A319198 -id:A319198 - cp's OEIS frontend

A322408 Compound sequence with a(n) = A319198(A278041(n)), for n >= 0.

3, 7, 11, 15, 18, 22, 26, 30, 34, 38, 42, 45, 49, 53, 57, 61, 65, 68, 72, 76, 80, 84, 88, 92, 95, 99, 103, 107, 110, 114, 118, 122, 126, 130, 134, 137, 141, 145, 149, 153, 157, 160, 164, 168, 172, 176, 180, 184, 187, 191, 195, 199, 203, 207, 211, 214, 218, 222, 226, 230, 234

Offset: 0

Wolfdieter Lang, Jan 02 2019

Old name was: Compound tribonacci sequence a(n) = A319198(A278041(n)), for n >= 0.
a(n) gives the sum of the entries of the tribonacci word sequence t = A080843 not exceeding t(C(n)), with C(n) = A278041(n).
The nine sequences A308199, A319967, A319968, A322410, A322409, A322411, A322413, A322412, A322414 are based on defining the tribonacci ternary word to start with index 0 (in contrast to the usual definition, in A080843 and A092782, which starts with index 1). As a result these nine sequences differ from the compound tribonacci sequences defined in  A278040, A278041, and A319966-A319972. - N. J. A. Sloane, Apr 05 2019
The difference sequence (a(n+1)-a(n)) is equal to a change of alphabet of the tribonacci word t = A092782. The alphabet is {4,4,3}. This follows from the formula a(n) = A278039(n) + 2*n + 3. - Michel Dekking, Oct 05 2019

			n = 2: C(2) = 16, t = {0, 1, 0, 2, 0, 1, 0, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, ...} which sums to 11 = a(2) = 4 + 7, because B(2) = 4.

Cf. A080843, A278039, A278041, A319198, A321333, A322407.

a(n) =  z(C(n)) = Sum_{j=0..C(n)} t(j), n >= 0, with z = A319198, C = A278041 and t = A080843.
a(n) = B(n) + 2*n + 3, where B(n) = A278039(n). For a proof see the W. Lang link in A080843, Proposition 8, eq. (47).
a(n) = 3 + Sum_{k=1..n-1} d(k), where d is the tribonacci sequence on the alphabet {4,4,3}. - Michel Dekking, Oct 05 2019

Name changed by Michel Dekking, Oct 08 2019

A321333 Compound sequence with a(n) = A319198(A278040(n)), for n >= 0.

Original entry on oeis.org

1, 4, 5, 8, 9, 12, 13, 16, 19, 20, 23, 24, 27, 28, 31, 32, 35, 36, 39, 40, 43, 46, 47, 50, 51, 54, 55, 58, 59, 62, 63, 66, 69, 70, 73, 74, 77, 78, 81, 82, 85, 86, 89, 90, 93, 96, 97, 100, 101, 104, 105, 108, 111, 112, 115, 116, 119, 120, 123, 124, 127

Offset: 0

Wolfdieter Lang, Dec 27 2018

Old name was: Compound tribonacci sequence a(n) = A319198(A278040(n)), for n >= 0.

a(n) gives the sum of the entries of the tribonacci word sequence t = A080843 not exceeding t(A(n)), with A(n) = A278040(n).

			n = 4, A(4) = 14, t = {0, 1, 0, 2, 0, 1, 0, 0, 1, 0, 2, 0, 1, 0, 1, ...}, which sums to  9  = a(4) = 2*(14 - 7) - 5, because B(4) = 7.

Cf. A080843, A278040, A278039, A319198, A322407, A322408.

a(n) =  z(A(n)) = Sum_{j=0..A(n)} t(j), n >= 0, with z = A319198, A = A278040 and t = A080843.
a(n) = 2*(A(n) - B(n)) - (n + 1), where B(n) = A278039(n). For a proof see the W. Lang link in A080843, Proposition 8, eq. (45).
a(n)= 1 + Sum_{k=1..n-1} d(k), where d is the tribonacci sequence on the alphabet {3,1,1}. - Michel Dekking, Oct 08 2019

Name changed by Michel Dekking, Oct 08 2019

A322407 Compound sequence a(n) = A319198(A278039(n)), for n >= 0.

Original entry on oeis.org

0, 1, 3, 4, 4, 5, 7, 8, 9, 11, 12, 12, 13, 15, 16, 18, 19, 19, 20, 22, 23, 24, 26, 27, 27, 28, 30, 31, 31, 32, 34, 35, 36, 38, 39, 39, 40, 42, 43, 45, 46, 46, 47, 49, 50, 51, 53, 54, 54, 55, 57, 58, 59, 61, 62, 62, 63, 65, 66, 68, 69

Offset: 0

Wolfdieter Lang, Jan 02 2019

Old name was: Compound tribonacci sequence a(n) = A319198(A278039(n)), for n >= 0.

a(n) gives the sum of the entries of the tribonacci word sequence t = A080843 not exceeding t(B(n)), with B(n) = A278039(n).

			n = 3: B(3) = 6, t = {0, 1, 0, 2, 0, 1, 0, ...} which sums to 4 = a(3) = -12 + 3*6 - 2, because A(3) = 12.

