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.

Showing 1-4 of 4 results.

A001590 Tribonacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3) with a(0)=0, a(1)=1, a(2)=0.

Original entry on oeis.org

0, 1, 0, 1, 2, 3, 6, 11, 20, 37, 68, 125, 230, 423, 778, 1431, 2632, 4841, 8904, 16377, 30122, 55403, 101902, 187427, 344732, 634061, 1166220, 2145013, 3945294, 7256527, 13346834, 24548655, 45152016, 83047505, 152748176, 280947697, 516743378, 950439251
Offset: 0

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Author

Keywords

Comments

The name "tribonacci number" is less well-defined than "Fibonacci number". The sequence A000073 (which begins 0, 0, 1) is probably the most important version, but the name has also been applied to A000213, A001590, and A081172. - N. J. A. Sloane, Jul 25 2024
Dimensions of the homogeneous components of the higher order peak algebra associated to cubic roots of unity (Hilbert series = 1 + 1*t + 2*t^2 + 3*t^3 + 6*t^4 + 11*t^5 ...). - Jean-Yves Thibon (jyt(AT)univ-mlv.fr), Oct 22 2006
Starting with offset 3: (1, 2, 3, 6, 11, 10, 37, ...) = row sums of triangle A145579. - Gary W. Adamson, Oct 13 2008
Starting (1, 2, 3, 6, 11, ...) = INVERT transform of the periodic sequence (1, 1, 0, 1, 1, 0, 1, 1, 0, ...). - Gary W. Adamson, May 04 2009
The comment of May 04 2009 is equivalent to: The numbers of ordered compositions of n using integers that are not multiples of 3 is equal to a(n+2) for n>=1, see [Hoggatt-Bicknell (1975) eq (2.7)]. - Gary W. Adamson, May 13 2013
Primes in the sequence are 2, 3, 11, 37, 634061, 7256527, ... in A231574. - R. J. Mathar, Aug 09 2012
Pisano period lengths: 1, 2, 13, 8, 31, 26, 48, 16, 39, 62,110,104,168, 48,403, 32, 96, 78, 360, 248, ... . - R. J. Mathar, Aug 10 2012
a(n+1) is the top left entry of the n-th power of any of 3 X 3 matrices [0, 1, 0; 1, 1, 1; 1, 0, 0], [0, 1, 1; 1, 1, 0; 0, 1, 0], [0, 1, 1; 0, 0, 1; 1, 0, 1] or [0, 0, 1; 1, 0, 0; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
a(n+3) equals the number of n-length binary words avoiding runs of zeros of lengths 3i+2, (i=0,1,2,...). - Milan Janjic, Feb 26 2015
Sums of pairs of successive terms of A000073. - N. J. A. Sloane, Oct 30 2016
The power Q^n, for n >= 0, of the tribonacci Q-matrix Q = matrix([1, 1, 1], [1, 0, 0], [0, 1, 0]) is, by the Cayley-Hamilton theorem, Q^n = matrix([a(n+2), a(n+1) + a(n), a(n+1)], [a(n+1), a(n) + a(n-1), a(n)], [a(n), a(n-1) + a(n-2), a(n-1)]), with a(-2) = -1 and a(-1) = 1. One can use a(n) = a(n-1) + a(n-2) + a(n-3) in order to obtain a(-1) and a(-2). - Wolfdieter Lang, Aug 13 2018
a(n+2) is the number of entries n, for n>=1, in the sequence {A278038(k)}A278038(0)%20=%201).%20-%20_Wolfdieter%20Lang">{k>=1} (without A278038(0) = 1). - _Wolfdieter Lang, Sep 11 2018
In terms of the tribonacci numbers T(n) = A000073(n) the nonnegative powers of the Q-matrix (from the Aug 13 2018 comment) are Q^n = T(n)*Q^2 + (T(n-1) + T(n-2))*Q + T(n-1)*1_3, for n >= 0, with T(-1) = 1, T(-2) = -1. This is equivalent to the powers t^n of the tribonacci constant t = A058255 (or also for powers of the complex solutions). - Wolfdieter Lang, Oct 24 2018

