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

0,4

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.

Links

Table of n, a(n) for n=0..44.

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

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).

Example

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.

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 *)

Crossrefs

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

Sequence in context: A079981 A117776 A298610 * A137448 A240606 A335889

Adjacent sequences: A186489 A186490 A186491 * A186493 A186494 A186495

Keyword

nonn,easy,tabl

Author

Peter Bala, Feb 22 2011

Status

approved