Cf. A080843, A278040, A278039, A319198, A321333, A322408.

a(n) =  z(B(n)) = Sum_{j=0..B(n)} t(j), n >= 0, with z = A319198, B = A278039 and t = A080843.
a(n) = -A(n) + 3*B(n) - (n - 1), where A(n) = A278040(n). For a proof see the W. Lang link in A080843, Proposition 8, eq. (46).
a(n) = Sum_{k=1..n-1} d(k), where d is the tribonacci sequence on the alphabet {1,2,0}. - Michel Dekking, Oct 08 2019

Name changed by Michel Dekking, Oct 07 2019

A080843 Tribonacci word: limit S(infinity), where S(0) = 0, S(1) = 0,1, S(2) = 0,1,0,2 and for n >= 0, S(n+3) = S(n+2) S(n+1) S(n).

Original entry on oeis.org

0, 1, 0, 2, 0, 1, 0, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 0, 1, 0, 2, 0, 1, 0, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 0, 0, 1, 0, 2, 0, 1, 0, 2, 0, 1, 0, 0, 1, 0, 2, 0, 1, 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

N. J. A. Sloane, Mar 29 2003

An Arnoux-Rauzy or episturmian word.
From N. J. A. Sloane, Jul 10 2018: (Start)
The initial terms in a form suitable for copying:
0,1,0,2,0,1,0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,0,2,0,1,0,2,0,1,0,0,1,0,2,0,1,
0,1,0,2,0,1,0,0,1,0,2,0,1,0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,0,2,0,1,0,2,0,1,
0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,0,2,0,1,0,2,0,1,
0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,0,2,0,1,0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,0,
2,0,1,0,2,0,1,0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,0,2,0,1,0,2,0,1,0,0,1,0,2,0,
1,0,1,0,2,0,1,0,0,1,0,2,0,1,0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,0,2,0,1,0,2,0,
1,0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,0,2,0,1,0,2,0,
1,0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,0,2,0,1,0,0,1,0,2,0,1,0,1,0,2,0,1,0,0,1,
...
Let TTW(a,b,c) denote this sequence written over the alphabet {a,b,c}. It begins:
a,b,a,c,a,b,a,a,b,a,c,a,b,a,b,a,c,a,b,a,a,b,a,c,a,b,a,c,a,b,a,a,b,a,c,a,b,
a,b,a,c,a,b,a,a,b,a,c,a,b,a,a,b,a,c,a,b,a,b,a,c,a,b,a,a,b,a,c,a,b,a,c,a,b,
a,a,b,a,c,a,b,a,b,a,c,a,b,a,a,b,a,c,a,b,a,b,a,c,a,b,a,a,b,a,c,a,b,a,c,a,b,
a,a,b,a,c,a,b,a,b,a,c,a,b,a,a,b,a,c,a,b,a,a,b,a,c,a,b,a,b,a,c,a,b,a,a,b,a,
c,a,b,a,c,a,b,a,a,b,a,c,a,b,a,b,a,c,a,b,a,a,b,a,c,a,b,a,c,a,b,a,a,b,a,c,a,
b,a,b,a,c,a,b,a,a,b,a,c,a,b,a,a,b,a,c,a,b,a,b,a,c,a,b,a,a,b,a,c,a,b,a,c,a,
b,a,a,b,a,c,a,b,a,b,a,c,a,b,a,a,b,a,c,a,b,a,b,a,c,a,b,a,a,b,a,c,a,b,a,c,a,
b,a,a,b,a,c,a,b,a,b,a,c,a,b,a,a,b,a,c,a,b,a,a,b,a,c,a,b,a,b,a,c,a,b,a,a,b,
... (End)
From Wolfdieter Lang, Aug 14 2018: (Start)
The substitution sequence 0 -> 0, 1; 1-> 0, 2; 2 -> 0  read as an irregular triangle with rows l >= 1 and length T(l+2), with the tribonacci numbers T = A000073, leads to the tribonacci tree TriT with level TriT(l) for l >= 1 given by a(0), a(1), ..., a(T(l+2)-1).
  E.g., l = 4:  0 1  0 2  0 1  0 with T(6) = 7 leaves (nodes). See the example below.
This tree can be used to find the tribonacci representation of nonnegative n given in A278038, call it ZTri(n) (Z for generalized Zeckendorf), by replacing every 2 by 1, and reading from bottom to top, omitting the final 0, except for n = 0 which is represented by 0. See the example below. (End)

