A082870 -id:A082870 - 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

A004396 One even number followed by two odd numbers.

Original entry on oeis.org

0, 1, 1, 2, 3, 3, 4, 5, 5, 6, 7, 7, 8, 9, 9, 10, 11, 11, 12, 13, 13, 14, 15, 15, 16, 17, 17, 18, 19, 19, 20, 21, 21, 22, 23, 23, 24, 25, 25, 26, 27, 27, 28, 29, 29, 30, 31, 31, 32, 33, 33, 34, 35, 35, 36, 37, 37, 38, 39, 39, 40, 41, 41, 42, 43, 43, 44, 45, 45, 46, 47, 47

Offset: 0

N. J. A. Sloane

Maximal number of points on a triangular grid of edge length n-1 with no 2 points on same row, column, or diagonal. See Problem 252 in The Inquisitive Problem Solver. - R. K. Guy [Comment revised by N. J. A. Sloane, Jul 01 2016]
See also Problem C2 of 2009 International Mathematical Olympiad. - Ruediger Jehn, Oct 19 2021
Dimension of the space of weight 2n+4 cusp forms for Gamma_0(3).
Starting at 3, 3, ..., gives maximal number of acute angles in an n-gon. - Takenov Nurdin (takenov_vert(AT)e-mail.ru), Mar 04 2003
Let b(1) = b(2) = 1, b(k) = b(k-1)+( b(k-2) reduced (mod 2)); then a(n) = b(n-1). - Benoit Cloitre, Aug 14 2002
(1+x+x^2+x^3 ) / ( (1-x^2)*(1-x^3)) is the Poincaré series [or Poincare series] (or Molien series) for Sigma_4.
For n > 6, maximum number of knight moves to reach any square from the corner of an (n-2) X (n-2) chessboard. Likewise for n > 6, the maximum number of knight moves to reach any square from the middle of an (2n-5) X (2n-5) chessboard. - Ralf Stephan, Sep 15 2004
A transform of the Jacobsthal numbers A001045 under the mapping of g.f.s g(x)->g(x/(1+x^2)). - Paul Barry, Jan 16 2005
For n >= 1; a(n) = number of successive terms of A040001 that add to n; or length of n-th term of A028359. - Jaroslav Krizek, Mar 28 2010
For n > 0: a(n) = length of n-th row in A082870. - Reinhard Zumkeller, Apr 13 2014
Also the independence number of the n-triangular honeycomb queen graph. - Eric W. Weisstein, Jul 14 2017
In a game of basketball points can be accumulated by making field goals (two or three points) or free throws (one point). a(n) is the number of different ways to score n-1 points. For example, a score of 4 can be achieved in 3 different ways, with 2 shots (3+1 or 2+2), 3 shots (2+1+1) or 4 shots (1+1+1+1), so a(5) = 3. - Ivan N. Ianakiev, Mar 31 2025

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

J. Kurschak, Hungarian Mathematical Olympiads, 1976, Mir, Moscow.
Paul Vanderlind, Richard K. Guy, and Loren C. Larson, The Inquisitive Problem Solver, MAA, 2002. See Problem 252.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
A. Adem and R. J. Milgram, Cohomology of Finite Groups, Springer-Verlag, 2nd. ed., 2004; p. 246.
Art of Problem Solving Forum, Ordered triples choosing. - _Joel B. Lewis_, May 21 2009
J. Choi and N. Pippenger, Counting the Angels and Devils in Escher's Circle Limit IV, arXiv preprint arXiv:1310.1357 [math.CO], 2013.
C. L. Mallows and N. J. A. Sloane, Weight enumerators of self-orthogonal codes, Discrete Math., 9 (1974), 391-400 (see proof of Theorem 1).
Gabriel Nivasch and Eyal Lev, Nonattacking Queens on a Triangle, Mathematics Magazine, Vol. 78, No. 5 (Dec., 2005), pp. 399-403. See Eq. (4).
John A. Pelesko, Generalizing the Conway-Hofstadter $10,000 Sequence, Journal of Integer Sequences, Vol. 7 (2004), Article 04.3.5.
William A. Stein, Dimensions of the spaces S_k(Gamma_0(N)).
William A. Stein, The modular forms database
Eric Weisstein's World of Mathematics, Independence Number.
50th International Mathematical Olympiad 2009, Problem Shortlist with Solutions.
Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).
Index to sequences related to Olympiads.

