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A186492 Recursive triangle for calculating A186491.

%I A186492 #14 Nov 13 2019 06:58:39
%S A186492 1,0,1,2,0,3,0,14,0,15,28,0,132,0,105,0,5556,0,1500,0,945,1112,0,
%T A186492 10668,0,1995,0,10395,0,43784,0,212940,0,304290,0,135135,87568,0,
%U A186492 1408992,0,4533480,0,5239080,0,2027025
%N A186492 Recursive triangle for calculating A186491.
%C A186492 The table entries are defined by a recurrence relation (see below).
%C A186492 This triangle can be used to calculate the entries of A186491: the nonzero entries of the first column of the triangle give A186491.
%C A186492 PRODUCTION MATRIX
%C A186492 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.
%H A186492 C. V. Sukumar and A. Hodges, <a href="https://doi.org/10.1098/rspa.2007.0003">Quantum algebras and parity-dependent spectra</a>, Proc. R. Soc. A (2007) 463, 2415-2427.
%F A186492 Recurrence relation
%F A186492 (1)... T(n,k) = (2*k-1)*T(n-1,k-1)+(2*k+2)*T(n-1,k+1).
%F A186492 GENERATING FUNCTION
%F A186492 E.g.f. (Compare with the e.g.f. of A104035):
%F A186492 (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!+....
%F A186492 ROW POLYNOMIALS
%F A186492 The row polynomials R(n,u) begin
%F A186492 ... R(1,u) = u
%F A186492 ... R(2,u) = 2+3*u^2
%F A186492 ... R(3,u) = 14*u+15*u^3
%F A186492 ... R(4,u) = 28+132*u^2+105u^4.
%F A186492 They satisfy the recurrence relation
%F A186492 (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.
%F A186492 Compare with Formula (1) of A104035 for the polynomials Q_n(u).
%F A186492 The polynomials R(n,u) are related to the shifted row polynomials A(n,u) of A142459 via
%F A186492 (4)... R(n,u) = ((u+I)/2)^n*A(n+1,(u-I)/(u+I))
%F A186492 with the inverse identity
%F A186492 (5)... A(n+1,u) = (-I)^n*(1-u)^n*R(n,I*(1+u)/(1-u)),
%F A186492 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).
%e A186492 Table begins
%e A186492 n\k|.....0.....1......2.....3......4.....5......6
%e A186492 =================================================
%e A186492 0..|.....1
%e A186492 1..|.....0.....1
%e A186492 2..|.....2.....0......3
%e A186492 3..|.....0....14......0....15
%e A186492 4..|....28.....0....132.....0....105
%e A186492 5..|.....0...556......0..1500......0...945
%e A186492 6..|..1112.....0..10668.....0..19950.....0..10395
%e A186492 ..
%e A186492 Examples of recurrence relation
%e A186492 T(4,2) = 3*T(3,1) + 6*T(3,3) = 3*14 + 6*15 = 132;
%e A186492 T(6,4) = 7*T(5,3) + 10*T(5,5) = 7*1500 + 10*945 = 19950.
%t A186492 R[0][_] = 1; R[1][u_] = u;
%t A186492 R[n_][u_] := R[n][u] = 2(1+u^2) R[n-1]'[u] + u R[n-1][u];
%t A186492 Table[CoefficientList[R[n][u], u], {n, 0, 8}] // Flatten (* _Jean-François Alcover_, Nov 13 2019 *)
%Y A186492 A104035, A142459, A144015 (row sums), A186491.
%K A186492 nonn,easy,tabl
%O A186492 0,4
%A A186492 _Peter Bala_, Feb 22 2011