			From _Joerg Arndt_, Mar 12 2013: (Start)
The first few steps of the substitution are
Start: 0
Rules:
  0 --> 01
  1 --> 02
  2 --> 0
-------------
0:   (#=1)
  0
1:   (#=2)
  01
2:   (#=4)
  0102
3:   (#=7)
  0102010
4:   (#=13)
  0102010010201
5:   (#=24)
  010201001020101020100102
6:   (#=44)
  01020100102010102010010201020100102010102010
7:   (#=81)
  010201001020101020100102010201001020101020100102010010201010201001020102010010201
(End)
From _Wolfdieter Lang_, Aug 14 2018: (Start)
The levels l of the tree TriT begin (the branches (edges) have been omitted):
Substitution rule: 0 -> 0 1; 1 -> 0 2; 2 -> 0.
l=1:                                 0
l=2:                  0                                 1
l=3:             0             1                  0             2
l=4:         0      1       0     2          0       1          0
l=5:      0    1  0   2   0   1   0        0   1   0   2      0    1
...
----------------------------------------------------------------------------------
n =       0    1  2   3   4   5   6        7   8   9  10     11   12
The tribonacci representation of n >= 0 (A278038; here at level 5 for n = 0.. 12) is obtained by reading from bottom to top (along the branches not shown) replacing 2 with 1, omitting the last 0 except for n = 0.
          0    1  0   1   0   1   0        0   1   0  1      0    1
                  1   1   0   0   1        0   0   1  1      0    0
                          1   1   1        0   0   0  0      1    1
                                           1   1   1  1      1    1
E.g., ZTri(9) = A278038(9) = 1010. (End)

The entry A092782 has a more complete list of references and links. - N. J. A. Sloane, Aug 17 2018
J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 246.

N. J. A. Sloane, Table of n, a(n) for n = 0..20000
Jean Berstel and J. Karhumaki, Combinatorics on words - a tutorial, Bull. EATCS, #79 (2003), pp. 178-228.
Nataliya Chekhova, Pascal Hubert, and Ali Messaoudi, Propriétés combinatoires, ergodiques et arithmétiques de la substitution de Tribonacci, Journal de théorie des nombres de Bordeaux, 13.2 (2001): 371-394.
D. Damanik and L. Q. Zamboni, Arnoux-Rauzy subshifts: linear recurrence, powers and palindromes, arXiv:math/0208137 [math.CO], 2002.
F. Michel Dekking, Morphisms, Symbolic Sequences, and Their Standard Forms, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.1.
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, Queen reflections: a modification of Wythoff Nim, Int'l J. Game Theory (2022).
Wolfdieter Lang, The Tribonacci and ABC Representations of Numbers are Equivalent, arXiv:1810.09787v1 [math.NT], 2018.
Joseph Meleshko, Pascal Ochem, Jeffrey Shallit, and Sonja Linghui Shan, Pseudoperiodic Words and a Question of Shevelev, arXiv:2207.10171 [math.CO], 2022.
Aayush Rajasekaran, Narad Rampersad and Jeffrey Shallit, Overpals, Underlaps, and Underpals, In: Brlek S., Dolce F., Reutenauer C., Vandomme É. (eds) Combinatorics on Words, WORDS 2017, Lecture Notes in Computer Science, vol 10432.
Gérard Rauzy, Nombres algébriques et substitutions, Bull. Soc. Math. France 110.2 (1982): 147-178. See page 148.
Bo Tan and Zhi-Ying Wen, Some properties of the Tribonacci sequence, European Journal of Combinatorics, 28 (2007) 1703-1719.
O. Turek, Abelian Complexity Function of the Tribonacci Word, J. Int. Seq. 18 (2015) # 15.3.4
Index entries for sequences that are fixed points of mappings

Cf. A003849 (the Fibonacci word), A092782.
See A092782 for a version over the alphabet {1,2,3}.
See A278045 for another construction.
Cf. A000073, A278038.
First differences: A317950. Partial sums: A319198.