Cf. A001045, A002620, A004523, A004773, A006369, A008620, A032766, A040001, A082870, A092200, A096777.

a004396 n = a004396_list !! n
a004396_list = 0 : 1 : 1 : map (+ 2) a004396_list
-- Reinhard Zumkeller, Nov 06 2012

[(Floor(n/3) + Ceiling(n/3)): n in [0..70]]; // Vincenzo Librandi, Aug 07 2011

A004396:=n->floor((2*n + 1)/3); seq(A004396(n), n=0..100); # Wesley Ivan Hurt, Nov 30 2013

Table[Floor[(2 n + 1)/3], {n, 0, 75}]
With[{n = 50}, Riffle[Range[0, n], Range[1, n, 2], {3, -1, 3}]] (* Harvey P. Dale, May 14 2015 *)
CoefficientList[Series[(x + x^3)/((1 - x) (1 - x^3)), {x, 0, 71}], x] (* Michael De Vlieger, Oct 27 2016 *)
a[ n_] := Quotient[2 n + 1, 3]; (* Michael Somos, Oct 23 2017 *)
a[ n_] := Sign[n] SeriesCoefficient[ (x + x^3) / ((1 - x) (1 - x^3)), {x, 0, Abs@n}]; (* Michael Somos, Oct 23 2017 *)
LinearRecurrence[{1, 0, 1, -1}, {1, 1, 2, 3}, {0, 20}] (* Eric W. Weisstein, Jul 14 2017 *)
f[-1]=0; f[n_]:=Length[Union[Plus@@@FrobeniusSolve[{1,2,3},n]]]; f/@Range[-1,100] (* Ivan N. Ianakiev, Mar 31 2025 *)

a(n)=2*n\/3 \\ Charles R Greathouse IV, Apr 17 2012

def a(n) : return( dimension_cusp_forms( Gamma0(3), 2*n+4) ); # Michael Somos, Jul 03 2014

G.f.: (x+x^3)/((1-x)*(1-x^3)).
a(n) = floor( (2*n + 1)/3 ).
a(n) = a(n-1) + (1/2)*((-1)^floor((4*n+2)/3) + 1), a(0) = 0. - Mario Catalani (mario.catalani(AT)unito.it), Oct 20 2003
a(n) = 2n/3 - cos(2*Pi*n/3 + Pi/3)/3 + sqrt(3)*sin(2*Pi*n/3 + Pi/3)/9. - Paul Barry, Mar 18 2004
a(n) = A096777(n+1) - A096777(n) for n > 0. - Reinhard Zumkeller, Jul 09 2004
From Paul Barry, Jan 16 2005: (Start)
G.f.: x*(1+x^2)/(1-x-x^3+x^4).
a(n) = a(n-1) + a(n-3) - a(n-4) for n>3.
a(n) = Sum_{k = 0..n} binomial(n-k-1, k)*(-1)^k*A001045(n-2k). (End)
a(n) = (A006369(n) - (A006369(n) mod 2) * (-1)^(n mod 3)) / (1 + A006369(n) mod 2). - Reinhard Zumkeller, Jan 23 2005
a(n) = A004773(n) - A004523(n). - Reinhard Zumkeller, Aug 29 2005
a(n) = floor(n/3) + ceiling(n/3). - Jonathan Vos Post, Mar 19 2006
a(n+1) = A008620(2n). - Philippe Deléham, Dec 14 2006
a(A032766(n)) = n. - Reinhard Zumkeller, Oct 30 2009
a(n) = floor((2*n^2+4*n+2)/(3*n+4)). - Gary Detlefs, Jul 13 2010
Euler transform of length 4 sequence [1, 1, 1, -1]. - Michael Somos, Jul 03 2014
a(n) = n - floor((n+1)/3). - Wesley Ivan Hurt, Sep 17 2015
a(n) = A092200(n) - floor((n+5)/3). - Filip Zaludek, Oct 27 2016
a(n) = -a(-n) for all n in Z. - Michael Somos, Oct 30 2016
E.g.f.: (2/9)*(3*exp(x)*x + sqrt(3)*exp(-x/2)*sin(sqrt(3)*x/2)). - Stefano Spezia, Sep 20 2022
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2)/2. - Amiram Eldar, Sep 29 2022

