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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=4;(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,-2,1,0,-4,1,2,7,-6,1,-4,-8,18,-8,1,6,5,-38,33,-10,1,-8,4,63,-96,52,

%T -12,1,10,-21,-84,222,-190,75,-14,1,-12,48,84,-432,550,-328,102,-16,1,

%U 14,-87,-36,726,-1342,1131,-518,133,-18,1,-16,140,-99,-1056,2860,-3276,2065,-768,168,-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=4;(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, -3, 4, -1, -3, 4, -1, -3, 4, -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=CharacteristicPolynomial[{{2, -4}, {-1, 2}}, x] or CharacteristicPolynomial[{{2, -1}, {-4, 2}}, x]

%e {1},

%e {-2, 1},

%e {0, -4, 1},

%e {2, 7, -6, 1},

%e {-4, -8, 18, -8, 1},

%e {6, 5, -38, 33, -10,1},

%e {-8, 4, 63, -96, 52, -12, 1},

%e {10, -21, -84, 222, -190, 75, -14, 1},

%e {-12, 48, 84, -432, 550, -328, 102, -16, 1},

%e {14, -87, -36, 726, -1342, 1131, -518, 133, -18, 1},

%e {-16, 140, -99, -1056, 2860, -3276, 2065, -768, 168, -20, 1}

%t Clear[p, a] p[x, 0] = 1; p[x, 1] = -2 + x; p[x, 2] = x^2 - 4*x ; 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