%I #20 Mar 25 2020 07:55:33
%S 1,1,1,2,6,1,6,38,15,1,24,272,188,28,1,120,2200,2340,580,45,1,720,
%T 19920,30280,11040,1390,66,1,5040,199920,413560,206920,37450,2842,91,
%U 1,40320,2204160,5989760,3931200,955920,102816,5208,120,1
%N Coefficients of a family of orthogonal polynomials. Triangle read by rows, T(n, k) for 0 <= k <= n.
%C The polynomials represent a family of orthogonal polynomials which obey a recurrence of the form p(n, x) = (x+r(n))*p(n-1, x) - s(n)*p(n-2, x) + t(n)*p(n-3, x) - u(n)*p(n-4, x). For the details see the Maple program.
%C We conjecture that the polynomials have only negative and simple real roots.
%H Peter Luschny, <a href="/A322944/a322944.jpg">Plot of the polynomials</a>.
%F Let R be the inverse of the Riordan square [see A321620] of (1 - 3*x)^(-1/3) then T(n, k) = (-1)^(n-k)*R(n, k).
%e Triangle starts:
%e [0] 1;
%e [1] 1, 1;
%e [2] 2, 6, 1;
%e [3] 6, 38, 15, 1;
%e [4] 24, 272, 188, 28, 1;
%e [5] 120, 2200, 2340, 580, 45, 1;
%e [6] 720, 19920, 30280, 11040, 1390, 66, 1;
%e [7] 5040, 199920, 413560, 206920, 37450, 2842, 91, 1;
%e Production matrix starts:
%e 1;
%e 1, 1;
%e 3, 5, 1;
%e 6, 18, 9, 1;
%e 6, 42, 45, 13, 1;
%e 0, 48, 132, 84, 17, 1;
%e 0, 0, 180, 300, 135, 21, 1;
%e 0, 0, 0, 480, 570, 198, 25, 1;
%p P := proc(n) option remember; local a, b, c, d;
%p a := n -> 4*n-3; b := n -> 3*(n-1)*(2*n-3);
%p c := n -> (n-1)*(n-2)*(4*n-9); d := n -> (n-2)*(n-1)*(n-3)^2;
%p if n = 0 then return 1 fi;
%p if n = 1 then return x + 1 fi;
%p if n = 2 then return x^2 + 6*x + 2 fi;
%p if n = 3 then return x^3 + 15*x^2 + 38*x + 6 fi;
%p expand((x+a(n))*P(n-1) - b(n)*P(n-2) + c(n)*P(n-3) - d(n)*P(n-4)) end:
%p seq(print(P(n)), n=0..9); # Computes the polynomials.
%t a[n_] := 4n - 3;
%t b[n_] := 3(n - 1)(2n - 3);
%t c[n_] := (n - 1)(n - 2)(4n - 9);
%t d[n_] := (n - 2)(n - 1)(n - 3)^2;
%t P[n_] := P[n] = Switch[n, 0, 1, 1, x + 1, 2, x^2 + 6x + 2, 3, x^3 + 15x^2 + 38x + 6, _, Expand[(x + a[n]) P[n - 1] - b[n] P[n - 2] + c[n] P[n - 3] - d[n] P[n - 4]]];
%t Table[CoefficientList[P[n], x], {n, 0, 9}] (* _Jean-François Alcover_, Jun 15 2019, from Maple *)
%o (Sage) # uses[riordan_square from A321620]
%o R = riordan_square((1 - 3*x)^(-1/3), 9, True).inverse()
%o for n in (0..8): print([(-1)^(n-k)*c for (k, c) in enumerate(R.row(n)[:n+1])])
%Y p(n, 1) = A322943(n) (row sums); p(n, 0) = n! = A000142(n).
%Y A321966 (m=2), this sequence (m=3).
%Y Cf. A321620.
%K nonn,tabl
%O 0,4
%A _Peter Luschny_, Jan 02 2019