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Positions of letter a in the tribonacci word abacabaabacababac... generated by a->ab, b->ac, c->a (cf. A092782).
(Formerly M2399)
+10
57
1, 3, 5, 7, 8, 10, 12, 14, 16, 18, 20, 21, 23, 25, 27, 29, 31, 32, 34, 36, 38, 40, 42, 44, 45, 47, 49, 51, 52, 54, 56, 58, 60, 62, 64, 65, 67, 69, 71, 73, 75, 76, 78, 80, 82, 84, 86, 88, 89, 91, 93, 95, 97, 99, 101, 102, 104, 106, 108, 110, 112, 113, 115, 117, 119, 121, 123, 125
COMMENTS
A003144, A003145, A003146 may be defined as follows. Consider the morphism psi: a -> ab, b -> ac, c -> a. The image (or trajectory) of a under repeated application of this map is the infinite ternary tribonacci word a, b, a, c, a, b, a, a, b, a, c, a, b, a, b, a, c, ... (setting a = 1, b = 2, c = 3 gives A092782). The indices of a, b, c give respectively A003144, A003145, A003146. (End) [For the word with a -> 0, b -> 1, c -> 2 with offset 0 see A080843. - Wolfdieter Lang, Aug 10 2018]
The infinite word may also be defined as the limit S_oo where S_1 = a, S_n = psi(S_{n-1}). Or, by S_1 = a, S_2 = ab, S_3 = abac, and thereafter S_n = S_{n-1} S_{n-2} S_{n-3}. It is the unique word such that S_oo = psi(S_oo).
Also, indices of a in the sequence closed under a -> abac, b -> aba, c -> ab; starting with a(1) = a. - Philippe Deléham, Apr 16 2004 Theorem: A number m is in this sequence iff the tribonacci representation of m-1 ends with 0. [Duchene and Rigo, Remark 2.5] - N. J. A. Sloane, Nov 18 2016; corrected Mar 02 2019.
REFERENCES
Eric Duchêne, Aviezri S. Fraenkel, Vladimir Gurvich, Nhan Bao Ho, Clark Kimberling, Urban Larsson, Wythoff Visions, Games of No Chance, Vol. 5; MSRI Publications, Vol. 70 (2017), pages 101-153.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
FORMULA
It appears that a(n) is always either floor(n*t) or floor(n*t)+1 for all n, where t is the tribonacci constant A058265. See A275926. - N. J. A. Sloane, Oct 28 2016. This is true - see the Dekking et al. paper. - N. J. A. Sloane, Jul 22 2019
MAPLE
M:=17; S[1]:=`a`; S[2]:=`ab`; S[3]:=`abac`;
for n from 4 to M do S[n]:=cat(S[n-1], S[n-2], S[n-3]); od:
t0:=S[M]: l:=length(t0); t1:=[];
for i from 1 to l do if substring(t0, i..i) = `a` then t1:=[op(t1), i]; fi; od: t1; # N. J. A. Sloane, Nov 01 2006
MATHEMATICA
A003144L = StringPosition[SubstitutionSystem[{"a" -> "ab", "b" -> "ac", "c" -> "a"}, "a", {#}][[1]], "a"][[All, 1]] &; A003144L[7] (* JungHwan Min, Dec 22 2016 *)
CROSSREFS
For tribonacci representations of numbers see A278038.
The tribonacci representation of a(n) is obtained by appending a 0 to the tribonacci representation of n (cf. A278038).
+10
29
0, 2, 4, 6, 7, 9, 11, 13, 15, 17, 19, 20, 22, 24, 26, 28, 30, 31, 33, 35, 37, 39, 41, 43, 44, 46, 48, 50, 51, 53, 55, 57, 59, 61, 63, 64, 66, 68, 70, 72, 74, 75, 77, 79, 81, 83, 85, 87, 88, 90, 92, 94, 96, 98, 100, 101, 103, 105, 107, 109, 111, 112, 114, 116, 118, 120, 122, 124, 125, 127, 129, 131, 132, 134, 136
COMMENTS
This sequence records the indices for the 0 values of A080843, ordered increasingly. In the W. Lang link a(n) = B(n). - Wolfdieter Lang, Dec 06 2018 Sequence gives the positions of letter a in the tribonacci word generated by a->ab, b->ac, c->a, when given offset 0. - Michel Dekking, Apr 03 2019
FORMULA
a(n) = A003144(n+1) - 1 = Sum_{k=1..n} A276788(k), n >= 0 (an empty sum is 0). a(n) = 2*n - ( A276798(n) - 1), n >= 0. For a proof see the link, Proposition 6 B). - Wolfdieter Lang, Dec 04 2018
EXAMPLE
The tribonacci representation of 7 is 1000 (see A278038), so a(7) has tribonacci representation 10000, which is 13, so a(7) = 13.
