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Triangle read by rows: coefficients of a Hermite-like set of recursive polynomials that appear by integration to be orthogonal using the substitution on the Hermite recursion of n->f(n) where f(n)=A000045[n] is the Fibonacci sequence.
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%I #3 Mar 30 2012 17:34:25

%S 1,0,1,-1,0,1,0,-3,0,1,3,0,-6,0,1,0,18,0,-11,0,1,-24,0,66,0,-19,0,1,0,

%T -258,0,209,0,-32,0,1,504,0,-1644,0,608,0,-53,0,1,0,9276,0,-8750,0,

%U 1696,0,-87,0,1,-27720,0,99696,0,-42190,0,4611,0,-142,0,1

%N Triangle read by rows: coefficients of a Hermite-like set of recursive polynomials that appear by integration to be orthogonal using the substitution on the Hermite recursion of n->f(n) where f(n)=A000045[n] is the Fibonacci sequence.

%C Note that Table[Integrate[Exp[ -x^2/2]*P[x,n]*P[x, n + 1], {x, -Infinity, Infinity}], {n, 0, 10}]; gives {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...}.

%C The row sums are: {1, 1, 0, -2, -2, 8, 24, -80, -584, 2136, 34256, ...}

%C Table[Apply[Plus, CoefficientList[P[x, n], x]], {n, 0, 10}];

%F a(n) = a(n-1)+a(n-2): A000045(n); p(x,0)=1;p(x,1)=x; p(x,n)=x*p(x,n-1)-a(n)*p(n,n-2)

%e {1},

%e {0, 1},

%e {-1, 0, 1},

%e {0, -3, 0, 1},

%e {3, 0, -6, 0, 1},

%e {0, 18, 0, -11, 0, 1},

%e {-24, 0, 66, 0, -19, 0, 1},

%e {0, -258, 0, 209, 0, -32, 0, 1},

%e {504, 0, -1644, 0, 608, 0, -53, 0, 1},

%e {0, 9276, 0, -8750, 0, 1696, 0, -87, 0, 1},

%e {-27720, 0, 99696, 0, -42190, 0, 4611, 0, -142, 0, 1}

%t f[0] = 0; f[1] = 1; f[n_] := f[n] = f[n - 1] + f[n - 2]; P[x, 0] = 1; P[x, 1] = x; P[x_, n_] := P[x, n] = x*P[x, n - 1] - f[n]*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. A000045.

%K uned,tabl,sign

%O 1,8

%A _Roger L. Bagula_, Mar 14 2008