cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A003144 Positions of letter a in the tribonacci word abacabaabacababac... generated by a->ab, b->ac, c->a (cf. A092782).

Original entry on oeis.org

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
Offset: 1

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Author

Keywords

Comments

From Philippe Deléham, Feb 27 2009: (Start)
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).

Crossrefs

Cf. A003145, A003146, A080843, A092782, A058265, A275926, A276793, A276796, A278039 (subtract 1 from each term, and use offset 0).
First differences are A276788.
For tribonacci representations of numbers see A278038.

Programs

  • 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 *)

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

Extensions

More terms from Philippe Deléham, Apr 16 2004
Entry revised by N. J. A. Sloane, Oct 13 2016

A003145 Positions of letter b in the tribonacci word abacabaabacababac... generated by a->ab, b->ac, c->a (cf. A092782).

Original entry on oeis.org

2, 6, 9, 13, 15, 19, 22, 26, 30, 33, 37, 39, 43, 46, 50, 53, 57, 59, 63, 66, 70, 74, 77, 81, 83, 87, 90, 94, 96, 100, 103, 107, 111, 114, 118, 120, 124, 127, 131, 134, 138, 140, 144, 147, 151, 155, 158, 162, 164, 168, 171, 175, 179, 182, 186, 188, 192, 195, 199, 202, 206, 208
Offset: 1

Views

Author

Keywords

Comments

A003144, A003145, A003146 may be defined as follows. Consider the map psi: a -> ab, b -> ac, c -> a. The image (or trajectory) of a under repeated application of this map is the infinite 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. - Philippe Deléham, Feb 27 2009
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 b 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 01. [Duchene and Rigo, Remark 2.5] - N. J. A. Sloane, 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).

Crossrefs

First differences give A276789. A278040 (subtract 1 from each term, and use offset 1).
For tribonacci representations of numbers see A278038.

Programs

  • 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) = `b` then t1:=[op(t1),i]; fi; od: # N. J. A. Sloane
  • Mathematica
    StringPosition[SubstitutionSystem[{"a" -> "ab", "b" -> "ac", "c" -> "a"}, "b", {#}][[1]], "b"][[All, 1]] &@9 (* Michael De Vlieger, Mar 30 2017, Version 10.2, after JungHwan Min at A003144 *)

Formula

It appears that a(n) = floor(n*t^2) + eps for all n, where t is the tribonacci constant A058265 and eps is 0, 1, or 2. See A276799. - N. J. A. Sloane, Oct 28 2016. This is true - see the Dekking et al. paper. - N. J. A. Sloane, Jul 22 2019

Extensions

More terms from Philippe Deléham, Apr 16 2004
Corrected by T. D. Noe and N. J. A. Sloane, Nov 01 2006
Entry revised by N. J. A. Sloane, Oct 13 2016

A003146 Positions of letter c in the tribonacci word abacabaabacababac... generated by a->ab, b->ac, c->a (cf. A092782).

Original entry on oeis.org

4, 11, 17, 24, 28, 35, 41, 48, 55, 61, 68, 72, 79, 85, 92, 98, 105, 109, 116, 122, 129, 136, 142, 149, 153, 160, 166, 173, 177, 184, 190, 197, 204, 210, 217, 221, 228, 234, 241, 247, 254, 258, 265, 271, 278, 285, 291, 298, 302, 309, 315, 322, 329, 335, 342, 346, 353, 359
Offset: 1

Views

Author

Keywords

Comments

Comment from Philippe Deléham, Feb 27 2009: A003144, A003145, A003146 may be defined as follows. Consider the map psi: a -> ab, b -> ac, c -> a. The image (or trajectory) of a under repeated application of this map is the infinite 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.
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 c 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 11. [Duchene and Rigo, Remark 2.5] - N. J. A. Sloane, 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).

Crossrefs

First differences are A276792. A278041 (subtract 1 from each term, and use offset 0).
For tribonacci representations of numbers see A278038.

Programs

  • 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) = `c` then t1:=[op(t1),i]; fi; od:
    # N. J. A. Sloane, Nov 01 2006
  • Mathematica
    StringPosition[SubstitutionSystem[{"a" -> "ab", "b" -> "ac", "c" -> "a"}, "c", {#}][[1]], "c"][[All, 1]] &@ 11 (* Michael De Vlieger, Mar 30 2017, Version 10.2, after JungHwan Min at A003144 *)

Formula

It appears that a(n) = floor(n*t^3) + eps for all n, where t is the tribonacci constant A058265 and eps is 0, 1, 2, or 3. See A277721. - N. J. A. Sloane, Oct 28 2016. This is true - see the Dekking et al. paper. - N. J. A. Sloane, Jul 22 2019

Extensions

More terms from Philippe Deléham, Apr 16 2004
Entry revised by N. J. A. Sloane, Oct 13 2016

A278038 Binary vectors not containing three consecutive 1's; or, representation of n in the tribonacci base.

