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A191897
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Coefficients of the Z(n,x) polynomials; Z(0,x) = 1, Z(1,x) = x and Z(n,x) = x*Z(n-1,x) - 2*Z(n-2,x), n >= 2.
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0
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1, 1, 0, 1, 0, -2, 1, 0, -4, 0, 1, 0, -6, 0, 4, 1, 0, -8, 0, 12, 0, 1, 0, -10, 0, 24, 0, -8, 1, 0, -12, 0, 40, 0, -32, 0, 1, 0, -14, 0, 60, 0, -80, 0, 16, 1, 0, -16, 0, 84, 0, -160, 0, 80, 0, 1, 0, -18, 0, 112, 0, -280, 0, 240, 0, -32
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OFFSET
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0,6
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COMMENTS
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The coefficients of the Z(n,x) polynomials by decreasing exponents, see the formulas, define this triangle.
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LINKS
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FORMULA
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Z(0,x) = 1, Z(1,x) = x and Z(n,x) = x*Z(n-1,x) - 2*Z(n-2,x), n >= 2.
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EXAMPLE
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The first few rows of the coefficients of the Z(n,x) are
1;
1, 0;
1, 0, -2;
1, 0, -4, 0;
1, 0, -6, 0, 4;
1, 0, -8, 0, 12, 0;
1, 0, -10, 0, 24, 0, -8;
1, 0, -12, 0, 40, 0, -32, 0;
1, 0, -14, 0, 60, 0, -80, 0, 16;
1, 0, -16, 0, 84, 0, -160, 0, 80, 0;
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MAPLE
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nmax:=10: Z(0, x):=1 : Z(1, x):=x: for n from 2 to nmax do Z(n, x) := x*Z(n-1, x) - 2*Z(n-2, x) od: for n from 0 to nmax do for k from 0 to n do T(n, k) := coeff(Z(n, x), x, n-k) od: od: seq(seq(T(n, k), k=0..n), n=0..nmax); # Johannes W. Meijer, Jun 27 2011, revised Nov 29 2012
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MATHEMATICA
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a[n_, k_] := If[OddQ[k], 0, 2^(k/2)*Coefficient[ ChebyshevU[n, x/2], x, n-k]]; Flatten[ Table[ a[n, k], {n, 0, 10}, {k, 0, n}]] (* Jean-François Alcover, Aug 02 2012, from 2nd formula *)
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CROSSREFS
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Row sum without sign: A113405(n+1).
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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