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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A001644 a(n) = a(n-1) + a(n-2) + a(n-3), a(0)=3, a(1)=1, a(2)=3.

Original entry on oeis.org

3, 1, 3, 7, 11, 21, 39, 71, 131, 241, 443, 815, 1499, 2757, 5071, 9327, 17155, 31553, 58035, 106743, 196331, 361109, 664183, 1221623, 2246915, 4132721, 7601259, 13980895, 25714875, 47297029, 86992799, 160004703, 294294531, 541292033, 995591267, 1831177831
Offset: 0

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Author

Keywords

Comments

For n >= 3, a(n) is the number of cyclic sequences consisting of n zeros and ones that do not contain three consecutive ones provided the positions of the zeros and ones are fixed on a circle. This is proved in Charalambides (1991) and Zhang and Hadjicostas (2015). For example, a(3)=7 because only the sequences 110, 101, 011, 001, 010, 100 and 000 avoid three consecutive ones. (For n=1,2 the statement is still true provided we allow the sequence to wrap around itself on a circle.) - Petros Hadjicostas, Dec 16 2016
For n >= 3, also the number of dominating sets on the n-cycle graph C_n. - Eric W. Weisstein, Mar 30 2017
For n >= 3, also the number of minimal dominating sets and maximal irredundant sets on the n-sun graph. - Eric W. Weisstein, Jul 28 and Aug 17 2017
For n >= 3, also the number of minimal edge covers in the n-web graph. - Eric W. Weisstein, Aug 03 2017
For n >= 1, also the number of ways to tile a bracelet of length n with squares, dominoes, and trominoes. - Ruijia Li and Greg Dresden, Sep 14 2019
If n is prime, then a(n)-1 is a multiple of n ; a counterexample for the converse is given by n = 182. - Robert FERREOL, Apr 03 2024

Examples

			G.f. = 3 + x + 3*x^2 + 7*x^3 + 11*x^4 + 21*x^5 + 39*x^6 + 71*x^7 + 131*x^8 + ...
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 500.
  • G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A000073, A073145, A106293 (Pisano periods), A073728 (partial sums).
Cf. A058265.

Programs

  • GAP
    a:=[3,1,3];; for n in [4..40] do a[n]:=a[n-1]+a[n-2]+a[n-3]; od; a; # Muniru A Asiru, Dec 18 2018
    
  • Haskell
    a001644 n = a001644_list !! n
    a001644_list = 3 : 1 : 3 : zipWith3 (((+) .) . (+))
                   a001644_list (tail a001644_list) (drop 2 a001644_list)
    -- Reinhard Zumkeller, Apr 13 2014
    
  • Magma
    I:=[3,1,3]; [n le 3 select I[n] else Self(n-1)+Self(n-2)+ Self(n-3): n in [1..40]]; // Vincenzo Librandi, Aug 04 2017
    
  • Maple
    A001644:=-(1+2*z+3*z**2)/(z**3+z**2+z-1); # Simon Plouffe in his 1992 dissertation; gives sequence except for the initial 3
    A001644 :=proc(n)
        option remember;
        if n <= 2 then
            1+2*modp(n+1,2)
        else
            procname(n-1)+procname(n-2)+procname(n-3);
        end if;
    end proc:
    seq(A001644(n),n=0..80) ;
  • Mathematica
    a[x_]:= a[x] = a[x-1] +a[x-2] +a[x-3]; a[0] = 3; a[1] = 1; a[2] = 3; Array[a, 40, 0]
    a[n_]:= n*Sum[Sum[Binomial[j, n-3*k+2*j]*Binomial[k, j], {j,n-3*k,k}]/k, {k, n}]; a[0] = 3; Array[a, 40, 0] (* Robert G. Wilson v, Feb 24 2011 *)
    LinearRecurrence[{1, 1, 1}, {3, 1, 3}, 40] (* Vladimir Joseph Stephan Orlovsky, Feb 08 2012 *)
    Table[RootSum[-1 - # - #^2 + #^3 &, #^n &], {n, 0, 40}] (* Eric W. Weisstein, Mar 30 2017 *)
    RootSum[-1 - # - #^2 + #^3 &, #^Range[0, 40] &] (* Eric W. Weisstein, Aug 17 2017 *)
  • PARI
    {a(n) = if( n<0, polsym(1 - x - x^2 - x^3, -n)[-n+1], polsym(1 + x + x^2 - x^3, n)[n+1])}; /* Michael Somos, Nov 02 2002 */
    
