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Search: id:A137298
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| A137298 |
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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)=A000931[n] is the Padovan sequence. |
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1
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| 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, 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, -270, 0, 1320, 0, -1197, 0, 340, 0, -34, 0, 1
(list; table; graph; listen)
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OFFSET
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1,8
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COMMENT
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The number-like behavior of the Padovan sequence made me think that I might get a orthogonal polynomial set by this substitution:
Table[Integrate[Exp[ -x2/2]*P[x,n]*P[x, n + 1], {x, -Infinity, Infinity}], {n, 0, 10}];
{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
The row sums are:
Table[Apply[Plus, CoefficientList[P[x, n], x]], {n, 0, 10}];
{1, 1, 0, -1, -1, 1, 4, 0, -20, -20, 160}
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.
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FORMULA
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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)
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EXAMPLE
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{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, 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},
{-270, 0, 1320, 0, -1197, 0, 340, 0, -34, 0, 1}
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MATHEMATICA
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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]
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CROSSREFS
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Cf. A000045, A000931.
Adjacent sequences: A137295 A137296 A137297 this_sequence A137299 A137300 A137301
Sequence in context: A025666 A025679 A071491 this_sequence A136487 A021501 A101662
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KEYWORD
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uned,tabl,sign
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AUTHOR
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Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Mar 14 2008
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