A092782 -id:A092782 - cp's OEIS frontend

A140100 Start with Y(0)=0, X(1)=1, Y(1)=2. For n > 1, choose least positive integers Y(n) > X(n) such that neither Y(n) nor X(n) appear in {Y(k), 1 <= k < n} or {X(k), 1 <= k < n} and such that Y(n) - X(n) does not appear in {Y(k) - X(k), 1 <= k < n} or {Y(k) + X(k), 1 <= k < n}; sequence gives X(n) (for Y(n) see A140101).

1, 3, 4, 6, 7, 9, 10, 12, 14, 15, 17, 18, 20, 21, 23, 24, 26, 27, 29, 30, 32, 34, 35, 37, 38, 40, 41, 43, 44, 46, 47, 49, 51, 52, 54, 55, 57, 58, 60, 61, 63, 64, 66, 67, 69, 71, 72, 74, 75, 77, 78, 80, 82, 83, 85, 86, 88, 89, 91, 92, 94, 95, 97, 98, 100, 102, 103, 105, 106

Offset: 1

Paul D. Hanna, Jun 04 2008

nonn

Sequence A140101 = {Y(n), n >= 1} is the complement of the current sequence, while the sequence of differences, A140102 = {Y(n) - X(n), n >= 1}, forms the complement of the sequence of sums, A140103 = {Y(n) + X(n), n >= 1}.
Compare with A140098(n) = floor(n*(1+1/t)), a Beatty sequence involving the tribonacci constant t = t^3 - t^2 - 1 = 1.83928675521416113255...
Conjecture: A140100(n) - A140098(n) = A276404(n) is always 0 or 1; see A276406 for the positions where a difference of 1 occurs.
This is the same problem as the "Greedy Queens in a spiral" problem described in A273059. See the Dekking et al. paper and comments in A140101. - N. J. A. Sloane, Aug 30 2016
The sequence is "tribonacci-synchronized"; this means there is a finite automaton recognizing the tribonacci representation of (n,a(n)) input in parallel, where a shorter input is padded with leading zeros.   This finite automaton has 22 states and was verified with Walnut.  In particular this finite automaton and a similar one for A140101 was used to verify that (conjecture of J. Cassaigne) either a(b(n)) = a(n)+b(n) or b(a(n)) = a(n)+b(n) for all n>=1, where b(n) = A140101(n). - Jeffrey Shallit, Oct 04 2022

			Start with Y(0)=0, X(1)=1, Y(1)=2; Y(1)-X(1)=1, Y(1)+X(1)=3.
Next choose X(2)=3 and Y(2)=5; Y(2)-X(2)=2, Y(2)+X(2)=8.
Next choose X(3)=4 and Y(3)=8; Y(3)-X(3)=4, Y(3)+X(3)=12.
Next choose X(4)=6 and Y(4)=11; Y(4)-X(4)=5, Y(4)+X(4)=17.
Continue to choose the least positive X and Y>X not appearing earlier such that Y-X and Y+X do not appear earlier as a difference or sum.
CONSTRUCTION: PLOT OF (A140100(n), A140101(n)).
This sequence gives the x-coordinates of the following construction.
Start with an x-y coordinate system and place an 'o' at the origin.
Define an open position as a point not lying in the same row, column, or diagonal (slope +1/-1) as any point previously given an 'o' marker.
From then on, place an 'o' marker at the first open position with integer coordinates that is nearest the origin and the y-axis in the positive quadrant, while simultaneously placing markers at rotationally symmetric positions in the remaining three quadrants.
Example: after the origin, begin placing markers at x-y coordinates:
n=1: (1,2),   (2,-1), (-1,-2),   (-2,1);
n=2: (3,5),   (5,-3), (-3,-5),   (-5,3);
n=3: (4,8),   (8,-4), (-4,-8),   (-8,4);
n=4: (6,11), (11,-6), (-6,-11), (-11,6);
n=5: (7,13), (13,-7), (-7,-13), (-13,7); ...
The result of this process is illustrated in the following diagram (see A273059 for an equivalent picture - _N. J. A. Sloane_, Aug 30 2016).
----------------+---o------------
--o-------------+----------------
----o-----------+----------------
----------------+--o-------------
--------o-------+----------------
-----------o----+----------------
----------------+o---------------
--------------o-+----------------
++++++++++++++++o++++++++++++++++
----------------+-o--------------
---------------o+----------------
----------------+----o-----------
----------------+-------o--------
-------------o--+----------------
----------------+------------o---
----------------+--------------o-
------------o---+----------------
Graph: no two points lie in the same row, column, diagonal, or antidiagonal.
Points in the positive quadrant are at (A140100(n), A140101(n)).
A140101 begins: [2,5,8,11,13,16,19,22,25,28,31,33,36,39,42,...].

