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 A049310 Triangle of coefficients of Chebyshev's S(n,x) := U(n,x/2) polynomials (exponents in increasing order). 439

%I

%S 1,0,1,-1,0,1,0,-2,0,1,1,0,-3,0,1,0,3,0,-4,0,1,-1,0,6,0,-5,0,1,0,-4,0,

%T 10,0,-6,0,1,1,0,-10,0,15,0,-7,0,1,0,5,0,-20,0,21,0,-8,0,1,-1,0,15,0,

%U -35,0,28,0,-9,0,1,0,-6,0,35,0,-56,0,36,0,-10,0,1,1,0,-21,0,70,0,-84,0

%N Triangle of coefficients of Chebyshev's S(n,x) := U(n,x/2) polynomials (exponents in increasing order).

%C G.f. for row polynomials S(n,x) (signed triangle): 1/(1-x*z+z^2). Unsigned triangle |a(n,m)| has Fibonacci polynomials F(n+1,x) as row polynomials with g.f. 1/(1-x*z-z^2). |a(n,m)| triangle has rows of Pascal's triangle A007318 in the even-numbered diagonals (odd-numbered ones have only 0's).

%C Row sums (unsigned triangle) A000045(n+1) (Fibonacci). Row sums (signed triangle) S(n,1) sequence = periodic(1,1,0,-1,-1,0) = A010892.

%C Alternating row sums A049347(n) = S(n,-1) = periodic(1,-1,0). - _Wolfdieter Lang_, Nov 04 2011

%C S(n,x) is the characteristic polynomial of the adjacency matrix of the n-path. - _Michael Somos_, Jun 24 2002

%C S(n,x) is also the matching polynomial of the n-path. - _Eric W. Weisstein_, Apr 10 2017

%C |T(n,k)| = number of compositions of n+1 into k+1 odd parts. Example: |T(7,3)| = 10 because we have (1,1,3,3), (1,3,1,3), (1,3,3,1), (3,1,1,3), (3,1,3,1), (3,3,1,1), (1,1,1,5), (1,1,5,1), (1,5,1,1) and (5,1,1,1). - _Emeric Deutsch_, Apr 09 2005

%C S(n,x)= R(n,x) + S(n-2,x), n >= 2, S(-1,x)=0, S(0,x)=1, R(n,x):=2*T(n,x/2) = Sum_{m=0..n} A127672(n,m)*x^m (monic integer Chebyshev T-Polynomials). This is the rewritten so-called trace of the transfer matrix formula for the T-polynomials. - _Wolfdieter Lang_, Dec 02 2010

%C In a regular N-gon inscribed in a unit circle, the side length is d(N,1) = 2*sin(Pi/N). The length ratio R(N,k):=d(N,k)/d(N,1) for the (k-1)-th diagonal, with k from {2,3,...,floor(N/2)}, N >= 4, equals S(k-1,x) = sin(k*Pi/N)/sin(Pi/N) with x=rho(N):=R(N,2) = 2*cos(Pi/N). Example: N=7 (heptagon), rho=R(7,2), sigma:=R(N,3) = S(2,rho) = rho^2 - 1. Motivated by the quoted paper by P. Steinbach. - _Wolfdieter Lang_, Dec 02 2010

%C From _Wolfdieter Lang_, Jul 12 2011: (Start)

%C In q- or basic analysis, q-numbers are [n]_q :=S(n-1,q+1/q) = (q^n-(1/q)^n})/(q-1/q), with the row polynomials S(n,x), n >= 0.

%C The zeros of the row polynomials S(n-1,x) are (from those of Chebyshev U-polynomials):

%C x(n-1;k) = +- t(k,rho(n)), k = 1..ceiling((n-1)/2), n >= 2, with t(n,x) the row polynomials of A127672 and rho(n):= 2*cos(Pi/n). The simple vanishing zero for even n appears here as +0 and -0.

%C Factorization of the row polynomials S(n-1,x), x >= 1, in terms of the minimal polynomials of cos(2 Pi/2), called Psi(n,x), with coefficients given by A181875/A181876:

%C S(n-1,x) = (2^(n-1))*Product_{n>=1}(Psi(d,x/2), 2 < d | 2n).

