A058265 -id:A058265 - cp's OEIS frontend

A000073 Tribonacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3) for n >= 3 with a(0) = a(1) = 0 and a(2) = 1.

0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, 5768, 10609, 19513, 35890, 66012, 121415, 223317, 410744, 755476, 1389537, 2555757, 4700770, 8646064, 15902591, 29249425, 53798080, 98950096, 181997601, 334745777, 615693474, 1132436852

Offset: 0

N. J. A. Sloane

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
Also (for n > 1) number of ordered trees with n+1 edges and having all leaves at level three. Example: a(4)=2 because we have two ordered trees with 5 edges and having all leaves at level three: (i) one edge emanating from the root, at the end of which two paths of length two are hanging and (ii) one path of length two emanating from the root, at the end of which three edges are hanging. - Emeric Deutsch, Jan 03 2004
a(n) is the number of compositions of n-2 with no part greater than 3. Example: a(5)=4 because we have 1+1+1 = 1+2 = 2+1 = 3. - Emeric Deutsch, Mar 10 2004
Let A denote the 3 X 3 matrix [0,0,1;1,1,1;0,1,0]. a(n) corresponds to both the (1,2) and (3,1) entries in A^n. - Paul Barry, Oct 15 2004
Number of permutations satisfying -k <= p(i)-i <= r, i=1..n-2, with k=1, r=2. - Vladimir Baltic, Jan 17 2005
Number of binary sequences of length n-3 that have no three consecutive 0's. Example: a(7)=13 because among the 16 binary sequences of length 4 only 0000, 0001 and 1000 have 3 consecutive 0's. - Emeric Deutsch, Apr 27 2006
Therefore, the complementary sequence to A050231 (n coin tosses with a run of three heads). a(n) = 2^(n-3) - A050231(n-3) - Toby Gottfried, Nov 21 2010
Convolved with the Padovan sequence = row sums of triangle A153462. - Gary W. Adamson, Dec 27 2008
For n > 1: row sums of the triangle in A157897. - Reinhard Zumkeller, Jun 25 2009
a(n+2) is the top left entry of the n-th power of any of the 3 X 3 matrices [1, 1, 1; 0, 0, 1; 1, 0, 0] or [1, 1, 0; 1, 0, 1; 1, 0, 0] or [1, 1, 1; 1, 0, 0; 0, 1, 0] or [1, 0, 1; 1, 0, 0; 1, 1, 0]. - R. J. Mathar, Feb 03 2014
a(n-1) is the top left entry of the n-th power of any of the 3 X 3 matrices [0, 0, 1; 1, 1, 1; 0, 1, 0], [0, 1, 0; 0, 1, 1; 1, 1, 0], [0, 0, 1; 1, 0, 1; 0, 1, 1] or [0, 1, 0; 0, 0, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
Also row sums of A082601 and of A082870. - Reinhard Zumkeller, Apr 13 2014
Least significant bits are given in A021913 (a(n) mod 2 = A021913(n)). - Andres Cicuttin, Apr 04 2016
The nonnegative powers of the tribonacci constant t = A058265 are t^n = a(n)*t^2 + (a(n-1) + a(n-2))*t + a(n-1)*1, for  n >= 0, with  a(-1) = 1 and a(-2) = -1. This follows from the recurrences derived from t^3 = t^2 + t + 1. See the example in A058265 for the first nonnegative powers. For the negative powers see A319200. - Wolfdieter Lang, Oct 23 2018
The term "tribonacci number" was coined by Mark Feinberg (1963), a 14-year-old student in the 9th grade of the Susquehanna Township Junior High School in Pennsylvania. He died in 1967 in a motorcycle accident. - Amiram Eldar, Apr 16 2021
Andrews, Just, and Simay (2021, 2022) remark that it has been suggested that this sequence is mentioned in Charles Darwin's Origin of Species as bearing the same relation to elephant populations as the Fibonacci numbers do to rabbit populations. - N. J. A. Sloane, Jul 12 2022

			G.f. = x^2 + x^3 + 2*x^4 + 4*x^5 + 7*x^6 + 13*x^7 + 24*x^8 + 44*x^9 + 81*x^10 + ...