N. J. A. Sloane, Table of n, a(n) for n = 1..50000, Sep 13 2016 (First 1001 terms from Reinhard Zumkeller)
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.
Eric Duchêne and Michel Rigo, A morphic approach to combinatorial games: the Tribonacci case. RAIRO - Theoretical Informatics and Applications, 42, 2008, pp 375-393. doi:10.1051/ita:2007039. [Also available here]
Robbert Fokkink and Dan Rust, A modification of Wythoff's Nim, arXiv:1904.08339 [math.CO], 2019.
Jeffrey Shallit, Some Tribonacci conjectures, arXiv:2210.03996 [math.CO], 2022.
Jeffrey Shallit, Automaton to be called xaut.txt in Walnut format, recognizing (n, X(n)) in parallel, in Tribonacci representation
N. J. A. Sloane, Maple program for A140100, A140101, A140102, A140103

Cf. A140101 (complement); A140102, A140103, A276404, A276405, A276406, A275926.
Cf. related Beatty sequences: A140098, A140099; A000201.
Cf. A058265 (tribonacci constant).
Cf. Greedy Queens in a spiral, A273059.
For first difference of A140100, A140101, A140102, A140103 see A305392, A305374, A305393, A305394.

Maple
```
See link.
```

y[0] = 0; x[1] = 1; y[1] = 2;
x[n_] := x[n] = For[yn = y[n - 1] + 1, True, yn++, For[xn = x[n - 1] + 1, xn < yn, xn++, xx = Array[x, n - 1]; yy = Array[y, n - 1]; If[FreeQ[xx, xn] && FreeQ[xx, yn] && FreeQ[yy, xn] && FreeQ[yy, yn] && FreeQ[yy - xx, yn - xn] && FreeQ[yy + xx, yn - xn], y[n] = yn; Return[xn]]]];
Table[x[n], {n, 1, 100}] (* Jean-François Alcover, Jun 17 2018 *)

/* Print (x,y) coordinates of the positive quadrant */
{X=[1]; Y=[2]; D=[1]; S=[3]; print1("["X[1]", "Y[1]"], "); for(n=1, 100, for(j=2, 2*n, if(setsearch(Set(concat(X, Y)), j)==0, Xt=concat(X, j); for(k=j+1, 3*n, if(setsearch(Set(concat(Xt, Y)), k)==0, if(setsearch(Set(concat(D, S)), k-j)==0, if(setsearch(Set(concat(D, S)), k+j)==0, X=Xt; Y=concat(Y, k); D=concat(D, k-j); S=concat(S, k+j); print1("["X[ #X]", "Y[ #Y]"], "); break); break))))))}

Conjecture: the limit of X(n)/n = 1+1/t and limit of Y(n)/n = 1+t where the limit of Y(n)/X(n) = t = tribonacci constant (A058265), and thus the limit of (Y(n) + X(n))/(Y(n) - X(n)) = t^2 and the limit of (Y(n)^2 + X(n)^2)/(Y(n)^2 - X(n)^2) = t.
From Michel Dekking, Mar 16 2019: (Start)
It is conjectured in A305392 that the first differences of (X(n)) as a word are given by 212121 delta(x), where x is the tribonacci word x = A092782, and delta is the morphism
      1 -> 2212121212121,
      2 -> 22121212121,
      3 -> 2212121.
This conjecture implies the frequency conjectures above: let N(i,n) be the number of letters i in x(1)x(2)...x(n). Then simple counting gives
       X(13*N(1,n)+11*N(2,n)+7*N(3,n)) = 20*N(1,n)+17*N(2,n)+11*N(3,n), where we neglected the first 6 symbols of X.
It is well known (see, e.g., A092782) that the frequencies of 1, 2 and 3 in x are respectively 1/t, 1/t^2 and 1/t^3. Dividing all the N(i,n) by n, and letting n tend to infinity, we then have to see that
     20*1/t + 17*1/t^2 + 11*1/t^3 = (1+1/t)*(13*1/t + 11*1/t^2 + 7*1/t^3).
This is a simple verification. (End)

