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%I #40 Aug 25 2023 01:52:35
%S 1,1,1,2,5,1,6,27,12,1,24,168,123,22,1,120,1200,1275,365,35,1,720,
%T 9720,13950,5655,855,51,1,5040,88200,163170,87465,18480,1722,70,1,
%U 40320,887040,2046240,1387680,383145,49476,3122,92,1
%N Triangle read by rows, coefficients of a family of orthogonal polynomials, 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 + alpha(n))*p(n-1, x) - beta(n)*p(n-2, x) + gamma(n)*p(n-3, x). For the details see the Maple program.
%C We conjecture that the polynomials have only negative and simple real roots.
%C From _Giuliano Cabrele_, Sep 09 2021: (Start)
%C Let He(n,x) define the probabilist's version of Hermite polynomials.
%C Then the terms of the triangle appear to be the connection coefficients in
%C x^n*He(n,x) = Sum_{k=0..n} T(n,k)*He(2k,x).
%C These are generated by the explicit formula
%C T(n,m) = 2^(n-m)*Sum_{j=0..floor(n/2)} C(n,2*j)*C(2*n-2*j,2*m)*Gamma(1/2 + n - m - j)/Gamma(1/2 - j).
%C A formal proof that they correspond to the original definition is needed. (End)
%H Peter Luschny, <a href="/A321966/a321966.jpg">Plot of the polynomials</a>
%H Robert S. Maier, <a href="https://arxiv.org/abs/2308.10332">Boson Operator Ordering Identities from Generalized Stirling and Eulerian Numbers</a>, arXiv:2308.10332 [math.CO], 2023. See p. 21.
%F Let R be the inverse of the Riordan square [see A321620] of (1 - 2*x)^(-1/2) then T(n, k) = (-1)^(n-k)*R(n, k).
%e p(0,x) = 1;
%e p(1,x) = x + 1;
%e p(2,x) = x^2 + 5*x + 2;
%e p(3,x) = x^3 + 12*x^2 + 27*x + 6;
%e p(4,x) = x^4 + 22*x^3 + 123*x^2 + 168*x + 24;
%e p(5,x) = x^5 + 35*x^4 + 365*x^3 + 1275*x^2 + 1200*x + 120;
%e p(6,x) = x^6 + 51*x^5 + 855*x^4 + 5655*x^3 + 13950*x^2 + 9720*x + 720;
%p P := proc(n) option remember; local a, b, c;
%p a := n -> 3*n-2; b := n -> (n-1)*(3*n-4); c := n -> (n-2)^2*(n-1);
%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 + 5*x + 2 fi;
%p expand((x+a(n))*P(n-1) - b(n)*P(n-2) + c(n)*P(n-3)) end:
%p seq(print(P(n)), n=0..6); # Computes the polynomials.
%t a[n_] := 3n-2; b[n_] := (n-1)(3n-4); c[n_] := (n-2)^2 (n-1);
%t P[n_] := P[n] = Switch[n, 0, 1, 1, x+1, 2, x^2 + 5x + 2, _, Expand[(x+a[n]) P[n-1] - b[n] P[n-2] + c[n] P[n-3]]];
%t Table[CoefficientList[P[n], x], {n, 0, 8}] // Flatten (* _Jean-François Alcover_, Jan 01 2019, from Maple *)
%o (Sage) # uses[RiordanSquare from A321620]
%o R = RiordanSquare((1 - 2*x)^(-1/2), 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) = A321965(n); p(n, 0) = n! = A000142(n).
%Y Cf. A321620.
%K nonn,tabl
%O 0,4
%A _Peter Luschny_, Dec 20 2018
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