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%I #3 Mar 30 2012 17:34:25
%S 1,0,1,-1,0,1,0,-2,0,1,2,0,-4,0,1,0,6,0,-6,0,1,-6,0,18,0,-9,0,1,0,-30,
%T 0,42,0,-13,0,1,30,0,-120,0,87,0,-18,0,1,0,240,0,-414,0,178,0,-25,0,1,
%U -270,0,1320,0,-1197,0,340,0,-34,0,1
%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)=A000931[n] is the Padovan sequence.
%C The number-like behavior of the Padovan sequence made me think that I might get a orthogonal polynomial set by this substitution:
%C Table[Integrate[Exp[ -x2/2]*P[x,n]*P[x, n + 1], {x, -Infinity, Infinity}], {n, 0, 10}];
%C {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
%C The row sums are:
%C Table[Apply[Plus, CoefficientList[P[x, n], x]], {n, 0, 10}];
%C {1, 1, 0, -1, -1, 1, 4, 0, -20, -20, 160}
%C The tiling property of the Fibonacci and Padovan sequences makes me think that other sequence of fundamental number theory "beta Integer-like" sequences might give orthogonal polynomials as well.
%F a(n) = a(n-2)+a(n-3): A000931(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, -2, 0, 1},
%e {2, 0, -4, 0, 1},
%e {0, 6, 0, -6, 0, 1},
%e {-6, 0, 18, 0, -9, 0, 1},
%e {0, -30, 0, 42, 0, -13, 0, 1},
%e {30, 0, -120, 0, 87, 0, -18, 0, 1},
%e {0, 240, 0, -414, 0, 178, 0, -25, 0, 1},
%e {-270, 0, 1320, 0, -1197, 0, 340, 0, -34, 0, 1}
%t f[0] = 0; f[1] = 1;f[2]=1; f[n_] := f[n] = f[n - 2] + f[n - 3]; 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, A000931.
%K uned,tabl,sign
%O 1,8
%A _Roger L. Bagula_, Mar 14 2008
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