Terms computed independently by Reinhard Zumkeller and Joshua Zucker

Edited by N. J. A. Sloane, Aug 30 2016

A275925 Trajectory of 3 under repeated application of the morphism sigma: 3 -> 3656, 5 -> 365656, 6 -> 3656656.

Original entry on oeis.org

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

Offset: 1

N. J. A. Sloane, Aug 29 2016

nonn

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]
From Michel Dekking, Dec 12 2018: (Start)
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)
From N. J. A. Sloane, Mar 03 2019: (Start)
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)
From Michel Dekking, Mar 11 2019: (Start)
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)
Lengths of runs of 2's in A276788. - John Keith, May 15 2022

			The first few generations of the iteration are:
3
3656
365636566563656563656656
3656365665636565636566563656365665636565636566563656656365656365665636563656656\
   3656563656656365656365665636563656656365656365665636566563656563656656
...

N. J. A. Sloane, Table of n, a(n) for n = 1..35890
J.-P. Allouche, M. Baake, J. Cassaigns, and D. Damanik, Palindrome complexity, arXiv:math/0106121 [math.CO], 2001; Theoretical Computer Science, 292 (2003), 9-31.
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.
Index entries for sequences that are fixed points of mappings

Cf. A140100, A140101, A273059, A000073.
See A276790 and A277745 for other versions. See also A276788 and A080843, A092782.
For partial sums see A305373, also A276796, A276797, A276798.

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 *)

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

A105083 Trajectory of 1 under the morphism 1 -> 12, 2 -> 3, 3 -> 1.

Original entry on oeis.org

1, 2, 3, 1, 1, 2, 1, 2, 3, 1, 2, 3, 1, 1, 2, 3, 1, 1, 2, 1, 2, 3, 1, 1, 2, 1, 2, 3, 1, 2, 3, 1, 1, 2, 1, 2, 3, 1, 2, 3, 1, 1, 2, 3, 1, 1, 2, 1, 2, 3, 1, 2, 3, 1, 1, 2, 3, 1, 1, 2, 1, 2, 3, 1, 1, 2, 1, 2, 3, 1, 2, 3, 1, 1, 2, 3, 1, 1, 2, 1, 2, 3, 1, 1, 2, 1, 2, 3, 1, 2, 3, 1, 1, 2, 1, 2, 3, 1, 2

Offset: 0

Roger L. Bagula, Apr 06 2005

nonn

Aresh Pourkavoos, Table of n, a(n) for n = 0..10000
P. Arnoux and E. Harriss, What is a Rauzy Fractal?, Notices Amer. Math. Soc., 61 (No. 7, 2014), 768-770, also p. 704 and front cover.
Marcy Barge and Jaroslaw Kwapisz, Geometric theory of unimodular Pisot substitutions, Amer. J. Math. 128 (2006), no. 5, 1219--1282. MR2262174 (2007m:37039). - _N. J. A. Sloane_, Aug 06 2014
Jeffrey Shallit, The Narayana Morphism and Related Words, arXiv:2503.01026 [math.CO], 2025.
Victor F. Sirvent and Yang Wang, Self-Affine Tiling via Substitution Dynamical Systems and Rauzy Fractals, Pac. J. Math., 206 (2002), 465-485. See Example 2.2 and Figure 2 pp. 474-475.
Index entries for sequences that are fixed points of mappings

Cf. A005374, A073058, A092782, A202340, A245553, A245554, A000930.