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Index entries for linear recurrences with constant coefficients, signature (1,1,1).

Cf. A000045, A000078, A000213, A000931, A001590 (first differences, also a(n)+a(n+1)), A001644, A008288 (tribonacci triangle), A008937 (partial sums), A021913, A027024, A027083, A027084, A046738 (Pisano periods), A050231, A054668, A062544, A063401, A077902, A081172, A089068, A118390, A145027, A153462, A230216.
A057597 is this sequence run backwards: A057597(n) = a(1-n).
Row 3 of arrays A048887 and A092921 (k-generalized Fibonacci numbers).
Partitions: A240844 and A117546.
Cf. also A092836 (subsequence of primes), A299399 = A092835 + 1 (indices of primes).

a:=[0,0,1];; 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

a000073 n = a000073_list !! n
a000073_list = 0 : 0 : 1 : zipWith (+) a000073_list (tail
                          (zipWith (+) a000073_list $ tail a000073_list))
-- Reinhard Zumkeller, Dec 12 2011

[n le 3 select Floor(n/3) else Self(n-1)+Self(n-2)+Self(n-3): n in [1..70]]; // Vincenzo Librandi, Jan 29 2016

a:= n-> (<<0|1|0>, <0|0|1>, <1|1|1>>^n)[1,3]:
seq(a(n), n=0..40);  # Alois P. Heinz, Dec 19 2016
# second Maple program:
A000073:=proc(n) option remember; if n <= 1 then 0 elif n=2 then 1 else procname(n-1)+procname(n-2)+procname(n-3); fi; end; # N. J. A. Sloane, Aug 06 2018

CoefficientList[Series[x^2/(1 - x - x^2 - x^3), {x, 0, 50}], x]
a[0] = a[1] = 0; a[2] = 1; a[n_] := a[n] = a[n - 1] + a[n - 2] + a[n - 3]; Array[a, 36, 0] (* Robert G. Wilson v, Nov 07 2010 *)
LinearRecurrence[{1, 1, 1}, {0, 0, 1}, 60] (* Vladimir Joseph Stephan Orlovsky, May 24 2011 *)
a[n_] := SeriesCoefficient[If[ n < 0, x/(1 + x + x^2 - x^3), x^2/(1 - x - x^2 - x^3)], {x, 0, Abs @ n}] (* Michael Somos, Jun 01 2013 *)
Table[-RootSum[-1 - # - #^2 + #^3 &, -#^n - 9 #^(n + 1) + 4 #^(n + 2) &]/22, {n, 0, 20}] (* Eric W. Weisstein, Nov 09 2017 *)

A000073[0]:0$
A000073[1]:0$
A000073[2]:1$
A000073[n]:=A000073[n-1]+A000073[n-2]+A000073[n-3]$
  makelist(A000073[n], n, 0, 40);  /* Emanuele Munarini, Mar 01 2011 */

{a(n) = polcoeff( if( n<0, x / ( 1 + x + x^2 - x^3), x^2 / ( 1 - x - x^2 - x^3) ) + x * O(x^abs(n)), abs(n))}; /* Michael Somos, Sep 03 2007 */

my(x='x+O('x^99)); concat([0, 0], Vec(x^2/(1-x-x^2-x^3))) \\ Altug Alkan, Apr 04 2016

a(n)=([0,1,0;0,0,1;1,1,1]^n)[1,3] \\ Charles R Greathouse IV, Apr 18 2016, simplified by M. F. Hasler, Apr 18 2018

def a(n, adict={0:0, 1:0, 2:1}):
    if n in adict:
        return adict[n]
    adict[n]=a(n-1)+a(n-2)+a(n-3)
    return adict[n] # David Nacin, Mar 07 2012
from functools import cache
@cache
def A000073(n: int) -> int:
    if n <= 1: return 0
    if n == 2: return 1
    return A000073(n-1) + A000073(n-2) + A000073(n-3) # Peter Luschny, Nov 21 2022

