A186492 Recursive triangle for calculating A186491.

1, 0, 1, 2, 0, 3, 0, 14, 0, 15, 28, 0, 132, 0, 105, 0, 5556, 0, 1500, 0, 945, 1112, 0, 10668, 0, 1995, 0, 10395, 0, 43784, 0, 212940, 0, 304290, 0, 135135, 87568, 0, 1408992, 0, 4533480, 0, 5239080, 0, 2027025

Offset: 0

Author

Peter Bala, Feb 22 2011

Keywords

• nonn
• easy
• tabl

Comments

The table entries are defined by a recurrence relation (see below).

This triangle can be used to calculate the entries of A186491: the nonzero entries of the first column of the triangle give A186491.

PRODUCTION MATRIX

The production matrix P for this triangle is the bidiagonal matrix with the sequence [2,4,6,...] on the main subdiagonal, the sequence [1,3,5,...] on the main superdiagonal and 0's elsewhere: the first row of P^n is the n-th row of this triangle.

Examples

			Table begins
n\k|.....0.....1......2.....3......4.....5......6
=================================================
0..|.....1
1..|.....0.....1
2..|.....2.....0......3
3..|.....0....14......0....15
4..|....28.....0....132.....0....105
5..|.....0...556......0..1500......0...945
6..|..1112.....0..10668.....0..19950.....0..10395
..
Examples of recurrence relation
T(4,2) = 3*T(3,1) + 6*T(3,3) = 3*14 + 6*15 = 132;
T(6,4) = 7*T(5,3) + 10*T(5,5) = 7*1500 + 10*945 = 19950.
		

Links

• C. V. Sukumar and A. Hodges, Quantum algebras and parity-dependent spectra, Proc. R. Soc. A (2007) 463, 2415-2427.

Crossrefs

A104035, A142459, A144015 (row sums), A186491.

Programs

Mathematica

R[0][] = 1; R[1][u] = u;
R[n_][u_] := R[n][u] = 2(1+u^2) R[n-1]'[u] + u R[n-1][u];
Table[CoefficientList[R[n][u], u], {n, 0, 8}] // Flatten (* Jean-François Alcover, Nov 13 2019 *)

Formula

Recurrence relation

(1)... T(n,k) = (2*k-1)*T(n-1,k-1)+(2*k+2)*T(n-1,k+1).

GENERATING FUNCTION

E.g.f. (Compare with the e.g.f. of A104035):

(2)... 1/sqrt(cos(2*t)-u*sin(2*t)) = sum {n = 0..inf } R(n,u)*t^n/n! = 1 + u*t + (2+3*u^2)*t^2/2! + (14*u+15*u^3)*t^3/3!+....

ROW POLYNOMIALS

The row polynomials R(n,u) begin

... R(1,u) = u

... R(2,u) = 2+3*u^2

... R(3,u) = 14*u+15*u^3

... R(4,u) = 28+132*u^2+105u^4.

They satisfy the recurrence relation

(3)... R(n+1,u) = 2*(1+u^2)*d/du(R(n,u))+u*R(n,u) with starting value R(0,u) = 1.

Compare with Formula (1) of A104035 for the polynomials Q_n(u).

The polynomials R(n,u) are related to the shifted row polynomials A(n,u) of A142459 via

(4)... R(n,u) = ((u+I)/2)^n*A(n+1,(u-I)/(u+I))

with the inverse identity

(5)... A(n+1,u) = (-I)^n*(1-u)^n*R(n,I*(1+u)/(1-u)),

where {A(n,u)}n>=1 begins [1,1+u,1+10*u+u^2,1+59*u+59*u^2+u^3,...] and I = sqrt(-1).