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

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

Views

Author

N. J. A. Sloane, Mar 29 2003

Keywords

Comments

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)

Examples

			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)
		

References

  • 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.

Crossrefs

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

Programs

  • Maple
    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
  • Mathematica
    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 *)
  • 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=[0];  for (k=1, 10, t=strsub( t, [[0,1], [0,2], [0]], 0 ) );  t
    \\ Joerg Arndt, Sep 14 2013

Formula

Fixed point of morphism 0 -> 0, 1; 1 -> 0, 2; 2 -> 0.
a(n) = A092782(n+1) - 1. - Joerg Arndt, Sep 14 2013

Extensions

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

A100683 a(n) = a(n-1) + a(n-2) + a(n-3); a(0) = -1, a(1) = 2, a(2) = 2.

Original entry on oeis.org

-1, 2, 2, 3, 7, 12, 22, 41, 75, 138, 254, 467, 859, 1580, 2906, 5345, 9831, 18082, 33258, 61171, 112511, 206940, 380622, 700073, 1287635, 2368330, 4356038, 8012003, 14736371, 27104412, 49852786, 91693569, 168650767, 310197122
Offset: 0

Views

Author

N. J. A. Sloane, Dec 08 2004

Keywords

Comments

A tribonacci sequence.
From Greg Dresden and Veda Garigipati, Jun 14 2022: (Start)
For n >= 2, a(n+2) is the number of ways to tile this figure of length n with squares, dominoes, and "trominoes" (of length 3):
_
|||___________
|||_|||_|||
As an example, here is one of the 254 possible tilings of this figure of length 8 with squares, dominoes, and trominoes:
_
||____|_|_|_|. (End)

Crossrefs

Programs

  • Maple
    a[0]:=-1:a[1]:=2:a[2]:=2:for n from 3 to 42 do a[n]:=a[n-1]+a[n-2]+a[n-3] od: seq(a[n],n=0..42);
  • Mathematica
    a[0] = -1; a[1] = a[2] = 2; a[n_] := a[n] = a[n - 1] + a[n - 2] + a[n - 3]; Table[ a[n], {n, 0, 35}] (* Robert G. Wilson v, Dec 09 2004 *)
    LinearRecurrence[{1,1,1},{-1,2,2},34] (* Ray Chandler, Dec 08 2013 *)
  • PARI
    Vec(-(1-3*x-x^2)/(1-x-x^2-x^3) + O(x^70)) \\ Michel Marcus, Sep 25 2015

Formula

a(n+1) = 2*A001590(n+1) + A020992(n). - Creighton Dement, May 02 2005
O.g.f.: -(1-3x-x^2)/(1-x-x^2-x^3). - R. J. Mathar, Aug 22 2008
a(n) = T(n-2) + T(n) + T(n+1) where T(n) = A000073(n) the tribonacci sequence, for n >= 2. - Greg Dresden and Veda Garigipati, Jun 14 2022

Extensions

More terms from Emeric Deutsch, Farideh Firoozbakht and Robert G. Wilson v, Dec 08 2004

A214727 a(n) = a(n-1) + a(n-2) + a(n-3) with a(0) = 1, a(1) = a(2) = 2.

Original entry on oeis.org

1, 2, 2, 5, 9, 16, 30, 55, 101, 186, 342, 629, 1157, 2128, 3914, 7199, 13241, 24354, 44794, 82389, 151537, 278720, 512646, 942903, 1734269, 3189818, 5866990, 10791077, 19847885, 36505952, 67144914, 123498751, 227149617, 417793282
Offset: 0

Views

Author

Abel Amene, Jul 27 2012

Keywords

Comments

Part of a group of sequences defined by a(0), a(1)=a(2), a(n) = a(n-1) + a(n-2) + a(n-3) which is a subgroup of sequences with linear recurrences and constant coefficients listed in the index.
Note: A000073 (with offset=1), 1 followed by A000073, A000213, A141523, A214727, A214825 to A214831 completely define possible sequences with a(0)=0,1,2...9 and a(1)=a(2)=0,1,2...9 excluding any multiples of these sequences and the trivial case of a(0)=a(1)=a(2)=0.
Note: allowing a(0)=0 and a(1)=a(2)=1,2,3....9 leads to A000073 (with offset=1) and its multiples.
Note: allowing a(0)=1,2,3....9 a(1)=a(2)=0 leads to 1 followed by A000073 and its multiples.
With offset of 6 this sequence is the 8th row of tribonacci array A136175.