G.f.: x^2/(1 - x - x^2 - x^3).
G.f.: x^2 / (1 - x / (1 - x / (1 + x^2 / (1 + x)))). - Michael Somos, May 12 2012
G.f.: Sum_{n >= 0} x^(n+2) *[ Product_{k = 1..n} (k + k*x + x^2)/(1 + k*x + k*x^2) ] = x^2 + x^3 + 2*x^4 + 4*x^5 + 7*x^6 + 13*x^7 + ... may be proved by the method of telescoping sums. - Peter Bala, Jan 04 2015
a(n+1)/a(n) -> A058265.  a(n-1)/a(n) -> A192918.
a(n) = central term in M^n * [1 0 0] where M = the 3 X 3 matrix [0 1 0 / 0 0 1 / 1 1 1]. (M^n * [1 0 0] = [a(n-1) a(n) a(n+1)].) a(n)/a(n-1) tends to the tribonacci constant, 1.839286755... = A058265, an eigenvalue of M and a root of x^3 - x^2 - x - 1 = 0. - Gary W. Adamson, Dec 17 2004
a(n+2) = Sum_{k=0..n} T(n-k, k), where T(n, k) = trinomial coefficients (A027907). - Paul Barry, Feb 15 2005
A001590(n) = a(n+1) - a(n); A001590(n) = a(n-1) + a(n-2) for n > 1; a(n) = (A000213(n+1) - A000213(n))/2; A000213(n-1) = a(n+2) - a(n) for n > 0. - Reinhard Zumkeller, May 22 2006
Let C = the tribonacci constant, 1.83928675...; then C^n = a(n)*(1/C) + a(n+1)*(1/C + 1/C^2) + a(n+2)*(1/C + 1/C^2 + 1/C^3). Example: C^4 = 11.444...= 2*(1/C) + 4*(1/C + 1/C^2) + 7*(1/C + 1/C^2 + 1/C^3). - Gary W. Adamson, Nov 05 2006
a(n) = j*C^n + k*r1^n + L*r2^n where C is the tribonacci constant (C = 1.8392867552...), real root of x^3-x^2-x-1=0, and r1 and r2 are the two other roots (which are complex), r1 = m+p*i and r2 = m-p*i, where i = sqrt(-1), m = (1-C)/2 (m = -0.4196433776...) and p = ((3*C-5)*(C+1)/4)^(1/2) = 0.6062907292..., and where j = 1/((C-m)^2 + p^2) = 0.1828035330..., k = a+b*i, and L = a-b*i, where a = -j/2 = -0.0914017665... and b = (C-m)/(2*p*((C-m)^2 + p^2)) = 0.3405465308... . - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 23 2007
a(n+1) = 3*c*((1/3)*(a+b+1))^n/(c^2-2*c+4) where a=(19+3*sqrt(33))^(1/3), b=(19-3*sqrt(33))^(1/3), c=(586+102*sqrt(33))^(1/3). Round to the nearest integer. - Al Hakanson (hawkuu(AT)gmail.com), Feb 02 2009
a(n) = round(3*((a+b+1)/3)^n/(a^2+b^2+4)) where a=(19+3*sqrt(33))^(1/3), b=(19-3*sqrt(33))^(1/3).. - Anton Nikonov