Nest[ Function[ l, {Flatten[(l /. {1 -> {1, 2}, 2 -> {3}, 3 -> {1}})] }], {1}, 12]

N_TERMS=10000
def a():
  # Index of the current term
  n = 0
  # Stores the place values of the greedy representation of n,
  # minus two since A000930 begins with duplicate ones.
  places = []
  # Edge case: a(0)=1.
  yield 0, 1
  while True:
    n += 1
    # Add A000930(2+0)=1 to the representation of n
    places.append(0)
    # Apply carryover rule for as long as necessary:
    # if places contains n+2 and n,
    # both terms are replaced by n+3.
    while len(places) > 1 and places[-2] <= places[-1]+2:
      places.pop()
      places[-1] += 1
    # Look at the smallest term to decide a(n)
    an = 1 if places[-1] > 1 else places[-1]+2
    yield n, an
# Asymptotic behavior is O(log(n)*log(log(n))) memory
# and O(n) time to generate the first n terms,
# although a term may take as long as O(log(n)).
for n, an in a():
  print(n, an)
  if (n >= N_TERMS):
    break
# Aresh Pourkavoos, Jan 26 2021

From Aresh Pourkavoos, Jan 26 2021: (Start)
Limit S(infinity) of the following strings: S(0) = 2, S(1) = 1, S(2) = 0, S(n+3) = S(n+2)S(n). S(n) has length A000930(n).
Individual terms of a(n) may also be found by greedily writing n as a sum of entries of A000930. a(n) is 2 if the smallest term is 1, 3 if the smallest term is 2, and 1 otherwise.
(End)
a(n) = A005374(n+1) - A005374(n) - 2*(A202340(n+1) - 2). - Alan Michael Gómez Calderón, Jul 19 2025

Edited by N. J. A. Sloane, Oct 10 2007 and Aug 03 2014

A305393 First differences of A140102.

Original entry on oeis.org

1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1

Offset: 1

N. J. A. Sloane, Jun 23 2018

nonn

Although initially this agrees with A293630, the sequences are distinct.
From Michel Dekking, Mar 18 2019: (Start)
Let x be the tribonacci word x = A092782 = 1,2,1,3,1,2,1,1,...
Consider the morphism delta:
      1 -> 1112,
      2 -> 112,
      3 -> 12.
Conjecture: (a(n)) =  12 delta(x).
(End)
Conjecture: This sequence (prefixed by 1 since A140102 should really begin with 0) is 1.TTW(1,2,1) where TTW is the ternary tribonacci word defined in A080843, or equally it is THETA(1,2,1), 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

N. J. A. Sloane, Table of n, a(n) for n = 1..49999
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.

For first differences of A140100, A140101, A140102, A140103 see A305392, A305374, A305393, A305394.

Cf. A293630.

a(n) = A140102(n+1)-A140102(n), n >= 1.

A276788 First differences of A003144.

Original entry on oeis.org

2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, 2

Offset: 1

N. J. A. Sloane, Oct 14 2016

In A276790, leave 2's unchanged, but replace 1's by 2's and 0's by 1's, and then omit the initial 1.
If we prefixed A003144 with an initial 0, then its first differences would be a' := 1 followed by a, that is, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2, ... If we now add 1 to every term of a' we get A305374 = first differences of A140101. - N. J. A. Sloane, Jul 17 2018
This relation between A003144 and A140101 is a conjecture - Michel Dekking, Mar 18 2019 [It has been a theorem since Mar 22 2019. - N. J. A. Sloane, Jun 25 2019. (See the Dekking et al. paper)]
(a(n)) is a morphic sequence: in the tribonacci word A092782 = 1,2,1,3,1,2,1,1,... map  1 -> 2, 2 -> 2, 3 -> 1. - Michel Dekking, Mar 21 2019

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 316.
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.

Cf. A003144, A003145, A003146, A080843, A276790, A276791, A275925, A305374, A140101.

See A278039 for partial sums.

M:= 10: # to use M generations of strings
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:
P:=select(t -> S[M][t]="a", [$1..length(S[M])]):
P[2..-1]-P[1..-2]; # Robert Israel, Nov 01 2016

a(n) = A003144(n+1) - A003144(n), n >= 1.

a(n+1) = 2 - t(n)*(t(n) - 1)/2 = 2 - A276791(n+1), for n >= 0, where t(n) = A080843(n). See the W. Lang link in A080843, eq. (38). - Wolfdieter Lang, Dec 06 2018

A305392 First differences of A140100.

