A321965 -id:A321965 - cp's OEIS frontend

A321966 Triangle read by rows, coefficients of a family of orthogonal polynomials, T(n, k) for 0 <= k <= n.

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

Offset: 0

Peter Luschny, Dec 20 2018

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.
From Giuliano Cabrele, Sep 09 2021: (Start)
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)

			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;

Peter Luschny, Plot of the polynomials
Robert S. Maier, Boson Operator Ordering Identities from Generalized Stirling and Eulerian Numbers, arXiv:2308.10332 [math.CO], 2023. See p. 21.

p(n, 1) = A321965(n); p(n, 0) = n! = A000142(n).

Cf. A321620.

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.

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 *)

# 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])])

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).

A322943 a(n) = n! [x^n] -exp(-1/(3*(x - 1)^3) - 1/3)/(x - 1).

Original entry on oeis.org

1, 2, 9, 60, 513, 5286, 63417, 865824, 13229505, 223336458, 4123226601, 82559530692, 1780580892609, 41125146159150, 1012187976013593, 26434618529133096, 729843662368002177, 21233024209964157714, 649022529915336217545, 20789723945673232443468, 696253958136289126229121

Offset: 0

Peter Luschny, Jan 02 2019

nonn

Row sums of A322944.

Cf. A322944, A321620, A321965.

a := proc(n) option remember; local e, b, c, d;
e := n -> 4*n-2; b := n -> 3*(n-1)*(2*n-3);
c := n -> (n-1)*(n-2)*(4*n-9); d := n -> (n-2)*(n-1)*(n-3)^2;
if n < 4 then return [1, 2, 9, 60][n+1] fi;
e(n)*a(n-1) - b(n)*a(n-2) + c(n)*a(n-3) - d(n)*a(n-4) end:
seq(a(n), n=0..20);

# uses[riordan_square from A321620]
R = riordan_square((1 - 3*x)^(-1/3), 24, True).inverse()
[sum([(-1)^(n-k)*c for k, c in enumerate(R.row(n))]) for n in (0..23)]

a(n) = (4*n - 2)*a(n-1) - 3*(n - 1)*(2*n - 3)*a(n-2) + (n - 1)*(n - 2)*(4*n - 9)*a(n-3) - (n - 2)*(n - 1)*(n - 3)^2*a(n-4) for n >= 4.

a(n) ~ exp(-1/4 + 5*n^(1/4)/24 + sqrt(n)/2 + 4*n^(3/4)/3 - n) * n^(n + 1/8) / 2 * (1 + 1357/(2560*n^(1/4))). - Vaclav Kotesovec, Jan 02 2019

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A321966 Triangle read by rows, coefficients of a family of orthogonal polynomials, T(n, k) for 0 <= k <= n.

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