Examples

			G.f. = 1 + 2*x + 2*x^2 + 5 x^3 + 9*x^4 + 16*x^5 + 30*x^6 + 55*x^7 + ...
		

Crossrefs

Programs

  • GAP
    a:=[1,2,2];; for n in [4..40] do a[n]:=a[n-1]+a[n-2]+a[n-3]; od; a; # G. C. Greubel, Apr 23 2019
  • Haskell
    a214727 n = a214727_list !! n
    a214727_list = 1 : 2 : 2 : zipWith3 (\x y z -> x + y + z)
       a214727_list (tail a214727_list) (drop 2 a214727_list)
    -- Reinhard Zumkeller, Jul 31 2012
    
  • Magma
    R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (1+x-x^2)/(1-x-x^2-x^3) )); // G. C. Greubel, Apr 23 2019
    
  • Mathematica
    LinearRecurrence[{1,1,1},{1,2,2},40] (* Ray Chandler, Dec 08 2013 *)
  • PARI
    a(n)=([0,1,0; 0,0,1; 1,1,1]^n*[1;2;2])[1,1] \\ Charles R Greathouse IV, Mar 22 2016
    
  • PARI
    my(x='x+O('x^40)); Vec((1+x-x^2)/(1-x-x^2-x^3)) \\ G. C. Greubel, Apr 23 2019
    
  • SageMath
    ((1+x-x^2)/(1-x-x^2-x^3)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Apr 23 2019
    

Formula

G.f.: (1+x-x^2)/(1-x-x^2-x^3).
a(n) = K(n) -2*T(n+1) + 3*T(n), where K(n) = A001644(n), T(n) = A000073(n+1). - G. C. Greubel, Apr 23 2019
a(n) = Sum_{r root of x^3-x^2-x-1} r^n/(-r^2+2*r+1). - Fabian Pereyra, Nov 20 2024

A141523 Expansion of (3-2*x-3*x^2)/(1-x-x^2-x^3).

Original entry on oeis.org

3, 1, 1, 5, 7, 13, 25, 45, 83, 153, 281, 517, 951, 1749, 3217, 5917, 10883, 20017, 36817, 67717, 124551, 229085, 421353, 774989, 1425427, 2621769, 4822185, 8869381, 16313335, 30004901, 55187617, 101505853, 186698371, 343391841
Offset: 0

Views

Author

Roger L. Bagula and Gary W. Adamson, Aug 11 2008

Keywords

Crossrefs

Programs

  • Magma
    I:=[3, 1, 1]; [n le 3 select I[n] else Self(n-1)+Self(n-2) +Self(n-3): n in [1..40]]; // Vincenzo Librandi, Oct 17 2012
    
  • Mathematica
    a[0]=3; a[1]=1; a[2]=1; a[n_]:= a[n]=a[n-1]+a[n-2]+a[n-3]; Table[a[n], {n, 0, 40}]
    LinearRecurrence[{1, 1, 1}, {3, 1, 1}, 40] (* Vincenzo Librandi, Oct 17 2012 *)
  • PARI
    a(n)=([0,1,0; 0,0,1; 1,1,1]^n*[3;1;1])[1,1] \\ Charles R Greathouse IV, Mar 22 2016
    
  • PARI
    my(x='x+O('x^40)); Vec((3-2*x-3*x^2)/(1-x-x^2-x^3)) \\ G. C. Greubel, Apr 22 2019
    
  • Sage
    ((3-2*x-3*x^2)/(1-x-x^2-x^3)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Apr 22 2019

Formula

a(0)=3; a(1)=1; a(2)=1; thereafter a(n) = a(n-1) + a(n-2) + a(n-3).
From R. J. Mathar, Aug 22 2008: (Start)
O.g.f.: (3-2*x-3*x^2)/(1-x-x^2-x^3).
a(n) = A001644(n) - 2*A000073(n). (End)

Extensions

Edited by N. J. A. Sloane, Oct 17 2012

A020992 a(n) = a(n-1) + a(n-2) + a(n-3), with a(0) = 0, a(1) = 2, a(2) = 1.

