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A triangle of coefficients based on the squares of the Chebyshev T and U polynomials: p(x,n)=If[Mod[n, 2] == 0, (ChebyshevT[n + 1, x]^2 + x^2*ChebyshevU[n, x]^2)/(2*x^2), (-1 + ChebyshevT[n + 1, x]^2 + x^2*ChebyshevU[n, x]^2)/(2*x^2)]
0

%I #2 Mar 30 2012 17:34:39

%S 1,-2,0,4,5,0,-16,0,16,-8,0,48,0,-96,0,64,13,0,-112,0,368,0,-512,0,

%T 256,-18,0,228,0,-1088,0,2432,0,-2560,0,1024,25,0,-416,0,2720,0,-8704,

%U 0,14592,0,-12288,0,4096,-32,0,704,0,-6016,0,25856,0,-61440,0,81920,0

%N A triangle of coefficients based on the squares of the Chebyshev T and U polynomials: p(x,n)=If[Mod[n, 2] == 0, (ChebyshevT[n + 1, x]^2 + x^2*ChebyshevU[n, x]^2)/(2*x^2), (-1 + ChebyshevT[n + 1, x]^2 + x^2*ChebyshevU[n, x]^2)/(2*x^2)]

%C Row sums are:A000982;

%C {1, 2, 5, 8, 13, 18, 25, 32, 41, 50, 61,...}.

%F p(x,n)=If[Mod[n, 2] == 0, (ChebyshevT[n + 1, x]^2 + x^2*ChebyshevU[n, x]^2)/(2*x^2),

%F (-1 + ChebyshevT[n + 1, x]^2 + x^2*ChebyshevU[n, x]^2)/(2*x^2)]

%e {1},

%e {-2, 0, 4},

%e {5, 0, -16, 0, 16},

%e {-8, 0, 48, 0, -96, 0, 64},

%e {13, 0, -112, 0, 368, 0, -512, 0, 256},

%e {-18, 0, 228, 0, -1088, 0, 2432, 0, -2560, 0, 1024},

%e {25, 0, -416, 0, 2720, 0, -8704, 0, 14592, 0, -12288, 0, 4096},

%e {-32, 0, 704, 0, -6016, 0, 25856, 0, -61440, 0, 81920, 0, -57344, 0, 16384},

%e {41, 0, -1120, 0, 12128, 0, -67072, 0, 211712, 0, -397312, 0, 438272, 0, -262144, 0, 65536},

%e {-50, 0, 1700, 0, -22720, 0, 156800, 0, -630272, 0, 1559552, 0, -2408448, 0, 2260992, 0, -1179648, 0, 262144},

%e {61, 0, -2480, 0, 40112, 0, -337408, 0, 1676032, 0, -5242880, 0, 10616832, 0, -13893632, 0, 11337728, 0, -5242880, 0, 1048576}

%t p[x_, n_] = If[Mod[n, 2] == 0, (ChebyshevT[n + 1, x]^2 + x^2* ChebyshevU[n, x]^2)/(2*x^2),

%t (-1 + ChebyshevT[n + 1, x]^2 + x^2*ChebyshevU[n, x]^2)/(2*x^2)];

%t Table[CoefficientList[p[x, n], x], {n, 0, 10}];

%t Flatten[%]

%Y A000982

%K sign,tabf,uned

%O 0,2

%A _Roger L. Bagula_, Feb 16 2010