Examples

			a(12)=a(11)+a(10)+a(9): 230=125+68+37.
For n=5 the partitions of 5 are 1+1+1+1+1 (1 composition), 1+1+1+2 (4 compositions), 1+2+2 (3 compositions), 1+1+3 (not contrib because 3 is a part), 2+3 (no contrib because 3 is a part), 1+4 (2 compositions) and 5 (1 composition), total 1+4+3+2+1=11 =a(5+2) - _R. J. Mathar_, Jan 13 2023
		

References

  • Kenneth Edwards and Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers squared, Part II, Fib. Q., 58:2 (2020), 169-177.
  • 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

Programs

  • GAP
    a:=[0,1,0];; for n in [4..40] do a[n]:=a[n-1]+a[n-2]+a[n-3]; od; a; # Muniru A Asiru, Oct 24 2018
  • Magma
    I:=[0,1,0]; [n le 3 select I[n]  else Self(n-1)+Self(n-2)+Self(n-3): n in [1..40]]; // Vincenzo Librandi, Apr 19 2018
    
  • Maple
    seq(coeff(series(x*(1-x)/(1-x-x^2-x^3),x,n+1), x, n), n = 0 .. 40); # Muniru A Asiru, Oct 24 2018
    # alternative
    A001590 := proc(n)
        option remember;
        if n <=2 then
            op(n+1,[0,1,0]) ;
        else
            procname(n-1)+procname(n-2)+procname(n-3) ;
        end if;
    end proc:
    seq(A001590(n),n=0..30) ;# R. J. Mathar, Nov 22 2024
  • Mathematica
    LinearRecurrence[{1,1,1}, {0,1,0}, 50] (* Vladimir Joseph Stephan Orlovsky, Jan 28 2012 *)
    RecurrenceTable[{a[0]==0, a[1]==1, a[2]==0, a[n]==a[n-1]+a[n-2]+a[n-3]}, a, {n, 40}] (* Vincenzo Librandi, Apr 19 2018 *)
  • PARI
    a(n)=([0,1,0; 0,0,1; 1,1,1]^n*[0;1;0])[1,1] \\ Charles R Greathouse IV, Jul 28 2015
    
  • Sage
    def A001590():
        W = [0, 1, 0]
        while True:
            yield W[0]
            W.append(sum(W))
            W.pop(0)
    a = A001590(); [next(a) for  in range(38)]  # _Peter Luschny, Sep 12 2016
    

Formula

G.f.: x*(1-x)/(1-x-x^2-x^3).
Limit a(n)/a(n-1) = t where t is the real solution of t^3 = 1 + t + t^2, t = A058265 = 1.839286755... . If T(n) = A000073(n) then t^n = T(n-1) + a(n)*t + T(n)*t^2, for n >= 0, with T(-1) = 1.
a(3*n) = Sum_{k+l+m=n} (n!/k!l!m!)*a(l+2*m). Example: a(12)=a(8)+4a(7)+10a(6)+16a(5)+19a(4)+16a(3)+10a(2)+4a(1)+a(0) The coefficients are the trinomial coefficients. T(n) and T(n-1) also satisfy this equation. (T(-1)=1)
From Reinhard Zumkeller, May 22 2006: (Start)
a(n) = A000073(n+1)-A000073(n);
a(n) = A000073(n-1)+A000073(n-2) for n>1;
A000213(n-2) = a(n+1)-a(n) for n>1. (End)
a(n) + a(n+1) = A000213(n). - Philippe Deléham, Sep 25 2006
If p[1]=0, p[i]=2, (i>1), and if A is Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n>=1, a(n+1)=det A. - Milan Janjic, May 02 2010
For n>=4, a(n)=2*a(n-1)-a(n-4). - Bob Selcoe, Feb 18 2014
a(-1-n) = -A078046(n). - Michael Somos, Jun 01 2014
a(n) = Sum_{r root of x^3-x^2-x-1} r^n/(3*r+1). - Fabian Pereyra, Nov 22 2024

Extensions

Additional comments from Miklos Kristof, Jul 03 2002

A090248 a(n) = 27a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 27.