Original entry on oeis.org

0, 2, 1, 3, 6, 10, 19, 35, 64, 118, 217, 399, 734, 1350, 2483, 4567, 8400, 15450, 28417, 52267, 96134, 176818, 325219, 598171, 1100208, 2023598, 3721977, 6845783, 12591358, 23159118, 42596259, 78346735, 144102112, 265045106, 487493953, 896641171, 1649180230
Offset: 0

Views

Author

Keywords

Comments

Tribonacci sequence beginning 0, 2, 1.
Pisano period lengths: 1, 4, 13, 8, 31, 52, 48, 16, 39, 124, 110, 104, 168, 48, 403, 32, 96, 156, 360, 248,.... - R. J. Mathar, Aug 10 2012
One bisection is 0, 1, 6, 19, 64, 217, 734, 2483, 8400,.. and the other 2, 3, 10, 35, 118, 399, 1350, 4567,... both with recurrence b(n)=3*b(n-1)+b(n-2)+b(n-3). - R. J. Mathar, Aug 10 2012
From Greg Dresden and Jiarui Zhou, Jun 30 2025: (Start)
For n >= 4, 2*a(n) is the number of ways to tile this shape of length n-2 with squares, dominos, and trominos (of length 3):
._
|||___________
|||_|||_|||
|_|.
As an example, here is one of the 2*a(10) = 434 ways to tile this shape of length 8:
._
| |_|_____|||
|_| (End)

Crossrefs

Programs

  • Magma
    I:=[0,2,1]; [n le 3 select I[n] else Self(n-1) + Self(n-2) + Self(n-3): n in [1..30]]; // G. C. Greubel, Feb 09 2018
  • Mathematica
    LinearRecurrence[{1,1,1},{0,2,1},100] (* Vladimir Joseph Stephan Orlovsky, Jun 07 2011 *)
  • PARI
    my(x='x+O('x^30)); concat([0], Vec(x*(2-x)/(1-x-x^2-x^3))) \\ G. C. Greubel, Feb 09 2018
    

Formula

G.f.: x*(2-x)/(1-x-x^2-x^3).
a(n) = 2*A000073(n+1)-A000073(n). - R. J. Mathar, Aug 22 2008
a(n) = 2*a(n-1) - a(n-4), n>3. - Vincenzo Librandi, Jun 08 2011

A214899 a(n) = a(n-1) + a(n-2) + a(n-3) with a(0)=2, a(1)=1, a(2)=2.

Original entry on oeis.org

2, 1, 2, 5, 8, 15, 28, 51, 94, 173, 318, 585, 1076, 1979, 3640, 6695, 12314, 22649, 41658, 76621, 140928, 259207, 476756, 876891, 1612854, 2966501, 5456246, 10035601, 18458348, 33950195, 62444144, 114852687, 211247026, 388543857
Offset: 0

Views

Author

Abel Amene, Jul 29 2012

Keywords

Comments

With offset of 5 this sequence is the 4th row of the tribonacci array A136175.
For n>0, a(n) is the number of ways to tile a strip of length n with squares, dominoes, and trominoes, such that there must be exactly one "special" square (say, of a different color) in the first three cells. - Greg Dresden and Emma Li, Aug 17 2024
From Greg Dresden and Jiarui Zhou, Jun 30 2025: (Start)
For n >= 3, a(n) is the number of ways to tile this shape of length n-1 with squares, dominos, and trominos (of length 3):
._
|||_|||_|||
|_|
As an example, here is one of the a(9) = 173 ways to tile this shape of length 8:
._
|| |__|_|___|
|_|. (End)

Crossrefs

Programs

  • GAP
    a:=[2,1,2];; for n in [4..40] do a[n]:=a[n-1]+a[n-2]+a[n-3]; od; a; # G. C. Greubel, Apr 23 2019
  • Magma
    R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (2-x-x^2)/(1-x-x^2-x^3) )); // G. C. Greubel, Apr 23 2019
    
  • Mathematica
    LinearRecurrence[{1,1,1},{2,1,2},34] (* Ray Chandler, Dec 08 2013 *)
  • PARI
    a(n)=([0,1,0;0,0,1;1,1,1]^n*[2;1;2])[1,1] \\ Charles R Greathouse IV, Jun 11 2015
    
  • PARI
    my(x='x+O('x^40)); Vec((2-x-x^2)/(1-x-x^2-x^3)) \\ G. C. Greubel, Apr 23 2019
    
  • Sage
    ((2-x-x^2)/(1-x-x^2-x^3)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Apr 23 2019
    

Formula

G.f.: (2-x-x^2)/(1-x-x^2-x^3).
a(n) = K(n) - T(n+1) + T(n), where K(n) = A001644(n), T(n) = A000073(n+1). - G. C. Greubel, Apr 23 2019

A214825 a(n) = a(n-1) + a(n-2) + a(n-3), with a(0) = 1, a(1) = a(2) = 3.

