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A053117
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Triangle read by rows of coefficients of Chebyshev's U(n,x) polynomials (exponents in increasing order).
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21
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1, 0, 2, -1, 0, 4, 0, -4, 0, 8, 1, 0, -12, 0, 16, 0, 6, 0, -32, 0, 32, -1, 0, 24, 0, -80, 0, 64, 0, -8, 0, 80, 0, -192, 0, 128, 1, 0, -40, 0, 240, 0, -448, 0, 256, 0, 10, 0, -160, 0, 672, 0, -1024, 0, 512, -1, 0, 60, 0, -560, 0, 1792, 0, -2304, 0, 1024, 0, -12, 0, 280, 0, -1792, 0, 4608, 0, -5120, 0, 2048, 1, 0, -84, 0, 1120, 0, -5376, 0, 11520, 0, -11264, 0, 4096
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OFFSET
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0,3
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COMMENTS
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G.f. for row polynomials U(n,x) (signed triangle): 1/(1-2*x*z+z^2). Unsigned triangle |a(n,m)| has Fibonacci polynomials F(n+1,2*x) as row polynomials with g.f. 1/(1-2*x*z-z^2).
Row sums (unsigned triangle) A000129(n+1) (Pell). Row sums (signed triangle) A000027(n+1) (natural numbers).
The o.g.f. for the Legendre polynomials L(n,x) is 1 / sqrt(1- 2x*z + z^2), and squaring it gives the o.g.f. of this entry, so Sum_{k=0..n} L(k,x) L(n-k,x) = U(n,x). This reduces to U(n,x) = L(n/2,x)^2 + 2*Sum_{k=0...n/2-1} L(k,x) L(n-k,x) for n even and U(n,x) = 2*Sum_{k=0..(n-1)/2} L(k,x) L(n-k.x) for odd n. (Cf. also Allouche et al.) For a connection through the Legendre polynomials to elliptic curves and modular forms, see the MathOverflow question below. For the normalized Legendre polynomials, see A100258. (Cf. A097610 with h1 = -2x and h2 = 1, A207538, A099089 and A133156.) - Tom Copeland, Feb 04 2016
The compositional inverse of the shifted o.g.f. x / (1 + 2xz + z^2) for differently signed row polynomials of this entry is the shifted o.g.f. of A121448. The unsigned, non-vanishing antidiagonals (top to bottom) of this triangle are the rows of A038207. - Tom Copeland, Feb 08 2016
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REFERENCES
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Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990.
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LINKS
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FORMULA
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a(n, m) := 0 if n<m or n+m odd, else ((-1)^((n+m)/2+m))*(2^m)*binomial((n+m)/2, m); a(n, m) = -a(n-2, m)+2*a(n-1, m-1), a(n, -1) := 0 =: a(-1, m), a(0, 0)=1, a(n, m)= 0 if n<m or n+m odd; G.f. for m-th column (signed triangle): (1/(1+x^2)^(m+1))*(2*x)^m.
If n and k are of the same parity then a(n,k)=(-1)^((n-k)/2)*sum(binomial((n+k)/2,i)*binomial((n+k)/2-i,(n-k)/2),i=0..k) and a(n,k)=0 otherwise. - Milan Janjic, Apr 13 2008
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EXAMPLE
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Triangle begins:
1;
0, 2;
-1, 0, 4;
0, -4, 0, 8;
1, 0, -12, 0, 16;
...
E.g., fourth row (n=3) {0,-4,0,8} corresponds to polynomial U(3,x) = -4*x + 8*x^3.
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MAPLE
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seq(seq(coeff(orthopoly[U](n, x), x, j), j=0..n), n=0..16); # Robert Israel, Feb 09 2016
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MATHEMATICA
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Flatten[ Table[ CoefficientList[ ChebyshevU[n, x], x], {n, 0, 12}]](* Jean-François Alcover, Nov 24 2011 *)
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PROG
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(PARI) T(n, k) = polcoeff(polchebyshev(n, 2), k); \\ Michel Marcus, Feb 10 2016
(Julia)
using Nemo
function A053117Row(n)
R, x = PolynomialRing(ZZ, "x")
p = chebyshev_u(n, x)
[coeff(p, j) for j in 0:n] end
for n in 0:6 A053117Row(n) |> println end # Peter Luschny, Mar 13 2018
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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