  • PARI
    my(x='x+O('x^40)); Vec((3-2*x-x^2)/(1-x-x^2-x^3)) \\ Altug Alkan, Apr 19 2018
    
  • SageMath
    ((3-2*x-x^2)/(1-x-x^2-x^3)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Mar 22 2019

Formula

Binet's formula: a(n) = r1^n + r2^n + r3^n, where r1, r2, r3 are the roots of the characteristic polynomial 1 + x + x^2 - x^3, see A058265.
a(n) = A000073(n) + 2*A000073(n-1) + 3*A000073(n-2).
G.f.: (3-2*x-x^2)/(1-x-x^2-x^3). - Miklos Kristof, Jul 29 2002
a(n) = n*Sum_{k=1..n} Sum_{j=n-3*k..k} binomial(j, n-3*k+2*j)*binomial(k,j)/k, n > 0, a(0)=3. - Vladimir Kruchinin, Feb 24 2011
a(n) = a(n-1) + a(n-2) + a(n-3), a(0)=3, a(1)=1, a(2)=3. - Harvey P. Dale, Feb 01 2015
a(n) = A073145(-n). for all n in Z. - Michael Somos, Dec 17 2016
Sum_{k=0..n} k*a(k) = (n*a(n+3) - a(n+2) - (n+1)*a(n+1) + 4)/2. - Yichen Wang, Aug 30 2020
a(n) = Trace(M^n), where M = [0, 0, 1; 1, 0, 1; 0, 1, 1] is the companion matrix to the monic polynomial x^3 - x^2 - x - 1. It follows that the sequence satisfies the Gauss congruences: a(n*p^r) == a(n*p^(r-1)) (mod p^r) for positive integers n and r and all primes p. See Zarelua. - Peter Bala, Dec 29 2022

Extensions

Edited by Mario Catalani (mario.catalani(AT)unito.it), Jul 17 2002
Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021

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

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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

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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

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

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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

A140101 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 Y(n) (for X(n) see A140100).

Original entry on oeis.org

0, 2, 5, 8, 11, 13, 16, 19, 22, 25, 28, 31, 33, 36, 39, 42, 45, 48, 50, 53, 56, 59, 62, 65, 68, 70, 73, 76, 79, 81, 84, 87, 90, 93, 96, 99, 101, 104, 107, 110, 113, 116, 118, 121, 124, 127, 130, 133, 136, 138, 141, 144, 147, 150, 153, 156, 158, 161, 164, 167, 170, 173
Offset: 0

Views

Author

Paul D. Hanna, Jun 04 2008

Keywords

Comments

Sequence A140100 = {X(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 A140099(n) = [n*(1+t)], a Beatty sequence involving the tribonacci constant t = t^3 - t^2 - 1 = 1.83928675521416113255...
Theorem: A140099(n) - A140101(n) is always in {-1,0,1} (see A275926). (See also A276385.)
Comments from N. J. A. Sloane, Aug 30 2016: (Start) This is the same problem as the "Greedy Queens in a spiral" problem described in A273059. In A273059 the queens come in sets of 4, each set of 4 being on the same shell around the central square.
a(n) specifies the shell number for the successive sets of 4 (taking the central square to be shell 0, the 8 squares around the center to be shell 1, etc.).
For example, the queens at squares 9, 13, 17, 21 in the spiral (terms A273059(2)-A273059(5)) are all on shell a(1) = 2. The next four queens, at squares 82, 92, 102, 112, are on shell a(2) = 5.
The four "spokes" in A273059 are given in A275916-A275919. The precise connection with the current sequence is that a(n) = nearest integer to (1 + sqrt(A275917(n-1)+1))/2.
This sequence links together the spokes A275916-A275919 in the sense that A275918(n) = A275917(n)+2*a(n+1), A275919(n) = A275917(n)+4*a(n+1), and A275916(n+1) = A275917(n)+6*a(n+1).
(End)
Conjecture: a(n) = A003144(n) + n. (This is from my notebook Lattices 115 page 20 from Oct 25 2016. It is now a theorem - see the Dekking et al. paper.) - N. J. A. Sloane, Jul 22 2019
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 23 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) = A140100(n). - Jeffrey Shallit, Oct 04 2022

Examples

			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.
This sequence gives the y-coordinates of the positive quadrant in the construction given in the examples for A140100.
		