%C (From the rewritten eq. (3) of the Watkins and Zeitlin reference, given under A181872.) [See the W. Lang ArXiv link, Proposition 9, eq. (62). - _Wolfdieter Lang_, Apr 14 2018]

%C (End)

%C The discriminants of the S(n,x) polynomials are found in A127670. - _Wolfdieter Lang_, Aug 03 2011

%C This is an example for a subclass of Riordan convolution arrays (lower triangular matrices) called Bell arrays. See the L. W. Shapiro et al. reference under A007318. If a Riordan array is named (G(z),F(z)) with F(z)=z*Fhat(z), the o.g.f. for the row polynomials is G(z)/(1-x*z*Fhat(z)), and it becomes a Bell array if G(z)=Fhat(z). For the present Bell type triangle G(z)=1/(1+z^2) (see the o.g.f. comment above). This leads to the o.g.f. for the column no. k, k >= 0, x^k/(1+x^2)^(k+1) (see the formula section), the one for the row sums and for the alternating row sums (see comments above). The Riordan (Bell) A- and Z-sequences (defined in a W. Lang link under A006232, with references) have o.g.f.s 1-x*c(x^2) and -x*c(x^2), with the o.g.f. of the Catalan numbers A000108. Together they lead to a recurrence given in the formula section. - _Wolfdieter Lang_, Nov 04 2011

%C The determinant of the N x N matrix S(N,[x, ..., x[N]]) with elements S(m-1,x[n]), for n, m = 1, 2, ..., N, and for any x[n], is identical with the determinant of V(N,[x, ..., x[N]]) with elements x[n]^(m-1) (a Vandermondian, which equals Product_{1 <= i < j<= N} (x[j] - x[i]). This is a special instance of a theorem valid for any N >= 1 and any monic polynomial system p(m,x), m>=0, with p(0,x) = 1. For this theorem see the Vein-Dale reference, p. 59. Thanks to _L. Edson Jeffery_ for an email asking for a proof of the non-singularity of the matrix S(N,[x, ...., x[N]]) if and only if the x[j], j = 1..N, are pairwise different. - _Wolfdieter Lang_, Aug 26 2013

%C These S polynomials also appear in the context of modular forms. The rescaled Hecke operator T*_n = n^((1-k)/2)*T_n acting on modular forms of weight k satisfies T*_(p^n) = S(n, T*_p), for each prime p and positive integer n. See the Koecher-Krieg reference, p. 223. - _Wolfdieter Lang_, Jan 22 2016

%C For a shifted o.g.f. (mod signs), its compositional inverse, and connections to Motzkin and Fibonacci polynomials, non-crossing partitions and other combinatorial structures, see A097610. - _Tom Copeland_, Jan 23 2016

%C From _M. Sinan Kul_, Jan 30 2016: (Start)

%C Solutions of the Diophantine equation u^2 + v^2 - k*u*v = 1 for integer k given by (u(k,n), v(k,n)) = (S(n,k), S(n-1,k)) because of the Cassini-Simson identity: S(n,x)^2 - S(n+1,x)*S(n-1, x) = 1, after use of the S-recurrence. Note that S(-n, x) = -S(-n-2, x), n >= 1, and the periodicity of some S(n, k) sequences.

%C Hence another way to obtain the row polynomials would be to take powers of the matrix [x, -1; 1,0]: S(n, x) = ([x, -1; 1, 0])^n)[1,1], n >= 0.

%C See also a Feb 01 2016 comment on A115139 for a well-known S(n, x) sum formula.

%C Then we have with the present T triangle

%C A039834(n) = -i^(n+1)*T(n-1, k) where i is the imaginary unit and n >= 0.

%C A051286(n) = Sum_{i=0..n} T(n,i)^2 (see the

%C _Philippe Deléham_, Nov 21 2005 formula),

%C A181545(n) = Sum_{i=0..n+1} abs(T(n,i)^3),

%C A181546(n) = Sum_{i=0..n+1} T(n,i)^4,

%C A181547(n) = Sum_{i=0..n+1} abs(T(n,i)^5).

%C S(n, 0) = A056594(n), and for k = 1..10 the sequences S(n-1, k) with offset n = 0 are A128834, A001477, A001906, A001353, A004254, A001109, A004187, A001090, A018913, A004189.

%C (End)

%C For more on the Diophantine equation presented by Kul, see the Ismail paper. - _Tom Copeland_, Jan 31 2016

%C 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 U(n,x), A053117, so Sum_{k=0..n} L(k,x/2) L(n-k,x/2) = S(n,x). This gives S(n,x) = L(n/2,x/2)^2 + 2*Sum_{k=0..n/2-1} L(k,x/2) L(n-k,x/2) for n even and S(n,x) = 2*Sum_{k=0..(n-1)/2} L(k,x/2) L(n-k,x/2) for odd n. For a connection to elliptic curves and modular forms, see A053117. For the normalized Legendre polynomials, see A100258. For other properties and relations to other polynomials, see Allouche et al. - _Tom Copeland_, Feb 04 2016