Original entry on oeis.org

0, 1, 10, 11, 100, 101, 110, 1000, 1001, 1010, 1011, 1100, 1101, 10000, 10001, 10010, 10011, 10100, 10101, 10110, 11000, 11001, 11010, 11011, 100000, 100001, 100010, 100011, 100100, 100101, 100110, 101000, 101001, 101010, 101011, 101100, 101101, 110000, 110001, 110010, 110011, 110100, 110101, 110110, 1000000
Offset: 0

Views

Author

N. J. A. Sloane, Nov 16 2016

Keywords

Comments

These are the nonnegative numbers written in the tribonacci numbering system.

Examples

			The tribonacci numbers (as in A000073(n), for n >= 3) are 1, 2, 4, 7, 13, 24, 44, 81, ... In terms of this base, 7 is written 1000, 8 is 1001, 11 is 1100, 12 is 1101, 13 is 10000, etc. Zero is 0.
		

Crossrefs

Cf. A000073, A080843 (tribonacci word, tribonacci tree).
See A003726 for the decimal representations of these binary strings.
Similar sequences: A014417 (Fibonacci), A130310 (Lucas).

Programs

  • Maple
    # maximum index in A73 such that A73 <= n.
    A73floorIdx := proc(n)
        local k ;
        for k from 3 do
            if A000073(k) = n then
                return k ;
            elif A000073(k) > n then
                return k -1 ;
            end if ;
        end do:
    end proc:
    A278038 := proc(n)
        local k,L,nres ;
        if n = 0 then
            0;
        else
            k := A73floorIdx(n) ;
            L := [1] ;
            nres := n-A000073(k) ;
            while k >= 4 do
                k := k-1 ;
                if nres >= A000073(k) then
                    L := [1,op(L)] ;
                    nres := nres-A000073(k) ;
                else
                    L := [0,op(L)] ;
                end if ;
            end do:
            add( op(i,L)*10^(i-1),i=1..nops(L)) ;
        end if;
    end proc:
    seq(A278038(n),n=0..40) ; # R. J. Mathar, Jun 08 2022
  • Mathematica
    t[1] = 1; t[2] = 2; t[3] = 4; t[n_] := t[n] = t[n - 1] + t[n - 2] + t[n - 3]; a[n_] := Module[{s = {}, m = n, k}, While[m > 0, k = 1; While[t[k] <= m, k++]; k--; AppendTo[s, k]; m -= t[k]; k = 1]; FromDigits @ IntegerDigits[Total[2^(s - 1)], 2]]; Array[a, 100, 0] (* Amiram Eldar, Mar 04 2022 *)

A092782 The ternary tribonacci word; also a Rauzy fractal sequence: fixed point of the morphism 1 -> 12, 2 -> 13, 3 -> 1, starting from a(1) = 1.

Original entry on oeis.org

1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 3
Offset: 1

Views

Author

Philippe Deléham, Apr 23 2004

Keywords

Comments

See A080843 for the {0,1,2} version, which in a sense is the most basic version.
See also A103269 for another version with further references and comments.
Also called a tribonacci word. In the limit the ratios #1's : #2's : #3's are t^2 : t : 1 where t is the tribonacci constant 1.839286755... (A058265). - Frank M Jackson, Mar 29 2018
a(n)-1 is the number of trailing 0's in the maximal tribonacci representation of n (A352103). - Amiram Eldar, Feb 29 2024

Examples

			From _Joerg Arndt_, Sep 14 2013: (Start)
The first few steps of the substitution are
Start: 1
Maps:
  1 --> 12
  2 --> 13
  3 --> 1
-------------
0:   (#=1)
  1
1:   (#=2)
  12
2:   (#=4)
  1213
3:   (#=7)
  1213121
4:   (#=13)
  1213121121312
5:   (#=24)
  121312112131212131211213
6:   (#=44)
  12131211213121213121121312131211213121213121
7:   (#=81)
  121312112131212131211213121312112131212131211213121121312121312112131213121121312
(End)
		

References

  • This entry has a fairly complete list of references and links concerning the ternary tribonacci word. - N. J. A. Sloane, Aug 17 2018
  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 246.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.

Crossrefs

See A080843 for a {0,1,2} version.
First differences: A317950.

