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A162514
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Triangle of coefficients of polynomials defined by the Binet form P(n,x) = U^n + L^n, where U = (x + d)/2, L = (x - d)/2, d = (4 + x^2)^(1/2). Decreasing powers of x.
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8
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2, 1, 0, 1, 0, 2, 1, 0, 3, 0, 1, 0, 4, 0, 2, 1, 0, 5, 0, 5, 0, 1, 0, 6, 0, 9, 0, 2, 1, 0, 7, 0, 14, 0, 7, 0, 1, 0, 8, 0, 20, 0, 16, 0, 2, 1, 0, 9, 0, 27, 0, 30, 0, 9, 0, 1, 0, 10, 0, 35, 0, 50, 0, 25, 0, 2, 1, 0, 11, 0, 44, 0, 77, 0, 55, 0, 11, 0, 1, 0, 12, 0, 54, 0, 112, 0, 105, 0, 36, 0, 2, 1, 0, 13, 0
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OFFSET
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0,1
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COMMENTS
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For a signed version of this triangle corresponding to the row reversed version of the triangle A127672 see A244422. - Wolfdieter Lang, Aug 07 2014
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LINKS
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FORMULA
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P(n,x) = x*P(n-1,x) + P(n-2,x) for n >= 2, P(0,x) = 2, P(1,x) = x.
T(n,m) = [x^(n-m)] P(n,x), m = 0, 1, ..., n and n >= 0.
G.f. of polynomials P(n,x): (2 - x*z)/(1 - x*z - z^2).
G.f. of row polynomials R(n,x) = Sum_{m=0..n} T(n,m)*x^m: (2 - z)/(1 - z - (x*z)^2) (rows for P(n,x) reversed).
(End)
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EXAMPLE
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Triangle begins
2; == 2
1, 0; == x + 0
1, 0, 2; == x^2 + 2
1, 0, 3, 0; == x^3 + 3*x + 0
1, 0, 4, 0, 2;
1, 0, 5, 0, 5, 0;
1, 0, 6, 0, 9, 0, 2;
1, 0, 7, 0, 14, 0, 7, 0;
1, 0, 8, 0, 20, 0, 16, 0, 2;
1, 0, 9, 0, 27, 0, 30, 0, 9, 0;
1, 0, 10, 0, 35, 0, 50, 0, 25, 0, 2;
...
The row polynomials R(n, x) are:
R(0, x) = 2, R(1, x) = 1 = x*P(1,1/x), R(2, x) = 1 + 2*x^2 = x^2*P(2,1/x), R(3, x) = 1 + 3*x^2 = x^3*P(3,1/x), ...
(End)
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MATHEMATICA
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Table[Reverse[CoefficientList[LucasL[n, x], x]], {n, 0, 12}]//Flatten (* G. C. Greubel, Nov 05 2018 *)
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PROG
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(PARI)
P(n)=
{
local(U, L, d, r, x);
if ( n<0, return(0) );
x = 'x+O('x^(n+1));
d=(4 + x^2)^(1/2);
U=(x+d)/2; L=(x-d)/2;
r = U^n+L^n;
r = truncate(r);
return( r );
}
for (n=0, 10, print(Vec(P(n))) ); /* show triangle */
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CROSSREFS
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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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