%C LG(x,h1,h2) = -log(1 - h1*x + h2*x^2) = Sum_{n>0} F(n,-h1,h2,0,..,0) x^n/n is a log series generator of the bivariate row polynomials of A127672 with A127672(0,0) = 0 and where F(n,b1,b2,..,bn) are the Faber polynomials of A263916. Exp(LG(x,h1,h2)) = 1 / (1 - h1*x + h2*x^2 ) is the o.g.f. of the bivariate row polynomials of this entry. - _Tom Copeland_, Feb 15 2016

%C For distinct odd primes p and q the Legendre symbol can be written as Legendre(q,p) = Product_{k=1..P} S(q-1, 2*cos(2*Pi*k/p)), with P = (p-1)/2. See the Lemmermeyer reference, eq. (8.1) on p. 236. Using the zeros of S(q-1, x) (see above) one has S(q-1, x) = Product_{l=1..Q} (x^2 - (2*cos(Pi*l/q))^2), with Q = (q-1)/2. Thus S(q-1, 2*cos(2*Pi*k/p)) = ((-4)^Q)*Product_{l=1..Q} (sin^2(2*Pi*k/p) - sin^2(Pi*l/q)) = ((-4)^Q)*Product_{m=1..Q} (sin^2(2*Pi*k/p) - sin^2(2*Pi*m/q)). For the proof of the last equality see a W. Lang comment on the triangle A057059 for n = Q and an obvious function f. This leads to Eisenstein's proof of the quadratic reciprocity law Legendre(q,p) = ((-1)^(P*Q)) * Legendre(p,q), See the Lemmermeyer reference, pp. 236-237. - _Wolfdieter Lang_, Aug 28 2016

%C For connections to generalized Fibonacci polynomials, compare their generating function on p. 5 of the Amdeberhan et al. link with the o.g.f. given above for the bivariate row polynomials of this entry. - _Tom Copeland_, Jan 08 2017

%C The formula for Ramanujan's tau function (see A000594) for prime powers is tau(p^k) = p^(11*k/2)*S(k, p^(-11/2)*tau(p)) for k >= 1, and p = A000040(n), n >= 1. See the Hardy reference, p. 164, eqs. (10.3.4) and (10.3.6) rewritten in terms of S. - _Wolfdieter Lang_, Jan 27 2017

%C From _Wolfdieter Lang_, May 08 2017: (Start)

%C The number of zeros Z(n) of the S(n, x) polynomials in the open interval (-1,+1) is 2*b(n) for even n >= 0 and 1 + 2*b(n) for odd n >= 1, where b(n) = floor(n/2) - floor((n+1)/3). This b(n) is the number of integers k in the interval (n+1)/3 < k <= floor(n/2). See a comment on the zeros of S(n, x) above, and b(n) = A008615(n-2), n >= 0. The numbers Z(n) have been proposed (with a conjecture related to A008611) by _Michel Lagneau_, Mar 2017, as the number of zeros of Fibonacci polynomials on the imaginary axis (-I,+I), with I=sqrt(-1). They are Z(n) = A008611(n-1), n >= 0, with A008611(-1) = 0. Also Z(n) = A194960(n-4), n >= 0. Proof using the A008611 version. A194960 follows from this.

%C In general the number of zeros Z(a;n) of S(n, x) for n >= 0 in the open interval (-a,+a) for a from the interval (0,2) (x >= 2 never has zeros, and a=0 is trivial: Z(0;n) = 0) is with b(a;n) = floor(n//2) - floor((n+1)*arccos(a/2)/Pi), as above Z(a;n) = 2*b(a;n) for even n >= 0 and 1 + 2*b(a;n) for odd n >= 1. For the closed interval [-a,+a] Z(0;n) = 1 and for a from (0,1) one uses for Z(a;n) the values b(a;n) = floor(n/2) - ceiling((n+1)*arccos(a/2)/Pi) + 1. (End)

%C The Riordan row polynomials S(n, x) (Chebyshev S) belong to the Boas-Buck class (see a comment and references in A046521), hence they satisfy the Boas-Buck identity: (E_x - n*1)*S(n, x) = (E_x + 1)*Sum_{p=0..n-1} (1 - (-1)^p)*(-1)^((p+1)/2)*S(n-1-p, x), for n >= 0, where E_x = x*d/dx (Euler operator). For the triangle T(n, k) this entails a recurrence for the sequence of column k, given in the formula section. - _Wolfdieter Lang_, Aug 11 2017