Original entry on oeis.org

1, 3, 3, 7, 13, 23, 43, 79, 145, 267, 491, 903, 1661, 3055, 5619, 10335, 19009, 34963, 64307, 118279, 217549, 400135, 735963, 1353647, 2489745, 4579355, 8422747, 15491847, 28493949, 52408543, 96394339, 177296831, 326099713, 599790883, 1103187427
Offset: 0

Views

Author

Abel Amene, Jul 28 2012

Keywords

Comments

Part of a group of sequences defined by a(0), a(1)=a(2), a(n) = a(n-1) + a(n-2) + a(n-3) which is a subgroup of sequences with linear recurrences and constant coefficients listed in the index. See Comments in A214727.

Crossrefs

Programs

  • GAP
    a:=[1,3,3];; for n in [4..40] do a[n]:=a[n-1]+a[n-2]+a[n-3]; od; a; # G. C. Greubel, Apr 23 2019
  • Magma
    R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (1+2*x-x^2)/(1-x-x^2-x^3) )); // G. C. Greubel, Apr 23 2019
    
  • Mathematica
    LinearRecurrence[{1,1,1},{1,3,3},40] (* Harvey P. Dale, Oct 05 2013 *)
  • PARI
    a(n)=([0,1,0; 0,0,1; 1,1,1]^n*[1;3;3])[1,1] \\ Charles R Greathouse IV, Mar 22 2016
    
  • PARI
    my(x='x+O('x^40)); Vec((1+2*x-x^2)/(1-x-x^2-x^3)) \\ G. C. Greubel, Apr 23 2019
    
  • SageMath
    ((1+2*x-x^2)/(1-x-x^2-x^3)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Apr 23 2019
    

Formula

G.f.: (1+2*x-x^2)/(1-x-x^2-x^3).
a(n) = K(n) - 2*T(n+1) + 4*T(n), where K(n) = A001644(n), and T(n) = A000073(n+1). - G. C. Greubel, Apr 23 2019

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 *)

A081172 Tribonacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3), with a(0) = 1, a(1) = 1, a(2) = 0.

Original entry on oeis.org

1, 1, 0, 2, 3, 5, 10, 18, 33, 61, 112, 206, 379, 697, 1282, 2358, 4337, 7977, 14672, 26986, 49635, 91293, 167914, 308842, 568049, 1044805, 1921696, 3534550, 6501051, 11957297, 21992898, 40451246, 74401441, 136845585, 251698272, 462945298, 851489155
Offset: 0

Views

Author

Harry J. Smith, Apr 19 2003

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
Completes the set of tribonacci numbers starting with 0's and 1's in the first three terms:
0,0,0 A000004;
0,0,1 A000073;
0,1,0 A001590;
0,1,1 A000073 starting at a(1);
1,0,0 A000073 starting at a(-1);
1,0,1 A001590;
1,1,0 this sequence;
1,1,1 A000213.

Crossrefs

Programs

  • GAP
    a:=[1,1,0];; for n in [4..40] do a[n]:=a[n-1]+a[n-2]+a[n-3]; od; a; # G. C. Greubel, Apr 23 2019
  • Magma
    R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (1-2*x^2)/(1-x-x^2-x^3) )); // G. C. Greubel, Apr 23 2019
    
  • Maple
    A081172 := proc(n)
        option remember;
        if n <= 2 then
            op(n+1,[1,1,0]) ;
        else
            add(procname(n-i),i=1..3) ;
        end if;
    end proc: # R. J. Mathar, Aug 09 2012
  • Mathematica
    LinearRecurrence[{1,1,1}, {1,1,0}, 40] (* Vladimir Joseph Stephan Orlovsky, Jun 07 2011 *)
  • PARI
    { a1=1; a2=1; a3=0; write("b081172.txt",0," ",a1); write("b081172.txt",1," ",a2); write("b081172.txt",2," ",a3); for(n=3,500, a=a1+a2+a3; a1=a2; a2=a3; a3=a; write("b081172.txt",n," ",a) ) } \\ Harry J. Smith, Mar 20 2009
    
  • PARI
    my(x='x+O('x^40)); Vec((1-2*x^2)/(1-x-x^2-x^3)) \\ G. C. Greubel, Apr 23 2019
    
  • Sage
    ((1-2*x^2)/(1-x-x^2-x^3)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Apr 23 2019
    

Formula

From R. J. Mathar, Mar 27 2009: (Start)
G.f.: (1-2*x^2)/(1 - x - x^2 - x^3).
a(n) = A000073(n+2) - 2*A000073(n). (End)
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