A082601 Tribonacci array: to get the next row, right-adjust the previous 3 rows and add them, then append a final 0.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 0, 0, 1, 4, 6, 2, 0, 0, 1, 5, 10, 7, 1, 0, 0, 1, 6, 15, 16, 6, 0, 0, 0, 1, 7, 21, 30, 19, 3, 0, 0, 0, 1, 8, 28, 50, 45, 16, 1, 0, 0, 0, 1, 9, 36, 77, 90, 51, 10, 0, 0, 0, 0, 1, 10, 45, 112, 161, 126, 45, 4, 0, 0, 0, 0, 1, 11, 55, 156, 266, 266, 141, 30, 1, 0

Offset: 0

Gary W. Adamson, May 24 2003

Coefficients of tribonacci polynomials: t_0 = 1, t_1 = x, t_2 = x^2 + x, t_n = x*(t_{n-1} + t_{n-2} + t_{n-3}).
Row sums are tribonacci numbers.
From Petros Hadjicostas, Jun 10 2020: (Start)
To prove a Swamy inequality for the above tribonacci polynomials, we use Guilfoyle's (1967) technique. We write t_n as the determinant of an n X n matrix and then apply Hadamard's inequality.
Since x*t_{n-3} + x*t_{n-2} + x*t_{n-1} - t_n = 0 (with the above initial conditions), we may prove that for n >= 3, t_n = det(A_n), where A_n is the n X n matrix A_n = [[x,-1,0,0,0,...,0,0,0,0,0], [x,x,-1,0,0,...,0,0,0,0,0], [x,x,x,-1,0,...,0,0,0,0,0], [0,x,x,x,-1,...,0,0,0,0,0], ..., [0,0,0,0,0,...,x,x,x,-1,0], [0,0,0,0,0,...,0,x,x,x,-1], [0,0,0,0,0,...,0,0,x,x,x]]).
Using Hadamard's inequality, we obtain t_n^2 <= 3*x^2*(2*x^2 + 1)*(x^2 + 1)*(3*x^2 + 1)^(n-3) for all integers n >= 3 and all real x. (Of course, it is not true for n = 0, 1, 2.)
Guilfoyle's technique can be applied for Werner Schulte's polynomial sequence below, i.e., for p^2*U(n) + p*q*U(n+1) + q^2*U(n+2) - U(n+3) = 0. The first three rows and first three columns of the matrix A_n depend on the initial conditions. We omit the details. (End)

			Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins:
  1;
  1,  0;
  1,  1,  0;
  1,  2,  1,  0;
  1,  3,  3,  0,  0;
  1,  4,  6,  2,  0,  0;
  1,  5, 10,  7,  1,  0,  0;
  ...
From _Petros Hadjicostas_, Jun 10 2020: (Start)
The n-th tribonacci polynomial is t_n = Sum_{k=0..n} T(n,k)*x^(n-k), so, for example:
t_4 = x^4 + 3*x^3 + 3*x^2;
t_5 = x^5 + 4*x^4 + 6*x^3 + 2*x^2;
t_6 = x^6 + 5*x^5 + 10*x^4 + 7*x^3 + x^2;
t_7 = x^7 + 6*x^6 + 15*x^5 + 16*x^4 + 6*x^3.
We have
t_4 = det([[x,-1,0,0]; [x,x,-1,0]; [x,x,x,-1]; [0,x,x,x]]);
t_5 = det([[x,-1,0,0,0]; [x,x,-1,0,0]; [x,x,x,-1,0]; [0,x,x,x,-1]; [0,0,x,x,x]]);
t_6 = det([[x,-1,0,0,0,0]; [x,x,-1,0,0,0]; [x,x,x,-1,0,0]; [0,x,x,x,-1,0]; [0,0,x,x,x,-1]; [0,0,0,x,x,x]]);
t_7 = det([[x,-1,0,0,0,0,0]; [x,x,-1,0,0,0,0]; [x,x,x,-1,0,0,0]; [0,x,x,x,-1,0,0]; [0,0,x,x,x,-1,0]; [0,0,0,x,x,x,-1]; [0,0,0,0,x,x,x]]). (End)

