

A186492


Recursive triangle for calculating A186491.


4



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
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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 nth row of this triangle.


LINKS

Table of n, a(n) for n=0..44.
C. V. Sukumar and A. Hodges, Quantum algebras and paritydependent spectra, Proc. R. Soc. A (2007) 463, 24152427.


FORMULA

Recurrence relation
(1)... T(n,k) = (2*k1)*T(n1,k1)+(2*k+2)*T(n1,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,(uI)/(u+I))
with the inverse identity
(5)... A(n+1,u) = (I)^n*(1u)^n*R(n,I*(1+u)/(1u)),
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[n1]'[u] + u R[n1][u];
Table[CoefficientList[R[n][u], u], {n, 0, 8}] // Flatten (* JeanFrançois Alcover, Nov 13 2019 *)


CROSSREFS

A104035, A142459, A144015 (row sums), A186491.
Sequence in context: A079981 A117776 A298610 * A137448 A240606 A324379
Adjacent sequences: A186489 A186490 A186491 * A186493 A186494 A186495


KEYWORD

nonn,easy,tabl


AUTHOR

Peter Bala, Feb 22 2011


STATUS

approved