Another form of the g.f.: f(z) = (z^2-z^3)/(1-2*z+z^4). Then we obtain a(n) as a sum: a(n) = Sum_{i=0..floor((n-2)/4)} ((-1)^i*binomial(n-2-3*i,i)*2^(n-2-4*i)) - Sum_{i=0..floor((n-3)/4)} ((-1)^i*binomial(n-3-3*i,i)*2^(n-3-4*i)) with natural convention: Sum_{i=m..n} alpha(i) = 0 for m > n. - Richard Choulet, Feb 22 2010
a(n+2) = Sum_{k=0..n} Sum_{i=k..n, mod(4*k-i,3)=0} binomial(k,(4*k-i)/3)*(-1)^((i-k)/3)*binomial(n-i+k-1,k-1). - Vladimir Kruchinin, Aug 18 2010
a(n) = 2*a(n-2) + 2*a(n-3) + a(n-4). - Gary Detlefs, Sep 13 2010
Sum_{k=0..2*n} a(k+b)*A027907(n,k) = a(3*n+b), b >= 0 (see A099464, A074581).
a(n) = 2*a(n-1) - a(n-4), with a(0)=a(1)=0, a(2)=a(3)=1. - Vincenzo Librandi, Dec 20 2010
Starting (1, 2, 4, 7, ...) is the INVERT transform of (1, 1, 1, 0, 0, 0, ...). - Gary W. Adamson, May 13 2013
G.f.: Q(0)*x^2/2, where Q(k) = 1 + 1/(1 - x*(4*k+1 + x + x^2)/( x*(4*k+3 + x + x^2) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 09 2013
a(n+2) = Sum_{j=0..floor(n/2)} Sum_{k=0..j} binomial(n-2*j,k)*binomial(j,k)*2^k. - Tony Foster III, Sep 08 2017
Sum_{k=0..n} (n-k)*a(k) = (a(n+2) + a(n+1) - n - 1)/2. See A062544. - Yichen Wang, Aug 20 2020
a(n) = A008937(n-1) - A008937(n-2) for n >= 2. - Peter Luschny, Aug 20 2020
From Yichen Wang, Aug 27 2020: (Start)
Sum_{k=0..n} a(k) = (a(n+2) + a(n) - 1)/2. See A008937.
Sum_{k=0..n} k*a(k) = ((n-1)*a(n+2) - a(n+1) + n*a(n) + 1)/2. See A337282. (End)
For n > 1, a(n) = b(n) where b(1) = 1 and then b(n) = Sum_{k=1..n-1} b(n-k)*A000931(k+2). - J. Conrad, Nov 24 2022
Conjecture: the congruence a(n*p^(k+1)) + a(n*p^k) + a(n*p^(k-1)) == 0 (mod p^k) holds for positive integers k and n and for all the primes p listed in A106282. - Peter Bala, Dec 28 2022
Sum_{k=0..n} k^2*a(k) = ((n^2-4*n+6)*a(n+1) - (2*n^2-2*n+5)*a(n) + (n^2-2*n+3)*a(n-1) - 3)/2. - Prabha Sivaramannair, Feb 10 2024
a(n) = Sum_{r root of x^3-x^2-x-1} r^n/(3*r^2-2*r-1). - Fabian Pereyra, Nov 23 2024