References

  • Robbert Fokkink, Gerard Francis Ortega, Dan Rust, Corner the Empress, arXiv:2204.11805. See Table 3.

Crossrefs

Cf. A140100 (complement); A140102, A140103, A275926 (deviation from A140099).
Cf. related Beatty sequences: A140098, A140099; A000201.
Cf. A058265 (tribonacci constant).
Cf. Greedy Queens in a spiral, A273059, A275916, A275917, A275918, A275919, A275925.
See also A276385.
For first differences of A140100, A140101, A140102, A140103 see A305392, A305374, A305393, A305394.
The indicator function of this sequence is A305386.

Programs

  • Maple
    See link.
  • Mathematica
    y[0] = 0; x[1] = 1; y[1] = 2;
    y[n_] := y[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], x[n] = xn; Return[yn]]]];
    Table[y[n], {n, 0, 100}] (* Jean-François Alcover, Jun 17 2018 *)
  • PARI
    /* 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))))))}

Formula

CONJECTURE: the limit of a(n)/n = 1+t and limit of X(n)/n = 1+1/t so that limit of a(n)/X(n) = t = tribonacci constant (A058265), and thus the limit of [a(n) + X(n)]/[a(n) - X(n)] = t^2 and the limit of [a(n)^2 + X(n)^2]/[a(n)^2 - X(n)^2] = t.
Conjectured recursion: Take first differences: 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, ... (appears to consist of only 3's and 2's); list the run lengths: 3, 1, 6, 1, 5, 1, 6, 1, 3, 1, 6, 1, 5, 1, 6, 1, ... (it appears that every second term is 1 and the other terms are 3, 5, and 6); and bisect, getting 3, 6, 5, 6, 3, 6, 5, 6, 6, 5, 6, 3, 6, ... This is (although I do not have a proof) the recursively defined A275925. Thanks to Alois P. Heinz for providing enough terms of A273059 to enable a (morally) convincing check of this conjecture. - N. J. A. Sloane, Aug 30 2016
From Michel Dekking, Mar 17 2019: (Start)
This conjecture can be reformulated as follows (cf. A140100).
The first differences of (a(n)) = (Y(n)) as a word are given by
3 delta(x),
where x is the tribonacci word x = A092782, and delta is the morphism
1 -> 3333332,
2 -> 333332,
3 -> 3332.
This conjecture implies the frequency conjecture above: let N(i,n) be the number of letters i in a(1)a(2)...a(n). Then simple counting gives
a(7*N(1,n)+6*N(2,n)+4*N(3,n)) = 20*N(1,n)+17*N(2,n)+11*N(3,n), where we neglected the first symbol of a = Y.
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/t + 17/t^2 + 11/t^3 = (1+t)*(7/t + 6/t^2 + 4/t^3).
This is a simple verification, using t^3 = t^2 + t + 1.
(End)

Extensions

Terms computed independently by Reinhard Zumkeller and Joshua Zucker
Edited and a(0)=0 added by N. J. A. Sloane, Aug 30 2016

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).

Original entry on oeis.org

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

Views

Author

Paul D. Hanna, Jun 04 2008

Keywords

Comments

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

Examples

			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,...].
		

Crossrefs

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.

Programs

  • Maple
    See link.
  • Mathematica
    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 *)
  • PARI
    /* 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))))))}

Formula

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)

Extensions

Terms computed independently by Reinhard Zumkeller and Joshua Zucker
Edited by N. J. A. Sloane, Aug 30 2016

A192918 Decimal expansion of the real root of r^3 + r^2 + r - 1.

