C(A(B(0))) = 10      1      1       1       3
%e A319195 n = 11:   0020         B(B(C(B(0)))) = 11      3      0       1       4
%e A319195 n = 12:    120            A(C(B(0))) = 12      1      1       1       3
%e A319195 n = 13: 000010   B(B(B(B(A(B(0)))))) = 13      5      1       0       6
%e A319195 n = 14:  10010      A(B(B(A(B(0))))) = 14      3      2       0       5
%e A319195 n = 15:  01010      B(A(B(A(B(0))))) = 15      3      2       0       5
%e A319195 n = 16:   2010         C(B(A(B(0)))) = 16      2      1       1       4
%e A319195 n = 17:  00110      B(B(A(A(B(0))))) = 17      3      2       0       5
%e A319195 n = 18:   1110         A(A(A(B(0)))) = 18      1      3       0       4
%e A319195 n = 19:   0210         B(C(A(B(0)))) = 19      2      1       1       4
%e A319195 n = 20:  00020      B(B(B(C(B(0))))) = 20      4      0       1       5
%e A319195 ...
%Y A319195 Cf. A000073, A000201, A001950, A189921, A003144, A003145, A003146, A278038, A278040, A278039, A278041, A316713, A316714, A316715, A316716, A316717, A317208.
%K A319195 nonn,tabf,easy
%O A319195 1,6
%A A319195 _Wolfdieter Lang_, Sep 13 2018