Minor edits by M. F. Hasler, Apr 18 2018

Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021

A000213 Tribonacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3) with a(0)=a(1)=a(2)=1.

Original entry on oeis.org

1, 1, 1, 3, 5, 9, 17, 31, 57, 105, 193, 355, 653, 1201, 2209, 4063, 7473, 13745, 25281, 46499, 85525, 157305, 289329, 532159, 978793, 1800281, 3311233, 6090307, 11201821, 20603361, 37895489, 69700671, 128199521, 235795681, 433695873, 797691075, 1467182629

Offset: 0

N. J. A. Sloane

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
Number of (n-1)-bit binary sequences with each one adjacent to a zero. - R. H. Hardin, Dec 24 2007
The binomial transform is A099216. The inverse binomial transform is (-1)^n*A124395(n). - R. J. Mathar, Aug 19 2008
Equals INVERT transform of (1, 0, 2, 0, 2, 0, 2, ...). a(6) = 17 = (1, 1, 1, 3, 5, 9) dot (0, 2, 0, 2, 0, 1) = (0 + 2 + 0 + 6 + 0 + 9) = 17. - Gary W. Adamson, Apr 27 2009
From John M. Campbell, May 16 2011: (Start)
Equals the number of tilings of a 2 X n grid using singletons and "S-shaped tetrominoes" (i.e., shapes of the form Polygon[{{0, 0}, {2, 0}, {2, 1}, {3, 1}, {3, 2}, {1, 2}, {1, 1}, {0, 1}}]).
Also equals the number of tilings of a 2 X n grid using singletons and "T-shaped tetrominoes" (i.e., shapes of the form Polygon[{{0, 0}, {3, 0}, {3, 1}, {2, 1}, {2, 2}, {1, 2}, {1, 1}, {0, 1}}]). (End)
Pisano period lengths: 1, 1, 13, 4, 31, 13, 48, 8, 39, 31, 110, 52, 168, 48, 403, 16, 96, 39, 360, 124, ... (differs from A106293). - R. J. Mathar, Aug 10 2012
a(n) is the number of compositions of n with no consecutive 1's. a(4) = 5 because we have: 4, 3+1, 1+3, 2+2, 1+2+1. Cf. A239791, A003242. - Geoffrey Critzer, Mar 27 2014
a(n+2) is the number of words of length n over alphabet {1,2,3} without having {11,12,22,23} as substrings. - Ran Pan, Sep 16 2015
Satisfies Benford's law [see A186190]. - N. J. A. Sloane, Feb 09 2017
a(n) is also the number of dominating sets on the (n-1)-path graph. - Eric W. Weisstein, Mar 31 2017
a(n) is also the number of maximal irredundant sets and minimal dominating sets in the (2n-3)-triangular snake graph. - Eric W. Weisstein, Jun 09 2019
a(n) is also the number of anti-palindromic compositions of n, where a composition (c(1), c(2),..., c(k)) is anti-palindromic if c(i) is not equal to c(k+1-i) whenever 1 <= i <= k/2. For instance, there are a(4) = 5 anti-palindromic compositions of 4: 4, 31, 13, 211, 112. - Jia Huang, Apr 08 2023

			G.f. = 1 + x + x^2 + 3*x^3 + 5*x^4 + 9*x^5+ 17*x^6 + 31*x^7 + 57*x^8 + ...

Kenneth Edwards, 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).

Indranil Ghosh, Table of n, a(n) for n = 0..3772 (terms 0..200 from T. D. Noe)
George E. Andrews, Matthew Just, and Greg Simay, Anti-palindromic compositions, arXiv:2102.01613 [math.CO], 2021. Also Fib. Q., 60:2 (2022), 164-176. See Table 1.
Joerg Arndt, Matters Computational (The Fxtbook), p.312
Jean-Luc Baril, Avoiding patterns in irreducible permutations, Discrete Mathematics and Theoretical Computer Science, Vol 17, No 3 (2016). See Table 4.
Jean-Luc Baril and Nathanaël Hassler, Intervals in a family of Fibonacci lattices, Univ. de Bourgogne (France, 2024). See p. 7.
B. G. Baumgart, Letter to the editor Part 1 Part 2 Part 3, Fib. Quart. 2 (1964), 260, 302.
Martin Burtscher, Igor Szczyrba and Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
Kenneth Edwards and Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers squared, arXiv:1907.06517 [math.CO], 2019.
Mark Feinberg, Fibonacci-Tribonacci, Fib. Quart. 1(#3) (1963), 71-74.
Nick Hobson, Python program for this sequence
Joanna Jaszunska and Jan Okninski, Structure of Chinese algebras, Journal of Algebra, Volume 346, Issue 1, 15 November 2011, Pages 31-81.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
I. Tasoulas, K. Manes, A. Sapounakis, and P. Tsikouras, Chains with Small Intervals in the Lattice of Binary Paths, arXiv:1911.10883 [math.CO], 2019.
Eric Weisstein's World of Mathematics, Dominating Set
Eric Weisstein's World of Mathematics, Irredundant Set
Eric Weisstein's World of Mathematics, Minimal Dominating Set
Eric Weisstein's World of Mathematics, Path Graph
Eric Weisstein's World of Mathematics, Triangular Snake Graph
Eric Weisstein's World of Mathematics, Tribonacci Number
Index entries for linear recurrences with constant coefficients, signature (1,1,1).
Index entries for sequences related to Benford's law