%C The e.g.f. E(x,t) := Sum_{n>=0} (t^n/n!)*S(n,x) for the row polynomials is obtained via inverse Laplace transformation from the above given o.g.f. as E(x,t) = ((1/xm)*exp(t/xm) - (1/xp)*exp(t/xp) )/(xp - xm) with xp = (x + sqrt(x^2-4))/2 and xm = (x - sqrt(x^2-4))/2. - _Wolfdieter Lang_, Nov 08 2017

%C From _Wolfdieter Lang_, Apr 12 2018: (Start)

%C Factorization of row polynomials S(n, x), for n >= 1, in terms of C polynomials (not Chebyshev C) with coefficients given in A187360. This is obtained from the factorization into Psi polynomials (see the Jul 12 2011 comment above) but written in terms of minimal polynomials of 2*cos(2*Pi/n) with coefficients in A232624:

%C S(2*k, x) = Product_{2 <= d | (2*k+1)} C(d, x)*(-1)^deg(d)*C(d, -x), with deg(d) = A055034(d) the degree of C(d, x).

%C S(2*k+1, x) = Product_{2 <= d | 2*(k+1)} C(d, x) * Product_{3 <= 2*d + 1 | (k+1)} (-1)^(deg(2*d+1))*C(2*d+1, -x).

%C Note that (-1)^(deg(2*d+1)*C(2*d+1, -x)*C(2*d+1 ,x) pairs always appear.

%C The number of C factors of S(2*k, x), for k >= 0, is 2*(tau(2*k+1) - 1) = 2*(A099774(k+1) - 1) = 2*A095374(k), and for S(2*k+1, x), for k >= 0, it is tau(2*(k+1)) + tau_{odd}(k+1) - 2 = A302707(k), with tau(2*k+1) = A099774(k+1), tau(n) = A000005 and tau(2*(k+1)) = A099777(k+1).

%C For the reverse problem, the factorization of C polynomials into S polynomials, see A255237. (End)

%C The S polynomials with general initial conditions S(a,b;n,x) = x*S(a,b;n-1,x) - S(a,b;n-2,x), for n >= 1, with S(a,b;-1,x) = a and S(a,b;0,x) = b are S(a,b;n,x) = b*S(n, x) - a*S(n-1, x), for n >= -1. Recall that S(-2, x) = -1 and S(-1, x) = 0. The o.g.f. is G(a,b;z,x) = (b - a*z)/(1 - x*z + z^2). - _Wolfdieter Lang_, Oct 18 2019

%D G. H. Hardy, Ramanujan: twelve lectures on subjects suggested by his life and work, AMS Chelsea Publishing, Providence, Rhode Island, 2002, p. 164.

%D Max Koecher and Aloys Krieg, Elliptische Funktionen und Modulformen, 2. Auflage, Springer, 2007, p. 223.

%D Franz Lemmermeyer, Reciprocity Laws. From Euler to Eisenstein, Springer, 2000.

%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.

%D R. Vein and P. Dale, Determinants and Their Applications in Mathematical Physics, Springer, 1999.

%H T. D. Noe, <a href="/A049310/b049310.txt">Rows 0 to 100 of the triangle, flattened.</a>

%H Wolfdieter Lang, <a href="/A049310/a049310.pdf">First rows of the triangle.</a>

%H M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy], Table 22.8, p.797.

%H J. Allouche and G. Skordev, <a href="http://dx.doi.org/10.1016/S0012-365X(99)00195-8">Schur congruences, Carlitz sequences of polynomials and automaticity</a>, Discrete Mathematics, Vol. 214, Issue 1-3, 21 March 2000, p.21-49.

%H T. Amdeberhan, X. Chen, V. Moll, and B. Sagan, <a href="http://arxiv.org/abs/1306.6511">Generalized Fibonacci polynomials and Fibonomial coefficients</a>, arXiv preprint arXiv:1306.6511 [math.CO], 2013.

%H P. Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL12/Barry4/barry64.html">Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays</a>, JIS 12 (2009) 09.8.6

%H P. Barry, A. Hennessy, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL13/Barry5/barry96s.html">Meixner-Type Results for Riordan Arrays and Associated Integer Sequences</a>, J. Int. Seq. 13 (2010) # 10.9.4, section 5.

%H J. R. Dias, <a href="http://www.stkpula.hr/ccacaa/CCA-PDF/cca2004/v77-n1_n2/CCA_77_2004_325-330_dias.pdf"> Properties and relationships of conjugated polyenes having a reciprocal eigenvalue spectrum - dendralene and radialene hydrocarbons </a>, Croatica Chem. Acta, 77 (2004), 325-330. [p. 328].

