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%I #54 Oct 29 2023 10:50:39
%S 1,1,0,1,0,-1,1,0,-2,0,1,0,-3,0,1,1,0,-4,0,3,0,1,0,-5,0,6,0,-1,1,0,-6,
%T 0,10,0,-4,0,1,0,-7,0,15,0,-10,0,1,1,0,-8,0,21,0,-20,0,5,0,1,0,-9,0,
%U 28,0,-35,0,15,0,-1,1,0,-10,0,36,0,-56,0,35,0,-6,0,1,0,-11,0,45,0,-84,0,70,0,-21,0,1
%N Triangle of coefficients of Chebyshev's S(n,x) polynomials (exponents in decreasing order).
%C These polynomials also give the determinant of the tridiagonal matrix having x on the diagonal and -1 next to these x. - _M. F. Hasler_, Oct 15 2019
%C The polynomial S(n,x) is the character of the irreducible (n+1) dimensional representation of the Lie algebra sl_2 when x is the character of irreducible 2-dimesional representation. - _Leonid Bedratyuk_, Oct 28 2023
%D D. S. Mitrinovic, Analytic Inequalities, Springer-Verlag, 1970; p. 232, Sect. 3.3.38.
%D Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990.
%H T. D. Noe, <a href="/A053119/b053119.txt">Rows n=0..100 of triangle, flattened</a>
%H Tom Copeland, <a href="http://tcjpn.wordpress.com/2015/10/12/">Addendum to Elliptic Lie Triad</a>
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%F a(n,m) = A049310(n,n-m).
%F G.f. for row polynomials S(n,x) (signed triangle): 1/(1-x*z+z^2).
%F Unsigned triangle |a(n,m)| has Fibonacci polynomials F(n+1,x) as row polynomials with G.f. 1/(1-x*z-z^2).
%F a(n, m) := 0 if n < m or m odd, else ((-1)^(3*m/2))*binomial(n-m/2, n-m); a(n, m) = a(n-1, m) - a(n-2, m-2), a(n, -2) := 0 =: a(n, -1), a(0, 0) = 1, a(n, m) = 0 if n < m or m odd.
%F G.f. for m-th column (signed triangle): (-1)^(3*m/2)*x^m/(1-x)^(m/2+1) if m >= 0 is even else 0.
%F Recurrence for the (unsigned) Fibonacci polynomials: F[1]=1, F[2]=x; for n>2, F[n] = x*F[n-1]+F[n-2].
%F a = 2*A192011 - 3*A192174. - _Thomas Baruchel_, Jun 02 2018
%F Recurrence for the polynomials S(n) = x S(n-1) - S(n-2); S(0) = 1, S(1) = x. - _M. F. Hasler_, Oct 15 2019
%e The triangle begins:
%e n\m 0 1 2 3 4 5 6 7 8 9 10 ...
%e 0: 1
%e 1: 1 0
%e 2: 1 0 -1
%e 3: 1 0 -2 0
%e 4: 1 0 -3 0 1
%e 5: 1 0 -4 0 3 0
%e 6: 1 0 -5 0 6 0 -1
%e 7: 1 0 -6 0 10 0 -4 0
%e 8: 1 0 -7 0 15 0 -10 0 1
%e 9: 1 0 -8 0 21 0 -20 0 5 0
%e 10: 1 0 -9 0 28 0 -35 0 15 0 -1
%e ... Reformatted. - _Wolfdieter Lang_, Dec 17 2013
%e E.g., fourth row (n=3) corresponds to polynomial S(3,x)= x^3-2*x.
%e Triangle of absolute values of coefficients (coefficients of Fibonacci polynomials) with exponents in increasing order begins:
%e [1]
%e [0, 1]
%e [1, 0, 1]
%e [0, 2, 0, 1]
%e [1, 0, 3, 0, 1]
%e [0, 3, 0, 4, 0, 1]
%e [1, 0, 6, 0, 5, 0, 1]
%e [0, 4, 0, 10, 0, 6, 0, 1]
%e [1, 0, 10, 0, 15, 0, 7, 0, 1]
%e [0, 5, 0, 20, 0, 21, 0, 8, 0, 1]
%e See A162515 for the Fibonacci polynomials with reversed row entries, starting there with row 1. - _Wolfdieter Lang_, Dec 16 2013
%t ChebyshevS[n_, x_] := ChebyshevU[n, x/2]; Flatten[ Table[ Reverse[ CoefficientList[ ChebyshevS[n, x], x]], {n, 0, 12}]] (* _Jean-François Alcover_, Nov 25 2011 *)
%o (PARI) tabl(nn) = for (n=0, nn, print(Vec(polchebyshev(n, 2, x/2)))); \\ _Michel Marcus_, Jan 14 2016
%Y Row sums give A000045. Reflection of A049310.
%Y Cf. A162515. - _Wolfdieter Lang_, Dec 16 2013
%K easy,nice,sign,tabl
%O 0,9
%A _Wolfdieter Lang_
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