Original entry on oeis.org

2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1

Offset: 1

N. J. A. Sloane, Jun 23 2018

nonn

a(n) seems to take only the values 1 or 2, where {a(n), a(n+1)} may be {2, 1} or {1, 2} or {2, 2}, but not {1, 1}, and where {a(n), a(n+1), a(n+2), a(n+3)} may be {2, 1, 2, 1} or {1, 2, 1, 2} or {1, 2, 2, 1}, but not {2, 1, 1, 2}. The second differences of A140100 (first differences of this sequence) thus seem to take only the values -1 or 0 or 1. - Daniel Forgues, Aug 17 2018
From Michel Dekking, Mar 16 2019: (Start)
Let x be the tribonacci word x = A092782 = 1,2,1,3,1,2,1,1,...
Consider the morphism delta:
      1 -> 2212121212121,
      2 -> 22121212121,
      3 -> 2212121.
Conjecture: (a(n)) =  212121 delta(x).
(End)
Conjecture: This sequence (prefixed by 1 since A140100 should really begin with 0) is 1.TTW(2,1,1) where TTW is the ternary tribonacci word defined in A080843,  or equally it is THETA(2,1,1), 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

N. J. A. Sloane, Table of n, a(n) for n = 1..49999
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.

For first differences of A140100, A140101, A140102, A140103 see A305392, A305374, A305393, A305394.

a(n) = A140100(n+1)-A140100(n).

A305394 First differences of A140103.

Original entry on oeis.org

5, 4, 5, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 5, 4, 5, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 5, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 5, 4, 5, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 5, 4, 5, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 5, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 5, 4, 5, 3, 5, 4, 5, 5, 4, 5, 3, 5, 4, 5, 4

Offset: 1

N. J. A. Sloane, Jun 23 2018

nonn

Conjecture: this sequence is the ternary tribonacci word on the alphabet {5,4,3}, i.e., (a(n)) is the unique fixed point of the  morphism 5 -> 54, 4 -> 53, 3 -> 5; see A092782. - Michel Dekking, Mar 13 2019
An equivalent conjecture: This sequence (prefixed by 3 since A140103 should really begin with 0) is 3.TTW(5,4,3) where TTW is the ternary tribonacci word defined in A080843, or equally it is THETA(5,4,3), where THETA is defined in A275925. There are similar conjectures for the first differences of A140100, A140101, A140102. - N. J. A. Sloane, Mar 14 2019 and Mar 19 2019
All these conjectures are now theorems - see the Dekking et al. paper. - N. J. A. Sloane, Jul 22 2019

N. J. A. Sloane, Table of n, a(n) for n = 1..49999F.
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.

For first differences of A140100, A140101, A140102, A140103 see A305392, A305374, A305393, A305394.

a(n) = A140103(n+1) - A140103(n).

A100619 Fixed point of the morphism 1 -> 12, 2 -> 31, 3 -> 1, starting from a(1) = 1.

Original entry on oeis.org

1, 2, 3, 1, 1, 1, 2, 1, 2, 1, 2, 3, 1, 1, 2, 3, 1, 1, 2, 3, 1, 1, 1, 2, 1, 2, 3, 1, 1, 1, 2, 1, 2, 3, 1, 1, 1, 2, 1, 2, 1, 2, 3, 1, 1, 2, 3, 1, 1, 1, 2, 1, 2, 1, 2, 3, 1, 1, 2, 3, 1, 1, 1, 2, 1, 2, 1, 2, 3, 1, 1, 2, 3, 1, 1, 2, 3, 1, 1, 1, 2, 1, 2, 3, 1, 1, 1, 2, 1, 2, 1, 2, 3, 1, 1, 2, 3, 1, 1, 2, 3, 1, 1, 1, 2

Offset: 1

N. J. A. Sloane, Dec 03 2004

Sirvent comments that in spite of the similarity of this map to the one in A092782, the two sequences have very different properties. They have different complexities, different Rauzy fractals, etc.

