A319968 a(n) = A003145(A003145(n)).

6, 19, 30, 43, 50, 63, 74, 87, 100, 111, 124, 131, 144, 155, 168, 179, 192, 199, 212, 223, 236, 249, 260, 273, 280, 293, 304, 317, 324, 337, 348, 361, 374, 385, 398, 405, 418, 429, 442, 453, 466, 473, 486, 497, 510, 523, 534, 547, 554, 567, 578, 591, 604, 615, 628, 635, 648, 659, 672, 683, 696, 703, 716, 727, 740, 753

Offset: 1

Author

N. J. A. Sloane, Oct 05 2018

Keywords

• nonn
• easy

Comments

By analogy with the Wythoff compound sequences A003622 etc., the nine compounds of A003144, A003145, A003146 might be called the tribonacci compound sequences. They are A278040, A278041, and A319966-A319972.

From Wolfdieter Lang, Oct 19 2018: (Start)

In another version with the tribonacci word TriWord = A080843 (written as a sequence which has offset 0) and the positions of 0, 1 and 2 given by the B = A278039, A = A278040 and C = A278041 numbers, respectively, the present sequence (with offset 0) gives the smaller of the B-number pairs (B(k), B(k+1)) with B(k+1) = B(k) + 1 for some k >= 0 (named tribonacci B0-numbers), ordered increasingly.

The B-numbers A278039 come in three disjoint and complementary types, called B0-, B1- and B2-numbers. They are defined by the indices k of pairs of consecutive entries TriWord(k), Triword(k+1) depending on their values 0, 0 or 0, 1 or 0, 2 for the B0- or B1- or B2-numbers, respectively.

The B0-numbers are a(n+1) = 2*C(n) - n = A(A(n)) + 1; the B1-numbers are B1(n) = A(n) - 1; and the B2-numbers are B2(n) = C(n) - 1, all for n >= 0.

B0(n) + 1 = B1(A(n)+1), B1(n) + 1 = A(n) and B2(n) + 1 = C(n).

(End)

(a(n)) equals the positions of the word baa in the tribonacci word t = abacabaa..., fixed point of the morphism a->ab, b->ac, c->a. This follows from the fact that the word aa is always preceded in t by the letter b, and the formula BB = AC-1, where A := A003144, B := A003145, C := A003146. - Michel Dekking, Apr 09 2019

Examples

			From _Wolfdieter Lang_, Oct 19 2018: (Start)
The TriWord A080843 starts: 0, 1, 0, 2, 0, 1, 0, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 0, 1, 0, 2, ... (offset 0)
The trisection of the B-numbers A278039 (indices for 0 in TriWord) begins:
n :  0   1   2   3   4   5   6   7    8    9   10   11   12   13   14   15   16 ...
B0:  6  19  30  43  50  63  74  87  100  111  124  131  144  155  168  179  192 ...
B1:  0   4   7  11  13  17  20  24   28   31   35   37   41   44   48   51   55 ...
B2:  2   9  15  22  26  33  39  46   53   59   66   70   77   83   90   96  103 ...
------------------------------------------------------------------------------------
(End)
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..10000
• Elena Barcucci, Luc Belanger and Srecko Brlek, On tribonacci sequences, Fib. Q., 42 (2004), 314-320. Compare page 318.
• L. Carlitz, R. Scoville and V. E. Hoggatt, Jr., Fibonacci representations of higher order, Fib. Quart., 10 (1972), 43-69, Theorem 13.
• Wolfdieter Lang, The Tribonacci and ABC Representations of Numbers are Equivalent, arXiv preprint arXiv:1810.09787 [math.NT], 2018.

Crossrefs

Cf. A003144, A003145, A003146, A003622, A080843, A278039, A278040, A278041, and A319966-A319972.

Formula

a(n) = A003145(A003145(n)), for n >= 1.

a(n) = B0(n-1) = 2*A003146(n) - (n+1) = 2*A278041(n-1) - (n-1) = A278040(A278040(n-1)) + 1, for n >= 1. For B0 see a comment above and the example. - Wolfdieter Lang, Oct 19 2018

a(n+1) = B(C(n)) = B(C(n) + 1) - 1 = 2*(A(n) + B(n)) + n + 4, for n >= 0, where B = A278039, C = A278041 and A = A278040. For a proof see the W. Lang link in A278040, Proposition 9, eq. (53). - Wolfdieter Lang, Dec 13 2018

a(n) = 2*(A003144(n) + A003145(n)) + n - 1, n >= 1. [Rewriting a formula of the precedimg entry]. - Wolfdieter Lang, Apr 11 2019

Extensions

More terms from Joerg Arndt, Oct 15 2018