%I #16 Jan 16 2019 11:13:14
%S 1,21,121,31,1121,221,131,11121,2121,1221,321,1131,231,111121,21121,
%T 12121,3121,11221,2221,1321,11131,2131,1231,331,1111121,211121,121121,
%U 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
%N Unique representation of nonnegative numbers by iterated tribonacci A, B and C sequences.
%C This representation is the tribonacci A000073 analog of the Wythoff representation of numbers (A189921 or A317208) for the Fibonacci case.
%C The complementary and disjoint sets A, B and C are given by the sequences A278040, A278039, and A278041, respectively.
%C 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)), ...
%C 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 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.
%H 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 The complementary and disjoint sequences A, B, C begin, for n >= 0:
%e 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 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 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 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 ---------------------------------------------------------------------------------
%e The ABC representations begin:
%e #(1) #(2) #(3) L(a(n))
%e a(n) A316715 A316716 A316717 A316714
%e n = 0: 1 B(0) = 0 1 0 0 1
%e n = 1: 21 A(B(0)) = 1 1 1 0 2
%e n = 2: 121 B(A(B(0))) = 2 2 1 0 3
%e n = 3: 31 C(B(0)) = 3 1 0 1 2
%e n = 4: 1121 B(B(A(B(0)))) = 4 3 1 0 4
%e n = 5: 221 A(A(B(0))) = 5 1 2 0 3
%e n = 6: 131 B(C(B(0))) = 6 2 0 1 3
%e n = 7: 11121 B(B(B(A(B(0))))) = 7 4 1 0 5
%e n = 8: 2121 A(B(A(B(0)))) = 8 2 2 0 4
%e n = 9: 1221 B(A(A(B(0)))) = 9 2 2 0 4
%e n = 10: 321 C(A(B(0))) = 10 1 1 1 3
%e n = 11: 1131 B(B(C(B(0)))) = 11 3 0 1 4
%e n = 12: 231 A(C(B(0))) = 12 1 1 1 3
%e n = 13: 111121 B(B(B(B(A(B(0)))))) = 13 5 1 0 6
%e n = 14: 21121 A(B(B(A(B(0))))) = 14 3 2 0 5
%e n = 15: 12121 B(A(B(A(B(0))))) = 15 3 2 0 5
%e n = 16: 3121 C(B(A(B(0)))) = 16 2 1 1 4
%e n = 17: 11221 B(B(A(A(B(0))))) = 17 3 2 0 5
%e n = 18: 2221 A(A(A(B(0)))) = 18 1 3 0 4
%e n = 19: 1321 B(C(A(B(0)))) = 19 2 1 1 4
%e n = 20: 11131 B(B(B(C(B(0))))) = 20 4 0 1 5
%e ...
%e ----------------------------------------------------------------------------
%Y Cf. A000073, A003144, A003145, A003146, A189921, A317208, A278040, A278039, A278041, A316714, A316715, A316716, A316717, A317208.
%K nonn,easy
%O 0,2
%A _Wolfdieter Lang_, Sep 11 2018