G. C. Greubel, Table of n, a(n) for n = 1..10000
Scott Balchin and Dan Rust, Computations for Symbolic Substitutions, Journal of Integer Sequences, Vol. 20 (2017), Article 17.4.1.
Victor F. Sirvent, Semigroups and the self-similar structure of the flipped tribonacci substitution, Applied Math. Letters, 12 (1999), 25-29.
Victor F. Sirvent, The common dynamics of the Tribonacci substitutions, Bulletin of the Belgian Mathematical Society-Simon Stevin 7.4 (2000): 571-582.
Index entries for sequences that are fixed points of mappings

Cf. A092782.

f(1):= (1, 2): f(2):= (3, 1): f(3):= (1): B:= [1]:
for i from 1 to 12 do B:= map(f, B) od:
B; # N. J. A. Sloane, Aug 17 2018

Nest[ Function[l, {Flatten[(l /. {1 -> {1, 2}, 2 -> {3, 1}, 3 -> {1}})]}], {1}, 7][[1]] (* Robert G. Wilson v, Feb 26 2005 *)

More terms from Robert G. Wilson v, Dec 05 2004

A103269 Apply the tribonacci morphism 1 -> {1, 2}, 2 -> {1, 3}, 3 -> {1} n times to 1, and concatenate the resulting string.

Original entry on oeis.org

1, 12, 1213, 1213121, 1213121121312, 121312112131212131211213, 12131211213121213121121312131211213121213121, 121312112131212131211213121312112131212131211213121121312121312112131213121121312

Offset: 0

Jonathan Vos Post and Robert G. Wilson v, May 16 2005

nonn

The number of letters in the n-th iteration is tribonacci(n+3) (that is, A000073(n+3)).

N. J. A. Sloane, Table of n, a(n) for n = 0..10 (shortened by _N. J. A. Sloane_, Jan 18 2019)
Elena Barcucci, Luc Belanger and Srecko Brlek, On tribonacci sequences, Fib. Q., 42 (2004), 314-320. See T on page 315.
Julien Cassaigne, Sebastien Ferenczi, and Luca Q. Zamboni, Imbalances in Arnoux-Rauzy sequences, Annales de l'institut Fourier, 50 (2000), 1265-1276.
David Damanik and Luca Q. Zamboni, Arnoux-Rauzy Subshifts: Linear Recurrence, Powers and Palindromes, arXiv:math/0208137 [math.CO], 2002.
C. Holton and L. Q. Zamboni, Directed Graphs and Substitutions, Theory Comput. Systems, 34 (2001) 545-564
S. Ito and M. Kimura, On Rauzy fractal, Japan J. Indust. Appl. Math. 8 (1991) 461-486.
A. Messaoudi, Propriétes arithmétiques et dynamiques du fractal de Rauzy, J. Th. Nombres Bordeaux, 10 (1998) 195-224.
Gérard Rauzy, Nombres algébriques et substitutions, Bull. Soc. Math. France, 110 (1982), 147-178.
Bo Tan and Zhi-Ying Wen, Some properties of the Tribonacci sequence, European Journal of Combinatorics, 28 (2007) 1703-1719. See A_n.
Index entries for sequences that are fixed points of mappings

A092782 is limit of these strings.

Cf. A000073, A106748, A106749, A003842, A106750, A092782, A317196.

FromDigits /@ NestList[ Flatten[ # /. {1 -> {1, 2}, 2 -> {1, 3}, 3 -> 1}] &, {1}, 7]
(* Second program: *)
FromDigits /@ SubstitutionSystem[{1 -> {1, 2}, 2 -> {1, 3}, 3 -> {1}}, {1}, 7] (* Jean-François Alcover, Nov 12 2018 *)

Definition edited by N. J. A. Sloane, Aug 06 2018

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A275925 Trajectory of 3 under repeated application of the morphism sigma: 3 -> 3656, 5 -> 365656, 6 -> 3656656.

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A105083 Trajectory of 1 under the morphism 1 -> 12, 2 -> 3, 3 -> 1.

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A305393 First differences of A140102.

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A276788 First differences of A003144.

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A305392 First differences of A140100.

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A305394 First differences of A140103.

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A100619 Fixed point of the morphism 1 -> 12, 2 -> 31, 3 -> 1, starting from a(1) = 1.

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