Cf. A000073, A000288, A000322, A000383, A046735, A060455, A180735, A186190.

a:=[1,1,1];; for n in [4..45] do a[n]:=a[n-1]+a[n-2]+a[n-3]; od; a; # G. C. Greubel, Jun 09 2019

a000213 n = a000213_list !! n
a000213_list = 1 : 1 : 1 : zipWith (+) a000213_list
   (tail $ zipWith (+) a000213_list (tail a000213_list))
-- Reinhard Zumkeller, Apr 07 2012

I:=[1,1,1]; [n le 3 select I[n] else Self(n-1) + Self(n-2) + Self(n-3): n in [1..45]]; // G. C. Greubel, Jun 09 2019

K:=(1-z^2)/(1-z-z^2-z^3): Kser:=series(K, z=0, 45): seq((coeff(Kser, z, n)), n= 0..34); # Zerinvary Lajos, Nov 08 2007
A000213:=(z-1)*(1+z)/(-1+z+z**2+z**3); # Simon Plouffe in his 1992 dissertation

LinearRecurrence[{1, 1, 1}, {1, 1, 1}, 45] (* Harvey P. Dale, May 23 2011 *)
Table[RootSum[-1 - # - #^2 + #^3 &, 2 #^n - 4 #^(n + 1) + 3 #^(n + 2) &]/11, {n, 0, 45}] (* Eric W. Weisstein, Apr 10 2018 *)
CoefficientList[Series[(1-x)(1+x)/(1-x-x^2-x^3), {x, 0, 45}], x] (* Eric W. Weisstein, Apr 10 2018 *)

a(n):=sum(sum(binomial(n-2*m+1,m-i)*binomial(n-2*m+i,n-2*m), i,0,m),m,0,(n)/2); /* Vladimir Kruchinin, Dec 17 2011 */

a(n)=tn=[1,1,1;1,0,0;0,1,0]^n;tn[3,1]+tn[3,2]+tn[3,3] \\ Charles R Greathouse IV, Feb 18 2011

alst = [1, 1, 1]
[alst.append(alst[n-1] + alst[n-2] + alst[n-3]) for n in range(3, 37)]
print(alst) # Michael S. Branicky, Sep 21 2021

((1-x^2)/(1-x-x^2-x^3)).series(x, 45).coefficients(x, sparse=False) # G. C. Greubel, Jun 09 2019

G.f.: (1-x)*(1+x)/(1-x-x^2-x^3). - Ralf Stephan, Feb 11 2004
G.f.: 1 / (1 - x / (1 - 2*x^2 / (1 + x^2))). - Michael Somos, May 12 2012
a(n) = rightmost term of M^n * [1 1 1], where M is the 3 X 3 matrix [1 1 1 / 1 0 0 / 0 1 0]. M^n * [1 1 1] = [a(n+2) a(n+1) a(n)]. a(n)/a(n-1) tends to the tribonacci constant, 1.839286755...; an eigenvalue of M and a root of x^3 - x^2 - x - 1 = 0. - Gary W. Adamson, Dec 17 2004
a(n) = A001590(n+3) - A001590(n+2); a(n+1) - a(n) = 2*A000073(n); a(n) = A000073(n+3) - A000073(n+1). - Reinhard Zumkeller, May 22 2006
a(n) = A001590(n) + A001590(n+1). - Philippe Deléham, Sep 25 2006
a(n) ~ (F - 1) * T^n, where F = A086254 and T = A058265. - Charles R Greathouse IV, Nov 09 2008
a(n) = 2*a(n-1) - a(n-4), n > 3. - Gary Detlefs, Sep 13 2010
a(n) = Sum_{m=0..n/2} Sum_{i=0..m} binomial(n-2*m+1,m-i)*binomial(n-2*m+i, n-2*m). - Vladimir Kruchinin, Dec 17 2011
a(n) = 2*A008937(n-2) + 1 for n > 1. - Reinhard Zumkeller, Apr 07 2012
G.f.: 1+x/(U(0) - x) where U(k) = 1 - x^2/(1 - 1/(1 + 1/U(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Nov 16 2012
G.f.: 1 + x + x^2/(G(0)-x) where G(k) = 1 - x*(2*k+1)/(1 - 1/(1 + (2*k+1)/G(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Nov 17 2012
G.f.: (1+x)*(1-x)*(1 + x*(G(0)-1)/(x+1)) where G(k) = 1 + (1+x+x^2)/(1-x/(x+1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 26 2013
G.f.: 1/(1+x-G(0)), where G(k) = 1 - 1/(1 - x/(x - 1/(1 + 1/(1 - x/(x + 1/G(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, Jun 20 2013
a(n) = (-1)^n * A180735(-1-n) for all n in Z. - Michael Somos, Aug 15 2015
a(n) = Sum_{r root of x^3-x^2-x-1} r^n/(-r^2+2*r+2). - Fabian Pereyra, Nov 21 2024