Thomas Koshy, Fibonacci and Lucas numbers with Applications, Vol. 2, Wiley, 2019; see p. 33. [He gives Swamy inequalities for the Fibonacci and the Lucas polynomials. Vol. 1 was published in 2001. - Petros Hadjicostas, Jun 10 2020]

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Richard Guilfoyle, Comment to the solution of Problem E1846, Amer. Math. Monthly, 74(5), 1967, 593.
Thomas Koshy, Fibonacci and Lucas Numbers with Applications, Wiley, 2001; Chapter 47: Tribonacci Polynomials: ("In 1973, V.E. Hoggatt, Jr. and M. Bicknell generalized Fibonacci polynomials to Tribonacci polynomials tx(x)"); Table 47.1, page 534: "Tribonacci Array".
M. N. S. Swamy and R. E. Giudici, Solution to Problem E1846, Amer. Math. Monthly, 74(5), 1967, 592-593.
Wikipedia, Hadamard's inequality.

Closely related to A078802. A better version of A082870. Cf. A000073.

Cf. A002426 (central terms).

a082601 n k = a082601_tabl !! n !! k
a082601_row n = a082601_tabl !! n
a082601_tabl = [1] : [1,0] : [1,1,0] : f [0,0,1] [0,1,0] [1,1,0]
   where f us vs ws = ys : f (0:vs) (0:ws) ys where
                      ys = zipWith3 (((+) .) . (+)) us vs ws ++ [0]
-- Reinhard Zumkeller, Apr 13 2014

G:=x*y/(1-x-x^2*y-x^3*y^2): Gs:=simplify(series(G,x=0,18)): for n from 1 to 16 do P[n]:=sort(coeff(Gs,x^n)) od: seq(seq(coeff(P[i],y^j),j=1..i),i=1..16);

Table[SeriesCoefficient[x/(1 - x - x^2*y - x^3*y^2), {x, 0, n}, {y, 0, k}], {n, 13}, {k, 0, n - 1}] // Flatten (* Michael De Vlieger, Feb 22 2017 *)

G.f.: x/(1 - x - x^2*y - x^3*y^2). - Vladeta Jovovic, May 30 2003
From Werner Schulte, Feb 22 2017: (Start)
T(n,k) = Sum_{j=0..floor(k/2)} binomial(k-j,j)*binomial(n-k,k-j) for 0 <= k and k <= floor(2*n/3) with binomial(i,j) = 0 for iDennis P. Walsh at A078802).
Based on two integers p and q define the integer sequence U(n) by U(0) = 0 and U(1) = 0 and U(n+2) = Sum_{k=0..floor(2*n/3)} T(n,k)*p^k*q^(2*n-3*k) for n >= 0. That yields the g.f. f(p,q,x) = x^2/(1 - q^2*x - p*q*x^2 - p^2*x^3) and the recurrence U(n+3) = q^2*U(n+2) + p*q*U(n+1) + p^2*U(n) for n >= 0 with initial values U(0) = U(1) = 0 and U(2) = 1. For p = q = +/-1, you'll get tribonacci numbers A000073. For p = -1 and q = 1, you'll get A021913. (End)

Edited by Anne Donovan and N. J. A. Sloane, May 27 2003

More terms from Emeric Deutsch, May 06 2004

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.

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A004396 One even number followed by two odd numbers.

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A082601 Tribonacci array: to get the next row, right-adjust the previous 3 rows and add them, then append a final 0.

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