Original entry on oeis.org

5, 4, 3, 6, 8, 9, 0, 1, 2, 6, 9, 2, 0, 7, 6, 3, 6, 1, 5, 7, 0, 8, 5, 5, 9, 7, 1, 8, 0, 1, 7, 4, 7, 9, 8, 6, 5, 2, 5, 2, 0, 3, 2, 9, 7, 6, 5, 0, 9, 8, 3, 9, 3, 5, 2, 4, 0, 8, 0, 4, 0, 3, 7, 8, 3, 1, 1, 6, 8, 6, 7, 3, 9, 2, 7, 9, 7, 3, 8, 6, 6, 4, 8, 5, 1, 5, 7, 9, 1, 4, 5, 7, 6, 0, 5, 9, 1, 2, 5, 4, 6, 2
Offset: 0

Views

Author

Frank M Jackson, Aug 26 2011

Keywords

Comments

The real solution r of the cubic equation r^3 + r^2 + r - 1 = 0 is the reciprocal of the tribonacci constant A058265. If the four sides of a quadrilateral form a geometric progression 1:r:r^2:r^3 where r is the common ratio then r is limited to the range 1/t < r < t where t is the tribonacci constant. More generally if f(n) is the n-th step Fibonacci constant then a polygon of n+1 sides can have sides in a geometric progression 1:r:r^2:...:r^n if the common ratio r is limited to the range 1/f(n) < r < f(n).
From Wolfdieter Lang, Aug 22 2022: (Start)
The roots of this cubic are obtained from the roots of y^3 + (2/3)*y - 34/27 after subtracting 1/3. The y-roots are y1 = (u_p^(1/3) + u_m^(1/3)*e_m)/3, y2 = (e_m*u_p^(1/3) + u_m^(1/3))/3 and y3 = e_p*(u_p^(1/3) + u_m^(1/3))/3. Here u_p = 17 + 3*sqrt(33), u_m = 17 - 3*sqrt(33), e_p = -(1 + sqrt(3)*i) and e_m = -(1 - sqrt(3)*i), where i = sqrt(-1).
The roots of the x-cubic are then x1, the present real solution, and x2 = y2 - 1/3 = -0.771844506... + 1.11514250...*i and the complex conjugate x3 = y3 - 1/3. (End)

Examples

			0.543689012692076361570855971801747986525203297650983935240...
		

Crossrefs

Reciprocal of A058265.
Cf. A376841.

Programs

  • Magma
    SetDefaultRealField(RealField(100)); (1/3)*(-1 -2/(17 +3*Sqrt(33))^(1/3) +(17+3*Sqrt(33))^(1/3)); // G. C. Greubel, Feb 06 2019
    
  • Mathematica
    N[Reduce[r+r^2+r^3==1, r], 100]
    RealDigits[(1/3)*(-1 -2/(17+3*Sqrt[33])^(1/3) +(17+3*Sqrt[33])^(1/3)), 10, 100][[1]] (* G. C. Greubel, Feb 06 2019 *)
    RealDigits[Root[r^3+r^2+r-1,1],10,120][[1]] (* Harvey P. Dale, May 18 2023 *)
  • PARI
    polrootsreal(r^3 + r^2 + r - 1)[1] \\ Charles R Greathouse IV, Apr 14 2014
    
  • Sage
    numerical_approx((1/3)*(-1 -2/(17+3*sqrt(33))^(1/3) +(17+ 3*sqrt(33))^(1/3)), digits=100) # G. C. Greubel, Feb 06 2019

Formula

Equals (1/3)*(-1-2/(17+3*sqrt(33))^(1/3) + (17+3*sqrt(33))^(1/3)).
Equals (1/3)*(u_p^(1/3) + u_m^(1/3)*e_m - 1), with u_p = 17 + 3*sqrt(33), u_m = 17 - 3*sqrt(33), and e_m = -(1 - sqrt(3)*i), with i = sqrt(-1). - Wolfdieter Lang, Aug 22 2022
Equals hypergeom([1/4,1/2,3/4],[2/3,4/3],16/27)/2. - Gerry Martens, Jul 13 2023

A086088 Decimal expansion of the limit of the ratio of consecutive terms in the tetranacci sequence A000078.