Trajectory of 3 under repeated application of the morphism sigma: 3 -> 3656, 5 -> 365656, 6 -> 3656656.
+10
14
3, 6, 5, 6, 3, 6, 5, 6, 6, 5, 6, 3, 6, 5, 6, 5, 6, 3, 6, 5, 6, 6, 5, 6, 3, 6, 5, 6, 3, 6, 5, 6, 6, 5, 6, 3, 6, 5, 6, 5, 6, 3, 6, 5, 6, 6, 5, 6, 3, 6, 5, 6, 6, 5, 6, 3, 6, 5, 6, 5, 6, 3, 6, 5, 6, 6, 5, 6, 3, 6, 5, 6, 3, 6, 5, 6, 6, 5, 6, 3, 6, 5, 6, 5, 6, 3, 6, 5, 6, 6, 5, 6, 3, 6, 5, 6, 5, 6, 3, 6
COMMENTS
Versions of this sequence arises in so many different ways in the analysis of the Lonely Queens problem described in A140100- A140103 that it is convenient to define THETA(a,b,c) to be the result of replacing {6,5,3} here by {a,b,c} respectively. - N. J. A. Sloane, Mar 19 2019 Conjecture 1: This sequence is a compressed version of A140101 (see that entry for details). [This was formerly stated as a theorem, but I am no longer sure I have a proof. - N. J. A. Sloane, Sep 29 2018. It is true: see the Dekking et al. paper. - N. J. A. Sloane, Jul 22 2019] Let tau be the tribonacci morphism from A092782, but on the alphabet {6,5,3}, i.e., tau(3)=6, tau(5)=63, tau(6)=65. Then tau^3 is given by 3 -> 6563, 5 -> 656365, 6 -> 6563656.
Let sigma be the morphism generating (a(n)). Then sigma is conjugate to tau^3 with conjugating word u = 656:
(656)^{-1} tau^3(3) 656 = 3656 = sigma(3)
(656)^{-1} tau^3(5) 656 = 365656 = sigma(5)
(656)^{-1} tau^3(6) 656 = 3656656 = sigma(6).
It follows that tau and sigma generate the same language, in particular the frequencies of corresponding letters are equal.
Added Mar 03 2019: Since tau and sigma are irreducible morphisms (which means that their incidence matrices are irreducible), all of their fixed points have the same collection of subwords, this is what is called the language of tau, respectively sigma. See Lemma 3 of Allouche et al. (2003) for background.
(End)
The tribonacci word A092782 is the limit S_oo where f is the morphism 1 -> 12, 2 -> 13, 3 -> 1; S_0 = 1, and S_n = f(S_{n-1}). The present sequence is the limit T_oo where
sigma: 3 -> 3656, 5 -> 365656, 6 -> 3656656; T_0 = 3, and T_n = sigma(T_{n-1}).
Conjecture 2: For all k=0,1,2,..., the following two finite words are identical:
S_{3k+2} with 1,2 mapped to 6,5 respectively, and 3 fixed,
T_{k+1} with its initial 3 moved to the end.
Example for k=1:
S_5 = 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 3,
T_2 = 3, 6, 5, 6, 3, 6, 5, 6, 6, 5, 6, 3, 6, 5, 6, 5, 6, 3, 6, 5, 6, 6, 5, 6,
Note that S_{3k+2} has length A000073(3k+5) and always ends with a 3. The conjecture would imply that if we omit the initial 3 here, and change 6 to 1, 5 to 2, and leave 3 fixed, we get A092782. Alternatively, if we omit the initial 3 here, and change 6 to 0, 5 to 1, and 3 to 2, we get A080843. (End)
Proof of Conjecture 2.
It is convenient to apply the letter to letter map 1->6, 2->5, 3->3 from the start, which changes f^3 to tau^3. Let alpha := tau^3.
