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
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
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
Comments