Original entry on oeis.org

2, 27, 727, 19602, 528527, 14250627, 384238402, 10360186227, 279340789727, 7531841136402, 203080369893127, 5475638145978027, 147639149571513602, 3980781400284889227, 107333458658120495527, 2894022602368968490002, 78031276805304028734527
Offset: 0

Views

Author

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 24 2004

Keywords

Comments

a(n+1)/a(n) converges to ((27+sqrt(725))/2) = 26.96291201...
Lim a(n)/a(n+1) as n approaches infinity = 0.03708798... = 2/(27+sqrt(725)) = (27-sqrt(725))/2.
Lim a(n+1)/a(n) as n approaches infinity = 26.96291201... = (27+sqrt(725))/2 = 2/(27-sqrt(725)).
Lim a(n)/a(n+1) = 27 - Lim a(n+1)/a(n).
A Chebyshev T-sequence with Diophantine property.
a(n) gives the general (nonnegative integer) solution of the Pell equation a^2 - 29*(5*b)^2 =+4 with companion sequence b(n)=A097781(n-1), n>=0.

Examples

			a(4) = 528527 = 27a(3) - a(2) = 27*19602 - 727 = ((27+sqrt(725))/2)^4 + ((27-sqrt(725))/2)^4 = 528526.999998107 + 0.000001892 = 528527.
(x;y) = (2;0), (27;1), (727;27), (19602;728), ... give the nonnegative integer solutions to x^2 - 29*(5*y)^2 = +4.
		

References

  • O. Perron, "Die Lehre von den Kettenbruechen, Bd.I", Teubner, 1954, 1957 (Sec. 30, Satz 3.35, p. 109 and table p. 108).

Crossrefs

a(n)=sqrt(4 + 29*(5*A097781(n-1))^2), n>=1.
Cf. A077428, A078355 (Pell +4 equations).
Cf. A090733 for 2*T(n, 25/2).
Cf. A087130.

Programs

  • Mathematica
    a[0] = 2; a[1] = 27; a[n_] := 27a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 15}] (* Robert G. Wilson v, Jan 30 2004 *)
    RecurrenceTable[{a[0]==2,a[1]==27,a[n]==27a[n-1]-a[n-2]},a,{n,20}] (* or *) LinearRecurrence[{27,-1},{2,27},20] (* Harvey P. Dale, Jan 03 2018 *)
  • PARI
    {a(n) = (-1)^n * subst(2 * poltchebi(2*n), 'x, -5/2 * I)}; /* Michael Somos, Nov 04 2008 */
    
  • Python
    def aupton(idx):
      alst = [2, 27]
      for n in range(2, idx+1): alst.append(27*alst[-1] - alst[-2])
      return alst
    print(aupton(16)) # Michael S. Branicky, Feb 27 2021
  • Sage
    [lucas_number2(n,27,1) for n in range(0,16)] # Zerinvary Lajos, Jun 27 2008
    

Formula

a(n) = 27a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 27. a(n) = ((27+sqrt(725))/2)^n + ((27-sqrt(725))/2)^n, (a(n))^2 = a(2n)+2.
a(n) = S(n, 27) - S(n-2, 27) = 2*T(n, 27/2) with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. S(n, 27)=A097781(n). U-, resp. T-, are Chebyshev's polynomials of the second, resp. first, kind. See A049310 and A053120.
a(n) = ap^n + am^n, with ap := (27+5*sqrt(29))/2 and am := (27-5*sqrt(29))/2.
G.f.: (2-27*x)/(1-27*x+x^2).
a(-n) = a(n). - Michael Somos, Nov 01 2008
A087130(2*n) = a(n). - Michael Somos, Nov 01 2008

Extensions

More terms from Robert G. Wilson v, Jan 30 2004
Chebyshev and Pell comments from Wolfdieter Lang, Aug 31 2004

A075298 Inverted (definition in A075193) generalized tribonacci numbers A001644.

