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A321966
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Triangle read by rows, coefficients of a family of orthogonal polynomials, T(n, k) for 0 <= k <= n.
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3
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1, 1, 1, 2, 5, 1, 6, 27, 12, 1, 24, 168, 123, 22, 1, 120, 1200, 1275, 365, 35, 1, 720, 9720, 13950, 5655, 855, 51, 1, 5040, 88200, 163170, 87465, 18480, 1722, 70, 1, 40320, 887040, 2046240, 1387680, 383145, 49476, 3122, 92, 1
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OFFSET
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0,4
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COMMENTS
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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.
We conjecture that the polynomials have only negative and simple real roots.
Let He(n,x) define the probabilist's version of Hermite polynomials.
Then the terms of the triangle appear to be the connection coefficients in
x^n*He(n,x) = Sum_{k=0..n} T(n,k)*He(2k,x).
These are generated by the explicit formula
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).
A formal proof that they correspond to the original definition is needed. (End)
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LINKS
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FORMULA
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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).
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EXAMPLE
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p(0,x) = 1;
p(1,x) = x + 1;
p(2,x) = x^2 + 5*x + 2;
p(3,x) = x^3 + 12*x^2 + 27*x + 6;
p(4,x) = x^4 + 22*x^3 + 123*x^2 + 168*x + 24;
p(5,x) = x^5 + 35*x^4 + 365*x^3 + 1275*x^2 + 1200*x + 120;
p(6,x) = x^6 + 51*x^5 + 855*x^4 + 5655*x^3 + 13950*x^2 + 9720*x + 720;
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MAPLE
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P := proc(n) option remember; local a, b, c;
a := n -> 3*n-2; b := n -> (n-1)*(3*n-4); c := n -> (n-2)^2*(n-1);
if n = 0 then return 1 fi;
if n = 1 then return x + 1 fi;
if n = 2 then return x^2 + 5*x + 2 fi;
expand((x+a(n))*P(n-1) - b(n)*P(n-2) + c(n)*P(n-3)) end:
seq(print(P(n)), n=0..6); # Computes the polynomials.
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MATHEMATICA
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a[n_] := 3n-2; b[n_] := (n-1)(3n-4); c[n_] := (n-2)^2 (n-1);
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]]];
Table[CoefficientList[P[n], x], {n, 0, 8}] // Flatten (* Jean-François Alcover, Jan 01 2019, from Maple *)
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PROG
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(Sage) # uses[RiordanSquare from A321620]
R = RiordanSquare((1 - 2*x)^(-1/2), 9, True).inverse()
for n in (0..8): print([(-1)^(n-k)*c for (k, c) in enumerate(R.row(n)[:n+1])])
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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