login
Triangular sequence of coefficients of a polynomial recursion for C_n and B_n Cartan matrices: p(x, n) = (-2 + x)*p(x, n - 1) - p(x, n - 2) p(x,n)=x2-4*x+4-m:m=5;(related sequence: A_n:m=1,G_n,m=3,B_n,C_n,m=2) This triangular sequence is an extension to the Cartan pattern of matrices.
0

%I #3 Mar 30 2012 17:34:22

%S 1,-2,1,-1,-4,1,4,6,-6,1,-7,-4,17,-8,1,10,-5,-32,32,-10,1,-13,24,42,

%T -88,51,-12,1,16,-56,-28,186,-180,74,-14,1,-19,104,-42,-312,495,-316,

%U 101,-16,1,22,-171,216,396,-1122,1053,-504,132,-18,1,-25,260,-561,-264,2145,-2912,1960,-752,167,-20,1

%N Triangular sequence of coefficients of a polynomial recursion for C_n and B_n Cartan matrices: p(x, n) = (-2 + x)*p(x, n - 1) - p(x, n - 2) p(x,n)=x2-4*x+4-m:m=5;(related sequence: A_n:m=1,G_n,m=3,B_n,C_n,m=2) This triangular sequence is an extension to the Cartan pattern of matrices.

%C Row sums are:

%C {1, -1, -4, 5, -1, -4, 5, -1, -4, 5, -1}

%C This sequence is also related to different p(x,2) start:

%C 1) A_n like sequence A053122 ( sign change)

%C 2) my G_n matrix A136674

%C 3) B_n,C_n A110162

%F p(x, n) = (-2 + x)*p(x, n - 1) - p(x, n - 2) Three start vectors necessary: p(x,0)=1;p(x,1)=2-x; p(x,2)=x^2-4*x-1=CharacteristicPolynomial[{{2, -5}, {-1, 2}}, x] or CharacteristicPolynomial[{{2, -1}, {-5, 2}}, x]

%e {1},

%e {-2, 1},

%e {-1, -4, 1},

%e {4, 6, -6, 1},

%e {-7, -4, 17, -8, 1},

%e {10, -5, -32, 32, -10, 1},

%e {-13, 24, 42, -88,51, -12, 1},

%e {16, -56, -28,186, -180, 74, -14, 1},

%e {-19, 104, -42, -312, 495, -316, 101, -16, 1},

%e {22, -171, 216, 396, -1122, 1053, -504, 132, -18, 1},

%e {-25, 260, -561, -264,2145, -2912, 1960, -752, 167, -20, 1}

%t Clear[p, a] p[x, 0] = 1; p[x, 1] = -2 + x; p[x, 2] = x^2 - 4*x - 1; p[x_, n_] := p[x, n] = (-2 + x)*p[x, n - 1] - p[x, n - 2]; Table[ExpandAll[p[x, n]], {n, 0, 10}]; a = Table[CoefficientList[p[x, n], x], {n, 0, 10}] Flatten[a]

%Y Cf. A053122, A136674, A110162.

%K tabl,uned,sign

%O 1,2

%A _Roger L. Bagula_, Apr 12 2008