A276789 First differences of A003145.

4, 3, 4, 2, 4, 3, 4, 4, 3, 4, 2, 4, 3, 4, 3, 4, 2, 4, 3, 4, 4, 3, 4, 2, 4, 3, 4, 2, 4, 3, 4, 4, 3, 4, 2, 4, 3, 4, 3, 4, 2, 4, 3, 4, 4, 3, 4, 2, 4, 3, 4, 4, 3, 4, 2, 4, 3, 4, 3, 4, 2, 4, 3, 4, 4, 3, 4, 2, 4, 3, 4, 2, 4, 3, 4, 4, 3, 4, 2, 4, 3, 4, 3, 4, 2, 4, 3, 4, 4, 3, 4, 2, 4, 3, 4, 3, 4, 2, 4, 3

Offset: 1

Author

N. J. A. Sloane, Oct 14 2016

Keywords

• nonn
• easy

Comments

The sequence of first differences of A003146 (the third of the trio A003144, A003145, A003146) is equal to A276788 + A276789 + 1.

Also first differences of A278040.- Wolfdieter Lang, Dec 05 2018

From Michel Dekking, Mar 21 2019: (Start)

(a(n)) is a fixed point of the tribonacci morphism on the alphabet {4,3,2}, i.e., the morphism given by 4 -> 43, 3 -> 42, 2 -> 4.

To see this, let U := baca, V := baa, W := ba be the three return words of the letter b in the tribonacci word

x = abacabaabacaba... = aUVUW...

[See Justin & Vuillon (2000) for definition of return word. - N. J. A. Sloane, Sep 23 2019]

Under the tribonacci morphism tau given by

tau(a) = ab, tau(b) = ac, tau(c) = a

one obtains

tau(U) = acabaab = b^{-1} UV b,

tau(V) = acabab = b^{-1} UW b,

tau(W) = acab = b^{-1} U b,

which is conjugate to the tribonacci morphism on the alphabet {U,V,W}.

Since these words have lengths 4, 3, and 2, the result follows.

(End)

Links

• Robert Israel, Table of n, a(n) for n = 1..10608
• Elena Barcucci, Luc Belanger and Srecko Brlek, On tribonacci sequences, Fib. Q., 42 (2004), 314-320. See page 317.
• F. Michel Dekking, Jeffrey Shallit, and N. J. A. Sloane, Queens in exile: non-attacking queens on infinite chess boards, Electronic J. Combin., 27:1 (2020), #P1.52.
• Jacques Justin and Laurent Vuillon, Return words in Sturmian and episturmian words, RAIRO-Theoretical Informatics and Applications 34.5 (2000): 343-356.
• Wolfdieter Lang, The Tribonacci and ABC Representations of Numbers are Equivalent, arXiv preprint arXiv:1810.09787 [math.NT], 2018.

Crossrefs

Cf. A003144, A003145, A003146, A080843, A276788, A278040.

Formula

a(n) = A003145(n+1) - A003145(n) = A278040(n) - A278040(n-1) = 4 - A080843(n-1), for n >= 1. See eq. (20) of the W. Lang link. - Wolfdieter Lang, Dec 04 2018