Original entry on oeis.org

1, 9, 2, 7, 5, 6, 1, 9, 7, 5, 4, 8, 2, 9, 2, 5, 3, 0, 4, 2, 6, 1, 9, 0, 5, 8, 6, 1, 7, 3, 6, 6, 2, 2, 1, 6, 8, 6, 9, 8, 5, 5, 4, 2, 5, 5, 1, 6, 3, 3, 8, 4, 7, 2, 7, 1, 4, 6, 6, 4, 7, 0, 3, 8, 0, 0, 9, 6, 6, 6, 0, 6, 2, 2, 9, 7, 8, 1, 5, 5, 5, 9, 1, 4, 9, 8, 1, 8, 2, 5, 3, 4, 6, 1, 8, 9, 0, 6, 5, 3, 2, 5
Offset: 1

Views

Author

Eric W. Weisstein, Jul 08 2003

Keywords

Comments

The tetranacci constant corresponds to the Golden Section in a quadripartite division 1 = u_1 + u_2 + u_3 + u_4 of a unit line segment, i.e., if 1/u_1 = u_1/u_2 = u_2/u_3 = u_3/u_4 = c, c is the tetranacci constant. - Seppo Mustonen, Apr 19 2005
The other 3 polynomial roots of 1+x+x^2+x^3-x^4 are -0.77480411321543385... and the complex-conjugated pair -0.07637893113374572508475 +- i * 0.814703647170386526841... - R. J. Mathar, Oct 25 2008
The continued fraction expansion starts 1, 1, 12, 1, 4, 7, 1, 21, 1, 2, 1, 4, 6, 1, 10, 1, 2, 2, 1, 7, 1, 1,... - R. J. Mathar, Mar 09 2012
For n>=4, round(c^prime(n)) == 1 (mod 2*prime(n)). Proof in Shevelev link. - Vladimir Shevelev, Mar 21 2014
Note that we have: c + c^(-4) = 2, and the k-nacci constant approaches 2 when k approaches infinity (Martin Gardner). - Bernard Schott, May 09 2022

Examples

			1.927561975...
		

References

  • Martin Gardner, The Second Scientific American Book Of Mathematical Puzzles and Diversions, "Phi: The Golden Ratio", Chapter 8, p. 101, Simon & Schuster, NY, 1961.

Crossrefs

Cf. A000078.
k-nacci constants: A001622 (Fibonacci), A058265 (tribonacci), this sequence (tetranacci), A103814 (pentanacci), A118427 (hexanacci), A118428 (heptanacci).

Programs

Formula

Equals 1/4 + sqrt(11/48 - s/72 + 7/s) + sqrt(11/24 + s/72 - 7/s + 1 / sqrt(704/507 - 128 * s/1521 + 7168 / (169 * s))) where s = (sqrt(177304464) + 7020)^(1/3). - Michal Paulovic, Oct 08 2022

A135491 Number of ways to toss a coin n times and not get a run of four.

Original entry on oeis.org

1, 2, 4, 8, 14, 26, 48, 88, 162, 298, 548, 1008, 1854, 3410, 6272, 11536, 21218, 39026, 71780, 132024, 242830, 446634, 821488, 1510952, 2779074, 5111514, 9401540, 17292128, 31805182, 58498850, 107596160, 197900192, 363995202, 669491554, 1231386948, 2264873704
Offset: 0

Views

Author

James R FitzSimons (cherry(AT)getnet.net), Feb 07 2008

Keywords

Crossrefs

Cf. A000073. Column 2 of A265624. Cf. A135492, A135493, A000213, A058265.

Programs

Formula

a(n) = 2*A000073(n+2) for n > 0.
a(n) = a(n-1) + a(n-2) + a(n-3) for n > 3.
G.f.: -(x+1)*(x^2+1)/(x^3+x^2+x-1).
a(n) = nearest integer to b*c^n, where b = 1.2368... and c = 1.839286755... is the real root of x^3-x^2-x-1 = 0. See A058265. - N. J. A. Sloane, Jan 06 2010
G.f.: (1-x^4)/(1-2*x+x^4) and generally to "not get a run of k" (1-x^k)/(1-2*x+x^k). - Geoffrey Critzer, Feb 01 2012
G.f.: Q(0)/x^2 - 2/x- 1/x^2, where Q(k) = 1 + (1+x)*x^2 + (2*k+3)*x - x*(2*k+1 +x+x^2)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 04 2013
a(n) = A000213(n+3) - A000213(n+2), n>=1. - Peter M. Chema, Jan 11 2017.

Extensions

More terms from Robert G. Wilson v, Feb 10 2008
a(0)=1 prepended by Alois P. Heinz, Dec 10 2015
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