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

N. J. A. Sloane

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

			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

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).

T. D. Noe, Table of n, a(n) for n = 0..200
Kassie Archer and Noel Bourne, Pattern avoidance in compositions and powers of permutations, arXiv:2505.05218 [math.CO], 2025. See p. 8.
Barry Balof, Restricted tilings and bijections, J. Integer Seq. 15 (2012), no. 2, Article 12.2.3, 17 pp.
Matthias Beck and Neville Robbins, Variations on a Generatingfunctional Theme: Enumerating Compositions with Parts Avoiding an Arithmetic Sequence, arXiv:1403.0665 [math.NT], 2014.
Martin Burtscher, Igor Szczyrba, and Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
M. Feinberg, Fibonacci-Tribonacci, Fib. Quart. 1(3) (1963), 71-74.
M. Feinberg, New slants, Fib. Quart. 2 (1964), 223-227.
Wojciech Florek, A class of generalized Tribonacci sequences applied to counting problems, Appl. Math. Comput., 338 (2018), 809-821.
Petros Hadjicostas, Cyclic Compositions of a Positive Integer with Parts Avoiding an Arithmetic Sequence, Journal of Integer Sequences, 19 (2016), Article 16.8.2.
V. E. Hoggatt, Jr. and Marjorie Bicknell, Palindromic compositions, Fib. Quart 13 (4) (1975) 357, eq (2.7)
Jia Huang, Partially Palindromic Compositions, J. Int. Seq. (2023) Vol. 26, Art. 23.4.1. See pp. 4, 9.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 401
Milan Janjic, Binomial Coefficients and Enumeration of Restricted Words, Journal of Integer Sequences, 2016, Vol 19, #16.7.3.
Tamara Kogan, Luba Sapir, Amir Sapir, and Ariel Sapir, The Fibonacci family of iterative processes for solving nonlinear equations, Applied Numerical Mathematics 110 (2016) 148-158.
Daniel Krob and Jean-Yves Thibon, Higher order peak algebras, arXiv:math/0411407 [math.CO], 2004.
Wolfdieter Lang, The Tribonacci and ABC Representations of Numbers are Equivalent, arXiv preprint arXiv:1810.09787 [math.NT], 2018.
Sepideh Maleki and Martin Burtscher, Automatic Hierarchical Parallelization of Linear Recurrences, Proceedings of the 23rd International Conference on Architectural Support for Programming Languages and Operating Systems, ACM, 2018.
M. A. Nyblom, Counting Palindromic Binary Strings Without r-Runs of Ones, J. Int. Seq. 16 (2013) #13.8.7. P_3(n) = 1,2,2,3,3,6,6,11,11,...
Helmut Prodinger, Counting Palindromes According to r-Runs of Ones Using Generating Functions, J. Int. Seq. 17 (2014) # 14.6.2, even length, r=2.
Karim Ritter von Merkl, Computing colored Khovanov homology, arXiv:2505.03916 [math.QA], 2025. See p. 13.
Neville Robbins, On Tribonacci Numbers and 3-Regular Compositions, Fibonacci Quart. 52 (2014), no. 1, 16-19. See Adamson's comments.
Bo Tan and Zhi-Ying Wen, Some properties of the Tribonacci sequence, European Journal of Combinatorics, 28 (2007) 1703-1719.
M. E. Waddill and L. Sacks, Another generalized Fibonacci sequence, Fib. Quart., 5 (1967), 209-222.
Eric Weisstein's World of Mathematics, Tribonacci Number
Index entries for linear recurrences with constant coefficients, signature (1,1,1).

