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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.

%I A082601 #45 Jun 12 2020 11:12:51
%S A082601 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,
%T A082601 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,
%U A082601 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
%N A082601 Tribonacci array: to get the next row, right-adjust the previous 3 rows and add them, then append a final 0.
%C A082601 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}).
%C A082601 Row sums are tribonacci numbers.
%C A082601 From _Petros Hadjicostas_, Jun 10 2020: (Start)
%C A082601 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.
%C A082601 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]]).
%C A082601 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.)
%C A082601 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)
%D A082601 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]
%H A082601 Reinhard Zumkeller, <a href="/A082601/b082601.txt">Rows n = 0..125 of triangle, flattened</a>
%H A082601 Richard Guilfoyle, <a href="https://www.jstor.org/stable/2314912">Comment to the solution of Problem E1846</a>, Amer. Math. Monthly, 74(5), 1967, 593.
%H A082601 Thomas Koshy, <a href="https://dx.doi.org/10.1002/9781118033067">Fibonacci and Lucas Numbers with Applications</a>, 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".
%H A082601 M. N. S. Swamy and R. E. Giudici, <a href="https://www.jstor.org/stable/2314912">Solution to Problem E1846</a>, Amer. Math. Monthly, 74(5), 1967, 592-593.
%H A082601 Wikipedia, <a href="https://en.wikipedia.org/wiki/Hadamard%27s_inequality">Hadamard's inequality</a>.
%F A082601 G.f.: x/(1 - x - x^2*y - x^3*y^2). - _Vladeta Jovovic_, May 30 2003
%F A082601 From _Werner Schulte_, Feb 22 2017: (Start)
%F A082601 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 i<j (see _Dennis P. Walsh_ at A078802).
%F A082601 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)
%e A082601 Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins:
%e A082601   1;
%e A082601   1,  0;
%e A082601   1,  1,  0;
%e A082601   1,  2,  1,  0;
%e A082601   1,  3,  3,  0,  0;
%e A082601   1,  4,  6,  2,  0,  0;
%e A082601   1,  5, 10,  7,  1,  0,  0;
%e A082601   ...
%e A082601 From _Petros Hadjicostas_, Jun 10 2020: (Start)
%e A082601 The n-th tribonacci polynomial is t_n = Sum_{k=0..n} T(n,k)*x^(n-k), so, for example:
%e A082601 t_4 = x^4 + 3*x^3 + 3*x^2;
%e A082601 t_5 = x^5 + 4*x^4 + 6*x^3 + 2*x^2;
%e A082601 t_6 = x^6 + 5*x^5 + 10*x^4 + 7*x^3 + x^2;
%e A082601 t_7 = x^7 + 6*x^6 + 15*x^5 + 16*x^4 + 6*x^3.
%e A082601 We have
%e A082601 t_4 = det([[x,-1,0,0]; [x,x,-1,0]; [x,x,x,-1]; [0,x,x,x]]);
%e A082601 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]]);
%e A082601 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]]);
%e A082601 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)
%p A082601 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);
%t A082601 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 *)
%o A082601 (Haskell)
%o A082601 a082601 n k = a082601_tabl !! n !! k
%o A082601 a082601_row n = a082601_tabl !! n
%o A082601 a082601_tabl = [1] : [1,0] : [1,1,0] : f [0,0,1] [0,1,0] [1,1,0]
%o A082601    where f us vs ws = ys : f (0:vs) (0:ws) ys where
%o A082601                       ys = zipWith3 (((+) .) . (+)) us vs ws ++ [0]
%o A082601 -- _Reinhard Zumkeller_, Apr 13 2014
%Y A082601 Closely related to A078802. A better version of A082870. Cf. A000073.
%Y A082601 Cf. A002426 (central terms).
%K A082601 nonn,tabl,easy
%O A082601 0,8
%A A082601 _Gary W. Adamson_, May 24 2003
%E A082601 Edited by Anne Donovan and _N. J. A. Sloane_, May 27 2003
%E A082601 More terms from _Emeric Deutsch_, May 06 2004