We prove by induction that 3 alpha^n(3) = sigma^n(3) 3.
This is true for n=1: 3 alpha(3) = 3 6563 = sigma(3) 3.
The conjugation observation in my comment from December 12 implies that for all words w from the language of tau:
alpha(w) 656 = 656 sigma(w).
Applying this with the word w = alpha^n(3) yields
3 alpha^{n+1}(3) 656 = 3 656 sigma(alpha^n(3)) =
sigma(3 alpha^n(3)) = sigma(sigma^n(3) 3) =
sigma^{n+1}(3) 3656,
where we used the induction hypothesis in the second line. Removing the 656's at the end completes the induction step.
(End)
FORMULA
Theorem: The partial sums of the generalized version THETA(r,s,t) (see Comments) are given by the following formula: Sum_{i=1..n} THETA(r,s,t)(i) = r* A276796(n-1) + s* A276797(n-1) + t* A276798(n-1). - N. J. A. Sloane, Mar 23 2019
EXAMPLE
The first few generations of the iteration are:
3
3656
365636566563656563656656
3656365665636565636566563656365665636565636566563656656365656365665636563656656\
3656563656656365656365665636563656656365656365665636566563656563656656
...
MATHEMATICA
SubstitutionSystem[{3 -> {3, 6, 5, 6}, 5 -> {3, 6, 5, 6, 5, 6}, 6 -> {3, 6, 5, 6, 6, 5, 6}}, {3}, 3] // Last (* Jean-François Alcover, Jan 21 2018 *)
2, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 2, 3, 3
COMMENTS
Or, prefix A276788 with a 1 and then add 1 to every term. This relation between A003144 and A140101 is a conjecture (Daniel Forgues remarks would trivially follow from this relation). - Michel Dekking, Mar 18 2019 The lengths of the successive runs of 3's are given by A275925. a(n) seems to take only the values 2 or 3, where {a(n), a(n+1)} may be {3, 2} or {2, 3} or {3, 3}, but not {2, 2}. The second differences of A140101 (first differences of this sequence) thus seem to take only the values -1 or 0 or 1. - Daniel Forgues, Aug 19 2018 Conjecture: This sequence is 2.TTW(3,3,2) where TTW is the ternary tribonacci word defined in A080843, or equally it is THETA(3,3,2), where THETA is defined in A275925. - N. J. A. Sloane, Mar 19 2019 All these conjectures are now theorems - see the Dekking et al. paper. - N. J. A. Sloane, Jul 22 2019
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
COMMENTS
(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)
Trajectory of 0 under repeated applications of the map psi: 0 -> 0121, 1 -> 0121121, 2 -> 012121.
+10
3
0, 1, 2, 1, 0, 1, 2, 1, 1, 2, 1, 0, 1, 2, 1, 2, 1, 0, 1, 2, 1, 1, 2, 1, 0, 1, 2, 1, 0, 1, 2, 1, 1, 2, 1, 0, 1, 2, 1, 2, 1, 0, 1, 2, 1, 1, 2, 1, 0, 1, 2, 1, 1, 2, 1, 0, 1, 2, 1, 2, 1, 0, 1, 2, 1, 1, 2, 1, 0, 1, 2, 1, 0, 1, 2, 1, 1, 2, 1, 0, 1, 2, 1, 2, 1, 0, 1, 2, 1, 1, 2, 1, 0, 1, 2, 1, 2, 1, 0, 1
COMMENTS
This is A275925 rewritten in a more canonical form using the alphabet 0,1,2.
7, 6, 7, 4, 7, 6, 7, 7, 6, 7, 4, 7, 6, 7, 6, 7, 4, 7, 6, 7, 7, 6, 7, 4, 7, 6, 7, 4, 7, 6, 7, 7, 6, 7, 4, 7, 6, 7, 6, 7, 4, 7, 6, 7, 7, 6, 7, 4, 7, 6, 7, 7, 6, 7, 4, 7, 6, 7, 6, 7, 4, 7, 6, 7, 7, 6, 7, 4, 7, 6, 7, 4, 7, 6, 7, 7, 6, 7, 4, 7, 6, 7, 6, 7, 4, 7, 6, 7, 7, 6, 7, 4, 7, 6, 7, 6, 7, 4, 7
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