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

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

Author

Gary W. Adamson, May 24 2003

Keywords

• nonn
• tabl
• easy

Comments

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)

Examples

			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)
		

References

• 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]

Links

• 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.

Crossrefs

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

Cf. A002426 (central terms).

Programs

Haskell

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

Maple

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);

Mathematica

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 *)

Formula

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)

Extensions

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

More terms from Emeric Deutsch, May 06 2004