M:=17; S[1]:=`0`; S[2]:=`01`; S[3]:=`0102`;
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); for i from 1 to l do lprint(i,substring(t0,i..i)); od:
# N. J. A. Sloane, Nov 01 2006
# A version that uses the letters a,b,c:
M:=10; B[1]:=`a`; B[2]:=`ab`; B[3]:=`abac`;
for n from 4 to M do B[n]:=cat(B[n-1], B[n-2], B[n-3]); od:
B[10]; # N. J. A. Sloane, Oct 30 2018

Nest[Flatten[ # /. {0 -> {0, 1}, 1 -> {0, 2}, 2 -> {0}}] &, {0}, 8] (* updated by Robert G. Wilson v, Nov 07 2010 *)
SubstitutionSystem[{0->{0,1},1->{0,2},2->{0}},{0},{8}]//Flatten (* Harvey P. Dale, Nov 21 2021 *)

strsub(s, vv, off=0)=
{
    my( nl=#vv, r=[], ct=1 );
    while ( ct <= #s,
        r = concat(r, vv[ s[ct] + (1-off) ] );
        ct += 1;
    );
    return( r );
}
t=[0];  for (k=1, 10, t=strsub( t, [[0,1], [0,2], [0]], 0 ) );  t
\\ Joerg Arndt, Sep 14 2013

Fixed point of morphism 0 -> 0, 1; 1 -> 0, 2; 2 -> 0.

a(n) = A092782(n+1) - 1. - Joerg Arndt, Sep 14 2013

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Apr 06 2003

A278040 The tribonacci representation of a(n) is obtained by appending 0,1 to the tribonacci representation of n (cf. A278038).

Original entry on oeis.org

1, 5, 8, 12, 14, 18, 21, 25, 29, 32, 36, 38, 42, 45, 49, 52, 56, 58, 62, 65, 69, 73, 76, 80, 82, 86, 89, 93, 95, 99, 102, 106, 110, 113, 117, 119, 123, 126, 130, 133, 137, 139, 143, 146, 150, 154, 157, 161, 163, 167, 170, 174, 178, 181, 185, 187, 191, 194, 198, 201, 205, 207, 211, 214, 218, 222, 225, 229, 231, 235

Offset: 0

N. J. A. Sloane, Nov 18 2016

This sequence gives the A(n) numbers of the W. Lang link. There the B(n) and C(n) numbers are A278039(n) and A278041(n), respectively. - Wolfdieter Lang, Dec 05 2018
Positions of letter b in the tribonacci word t generated by a->ab, b->ac, c->a, when given offset 0. - Michel Dekking, Apr 03 2019
This sequence gives the positions of the word ab in the tribonacci word t. This follows from the fact that the letter b is always preceded in t by the letter a, and the formula AA = B-1, where A := A003144, B := A003145, C := A003146. - Michel Dekking, Apr 09 2019

			The tribonacci representation of 7 is 1000 (see A278038), so a(7) has tribonacci representation 100001, which is 24+1 = 25, so a(7) = 25.

N. J. A. Sloane, Table of n, a(n) for n = 0..20000
L. Carlitz, R. Scoville and V. E. Hoggatt, Jr., Fibonacci representations of higher order, Fib. Quart., 10 (1972), 43-69.
Wolfdieter Lang, The Tribonacci and ABC Representations of Numbers are Equivalent, arXiv:1810.09787v1 [math.NT], 2018.

Cf. A003145, A276789, A276796, A278038, A278039, A278041, A319198, A319968.

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.

a(n) = A003145(n+1) - 1.
a(n) = A003144(A003144(n)). - N. J. A. Sloane, Oct 05 2018
See Theorem 13 in the Carlitz, Scoville and Hoggatt paper. - Michel Dekking, Mar 20 2019
From Wolfdieter Lang, Dec 13 2018: (Start)
This sequence gives the indices k with A080843(k) = 1, ordered increasingly with offset 0.
a(n) = 1 + 4*n - A319198(n-1), n >= 0, with A319198(-1) = 0.
a(n) = A276796(C(n)) - 1, with C(n) = A278041(n).
For a proof see the W. Lang link, Proposition 5, and eq. (58).
a(n) - 1 = B1(n), where B1-numbers are B-numbers from A278039 followed by an A-number from A278040. See a comment and example in A319968.
a(n) - 1 = B(B(n)) = B(B(n) + 1) - 2, for n > = 0, where B = A278039.
(End)

cp's OEIS Frontend

A322408 Compound sequence with a(n) = A319198(A278041(n)), for n >= 0.

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula

Extensions

A321333 Compound sequence with a(n) = A319198(A278040(n)), for n >= 0.

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula

Extensions

A322407 Compound sequence a(n) = A319198(A278039(n)), for n >= 0.

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula

Extensions

A080843 Tribonacci word: limit S(infinity), where S(0) = 0, S(1) = 0,1, S(2) = 0,1,0,2 and for n >= 0, S(n+3) = S(n+2) S(n+1) S(n).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A278040 The tribonacci representation of a(n) is obtained by appending 0,1 to the tribonacci representation of n (cf. A278038).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

A324470 Partial sums of ternary tribonacci word A092782.

Views

Author

Keywords

Crossrefs