A186491 -id:A186491 - cp's OEIS frontend

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

Peter Bala, Feb 22 2011

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.

			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.

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

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

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 *)

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).

A190195 Numerators of a Taylor series expansion of 1/sqrt(cosh(x)) (even powers only).

Original entry on oeis.org

1, -1, 7, -139, 5473, -51103, 34988647, -4784061619, 17782347217, -203906055033841, 4586025046220899, -234038275571853889, 9127322584507530151393, -4621897483978366951337161, 390009953658229908025520161, -1860452328661957054823447670979, 111446346975327291562408943638981, -14050053632877769956552601074149491, 1269258883676324618437848731917951368967, -1408182090109327874242950762763137949746859

Offset: 0

N. J. A. Sloane, May 05 2011

			1/sqrt(cosh(x)) = 1 - (1/4)*x^2 + (7/96)*x^4 - (139/5760)*x^6 + (5473/645120)*x^8 - (51103/16588800)*x^10 + ...

Philippe Flajolet, Xavier Gourdon, and Philippe Dumas, Mellin transforms and asymptotics: harmonic sums, Special volume on mathematical analysis of algorithms. Theoret. Comput. Sci. 144 (1995), no. 1-2, 3-58.

Cf. A190196 (denominators), A186491.

a:= n-> numer(coeff(series(1/sqrt(cosh(x)),x,2*n+1),x,2*n)):
seq(a(n), n=0..19);  # Alois P. Heinz, Sep 19 2023

b[n]:=if n=0 then 1 else sum(b[n-k]*(k/n/2-1)/(2*k)!,k,1,n)$ a[n]:=num(b[n])$
makelist(a[n],n,0,20); /* Tani Akinari, Sep 17 2023 */

a(n) = numerator(b(n)), where b(n) = Sum_{k=1..n} b(n-k)*(k/(2*n)-1)/(2*k)!, with b(0)=1. - Tani Akinari, Sep 17 2023

a(n) = numerator((-1)^n*A186491(n)/(4^n*(2*n)!)). - Andrew Howroyd, Sep 19 2023

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A186492 Recursive triangle for calculating A186491.

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