Cf. A000045, A000073, A027907, A027053, A078042, A145579, A278038.

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

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

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

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 *)

a(n)=([0,1,0; 0,0,1; 1,1,1]^n*[0;1;0])[1,1] \\ Charles R Greathouse IV, Jul 28 2015

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

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

Additional comments from Miklos Kristof, Jul 03 2002

Because the tribonacci constant t = A058265 > 1, with Beatty sequence At(n) := floor(n*t), n >= 1 (with At(0) = 0) given in A158919, has the companion sequence Bt := floor(n*s), n >= 1, (with Bt(0) = 0), with 1/t + 1/s = 1, and At and Bt are complementary, disjoint sequences for the positive integers. Note that Bt is not A172278. The first entries n = 0..161 coincide. A172278(162) = 354 but At(193) = A158919(193) = 354, hence A172278 is not complementary together with At. In fact, Bt(162) = 355, which is not a member of At.

s-1 = 1/(t-1) equals the real root of 2*x^3 - 2*x - 1. See the formulas below. - Wolfdieter Lang, Sep 15 2022

This is the Beatty sequence for tau_prime = 2.191487883953118747061354268227517294...,
defined by 1/tau + 1/tau_prime = 1.
Differs from A172278 at n = 162, 209, 256, 303, 324, ...
Note that Beatty sequences do not normally include 0 - see the classic pair A000201, A001950. - N. J. A. Sloane, Oct 19 2018
Note that the tribonacci numbers T = A000073 related to the ternary sequence A080843 lead to the three complementary sequences for the nonnegative integers AT(n) = A278040(n), BT(n) = A278039(n) and CT(n) = A278041(n). - Wolfdieter Lang, Sep 08 2018

This is a measure of the quality of the n-th convergent to the tribonacci constant A058265 if the convergent and the exact value are compared rounded to an increasing number of digits. The sequence of rounded values of A058265 is 2, 1.8, 1.84, 1.839, 1.8393, 1.83929, 1.839287, 1.8392868, etc. The n-th convergents are 2 (n=1), 11/6 (n=2), 46/25 (n=3), 103/56 (n=4), 31451/17105 (n=5) etc., each with associated rounded decimal expansions.

a(n) is the maximum number of post-period digits of the two expansions if compared at the same level of rounding. Counting only post-period digits (which is one less than the full number of decimal digits) is just a convention taken from A084407.

cp's OEIS Frontend

A277721 a(n) = floor(n*t^3) - A003146(n), where t = 1.8392867... is the tribonacci constant A058265.

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A316711 Decimal expansion of s:= t/(t - 1), with the tribonacci constant t = A058265.

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A276383 Complement of A158919: complementary Beatty sequence to the Beatty sequence defined by the tribonacci constant tau = A058265.

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A319428 Numerators of convergents to continued fraction expansion of tribonacci constant (A058265, A019712).

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A319429 Denominators of convergents to continued fraction expansion of tribonacci constant (A058265, A019712).

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A108173 Let beta = A058265. Sequence gives a(n) = 1 + ceiling((n-1)*beta^2).

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A138371 Count of post-period decimal digits up to which the rounded n-th convergent to A058265 agrees with the exact value.

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A000073 Tribonacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3) for n >= 3 with a(0) = a(1) = 0 and a(2) = 1.

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