Original entry on oeis.org

1, 1, -5, 5, 1, -11, 15, -3, -23, 41, -21, -43, 105, -83, -65, 253, -271, -47, 571, -795, 177, 1189, -2161, 1149, 2201, -5511, 4459, 3253, -13223, 14429, 2047, -29699, 42081, -10335, -61445, 113861, -62751, -112555, 289167, -239363, -162359, 690889, -767893, -85355, 1544137, -2226675, 597183, 3173629
Offset: 0

Views

Author

Mario Catalani (mario.catalani(AT)unito.it), Sep 13 2002

Keywords

Comments

a(n) = -C(n+1), C(n)=reflected generalized tribonacci numbers A073145.

Crossrefs

Programs

  • GAP
    a:=[1,1,-5];; for n in [4..50] do a[n]:=-a[n-1]-a[n-2]+a[n-3]; od; a; # G. C. Greubel, Apr 09 2019
  • Magma
    R:=PowerSeriesRing(Integers(), 50); Coefficients(R!( (1+2*x-3*x^2)/(1+x+x^2-x^3) )); // G. C. Greubel, Apr 09 2019
    
  • Mathematica
    CoefficientList[Series[(1+2x-3x^2)/(1+x+x^2-x^3), {x, 0, 50}], x]
  • PARI
    my(x='x+O('x^50)); Vec((1+2*x-3*x^2)/(1+x+x^2-x^3)) \\ G. C. Greubel, Apr 09 2019
    
  • Sage
    ((1+2*x-3*x^2)/(1+x+x^2-x^3)).series(x, 50).coefficients(x, sparse=False) # G. C. Greubel, Apr 09 2019
    

Formula

a(n) = -a(n-1) - a(n-2) + a(n-3), a(0)=1, a(1)=1, a(2)=-5.
G.f.: (1+2*x-3*x^2)/(1+x+x^2-x^3).
a(n) = A078046(n) + 3*A078046(n-1). - R. J. Mathar, Sep 20 2020

A105580 a(n+3) = a(n) - a(n+1) - a(n+2); a(0) = -5, a(1) = 6, a(2) = 0.

Original entry on oeis.org

-5, 6, 0, -11, 17, -6, -22, 45, -29, -38, 112, -103, -47, 262, -318, 9, 571, -898, 336, 1133, -2367, 1570, 1930, -5867, 5507, 2290, -13664, 16881, -927, -29618, 47426, -18735, -58309, 124470, -84896, -97883, 307249, -294262, -110870, 712381, -895773, 72522, 1535632, -2503927, 1040817, 2998742
Offset: 0

Views

Author

Creighton Dement, Apr 14 2005

Keywords

Comments

Floretion Algebra Multiplication Program, FAMP Code: 2tesforseq[.5'j + .5'k + .5j' + .5k' + .5'ii' + .5e], 1vesforseq = A000004, ForType: 1A.

Examples

			This sequence was generated using the same floretion which generated the sequences A105577, A105578, A105579, etc.. However, in this case a force transform was applied. [Specifically, (a(n)) may be seen as the result of a tesfor-transform of the zero-sequence A000004 with respect to the floretion given in the program code.]
		

Crossrefs

Programs

  • Mathematica
    Transpose[NestList[Join[Rest[#],ListCorrelate[ {1,-1,-1}, #]]&,{-5,6,0},50]][[1]]  (* Harvey P. Dale, Mar 14 2011 *)
    CoefficientList[Series[(5-x-x^2)/(x^3-x^2-x-1),{x,0,50}],x]  (* Harvey P. Dale, Mar 14 2011 *)

Formula

G.f. (5-x-x^2)/(x^3-x^2-x-1)
a(n) = A078046(n-1) - A073145(n+3).
a(n) = -5*A057597(n+2) + A057597(n+1)+A057597(n). - R. J. Mathar, Oct 25 2022
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