Programs

  • Maple
    f(1):= (1, 2): f(2):= (1, 3): f(3):= (1): A:= [1]:
    for i from 1 to 16 do A:= map(f, A) od:
    A; # 19513 terms of A092782; A103269; from N. J. A. Sloane, Aug 06 2018
  • Mathematica
    Nest[ Flatten[# /. {1 -> {1, 2}, 2 -> {1, 3}, 3 -> 1}] &, {1}, 8] (* Robert G. Wilson v, Mar 04 2005 and updated Apr 29 2018 *)
  • PARI
    w=vector(9,x,[]); w[1]=[1];
    for(n=2,9,for(k=1,#w[n-1],m=w[n-1][k];v=[];if(m-1,if(m-2,v=[1],v=[1,3]),v=[1,2]);w[n]=concat(w[n],v)));
    w[9] \\ Gerald McGarvey, Dec 18 2009
    
  • PARI
    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=[1];  for (k=1, 10, t=strsub( t, [[1,2], [1,3], [1]], 1 ) );  t
    \\ Joerg Arndt, Sep 14 2013
    
  • PARI
    A092782_vec(N,s=[[1,2],[1,3],1],A=[1])={while(#AM. F. Hasler, Dec 14 2018

Formula

a(n) = 1 for n in A003144; a(n) = 2 for n in A003145; a(n) = 3 for n in A003146.
a(n) = A080843(n-1) + 1. - Joerg Arndt, Sep 14 2013

Extensions

Additional references and links added by N. J. A. Sloane, Aug 17 2018

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

Views

Author

N. J. A. Sloane, Nov 18 2016

Keywords

Comments

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

Examples

			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.
		

Crossrefs

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.

Formula

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)

A278039 The tribonacci representation of a(n) is obtained by appending a 0 to the tribonacci representation of n (cf. A278038).

Original entry on oeis.org

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
Offset: 0

Views

Author

N. J. A. Sloane, Nov 18 2016

Keywords

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

Examples

			The tribonacci representation of 7 is 1000 (see A278038), so a(7) has tribonacci representation 10000, which is 13, so a(7) = 13.
		

Crossrefs

Partial sums of A276788.

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

A278041 The tribonacci representation of a(n) is obtained by appending 0,1,1 to the tribonacci representation of n (cf. A278038).

Original entry on oeis.org

3, 10, 16, 23, 27, 34, 40, 47, 54, 60, 67, 71, 78, 84, 91, 97, 104, 108, 115, 121, 128, 135, 141, 148, 152, 159, 165, 172, 176, 183, 189, 196, 203, 209, 216, 220, 227, 233, 240, 246, 253, 257, 264, 270, 277, 284, 290, 297, 301, 308, 314, 321, 328, 334, 341, 345, 352, 358, 365, 371, 378, 382, 389, 395, 402, 409, 415
Offset: 0

Views

Author

N. J. A. Sloane, Nov 18 2016

Keywords

Comments

This sequence gives the indices k for which A080843(k) = 2, sorted increasingly with offset 0. In the W. Lang link a(n) = C(n). - Wolfdieter Lang, Dec 06 2018
Positions of letter c 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 ac in the tribonacci word t. This follows from the fact that the letter c is always preceded in t by the letter a, and the formula AB = C-1, where A := A003144, B := A003145, C := A003146. - Michel Dekking, Apr 09 2019

Examples

			The tribonacci representation of 7 is 1000 (see A278038), so a(7) has tribonacci representation 1000011, which is 44+2+1 = 47, so a(7) = 47.
		

Crossrefs

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.

Formula

a(n) = A003146(n+1) - 1.
a(n) = A003144(A003145(n)). - N. J. A. Sloane, Oct 05 2018
From Wolfdieter Lang, Dec 06 2018: (Start)
a(n) = n + 2 + A(n) + B(n), where A(n) = A278040(n) and B = A278039(n).
a(n) = 7*n + 3 - (z_A(n-1) + 3*z_C(n-1)), where z_A(n) = A276797(n+1) and z_C(n) = A276798(n+1) - 1, n >= 0.
For proofs see the W. Lang link in A080843, eqs. 37 and 40.
a(n) - 1 = B2(n), where B2-numbers are B-numbers from A278039 followed by a C-number from A278041. See a comment and example in A319968.
(End)

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

Views

Author

N. J. A. Sloane, Aug 29 2016

Keywords

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]
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

Examples

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

Crossrefs

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

Programs

  • 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 *)

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

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

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane, Oct 05 2018

Keywords

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)
		

Crossrefs

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
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