%H S. R. Finch, P. Sebah and Z.-Q. Bai, <a href="http://arXiv.org/abs/0802.2654">Odd Entries in Pascal's Trinomial Triangle</a>, arXiv:0802.2654 [math.NT], 2008.

%H Aoife Hennessy, <a href="http://repository.wit.ie/1693">A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths</a>, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.

%H M. Ismail, <a href="http://arxiv.org/abs/math/0606743v1">One parameter generalizations of the Fibonacci and Lucas numbers</a>, arXiv preprint arXiv:0606743v1 [math.CA], 2006.

%H W. Lang, <a href="/A049310/a049310appl.pdf ">Chebyshev S-polynomials: ten applications.</a>

%H Wolfdieter Lang, <a href="http://arxiv.org/abs/1210.1018">The field Q(2cos(pi/n)), its Galois group and length ratios in the regular n-gon</a>, arXiv:1210.1018 [math.GR], 2012-2017.

%H R. Sazdanovic, <a href="http://www.math.toronto.edu/~drorbn/SK11/Sazdanovic.pdf ">A categorification of the polynomial ring</a>, slide presentation, 2011 (From Tom Copeland, Dec 27 2015)

%H P. Steinbach, <a href="http://www.jstor.org/stable/2691048">Golden fields: a case for the heptagon</a>, Math. Mag. 70 (1997), no. 1, 22-31.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/AdjacencyMatrix.html">Adjacency Matrix</a>, <a href="http://mathworld.wolfram.com/CharacteristicPolynomial.html">Characteristic Polynomial</a>, <a href="http://mathworld.wolfram.com/ChebyshevPolynomialoftheSecondKind.html">Chebyshev Polynomial of the Second Kind</a>, <a href="http://mathworld.wolfram.com/FibonacciPolynomial.html">Fibonacci Polynomial</a>, <a href="http://mathworld.wolfram.com/MatchingPolynomial.html">Matching Polynomial</a>, and <a href="http://mathworld.wolfram.com/PathGraph.html">Path Graph</a>

%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>

%H <a href="/index/Cor#core">Index entries for "core" sequences</a>

%F T(n,k) := 0 if n < k or n+k odd, else ((-1)^((n+k)/2+k))*binomial((n+k)/2, k); T(n, k) = -T(n-2, k)+T(n-1, k-1), T(n, -1) := 0 =: T(-1, k), T(0, 0)=1, T(n, k)= 0 if n < k or n+k odd; G.f. k-th column: (1 / (1 + x^2)^(k + 1)) * x^k. - _Michael Somos_, Jun 24 2002

%F T(n,k) = binomial((n+k)/2, (n-k)/2)*cos(Pi*(n-k)/2)*(1+(-1)^(n-k))/2. - _Paul Barry_, Aug 28 2005

%F Sum_{k=0..n} T(n,k)^2 = A051286(n). - _Philippe Deléham_, Nov 21 2005

%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 From _Wolfdieter Lang_, Nov 04 2011: (Start)

%F The Riordan A- and Z-sequences, given in a comment above, lead together to the recurrence:

%F T(n,k) = 0 if n < k, if k=0 then T(0,0)=1 and

%F T(n,0)= -Sum_{i=0..floor((n-1)/2)} C(i)*T(n-1,2*i+1), else T(n,k) = T(n-1,k-1) - Sum_{i=1..floor((n-k)/2)} C(i)*T(n-1,k-1+2*i), with the Catalan numbers C(n)=A000108(n).

%F (End)

%F The row polynomials satisfy also S(n,x) = 2*(T(n+2, x/2) - T(n, x/2))/(x^2-4) with the Chebyshev T-polynomials. Proof: Use the trace formula 2*T(n, x/2) = S(n, x) - S(n-2, x) (see the Dec 02 2010 comment above) and the S-recurrence several times. This is a formula which expresses the S- in terms of the T-polynomials. - _Wolfdieter Lang_, Aug 07 2014

%F From _Tom Copeland_, Dec 06 2015: (Start)

%F The non-vanishing, unsigned subdiagonals Diag_(2n) contain the elements D(n,k) = Sum_{j=0..k} D(n-1,j) = (k+1) (k+2) ... (k+n) / n! = binomial(n+k,n), so the o.g.f. for the subdiagonal is (1-x)^(-(n+1)). E.g., Diag_4 contains D(2,3) = D(1,0) + D(1,1) + D(1,2) + D(1,3) = 1 + 2 + 3 + 4 = 10 = binomial(5,2). Diag_4 is shifted A000217; Diag_6, shifted A000292: Diag_8, shifted A000332; and Diag_10, A000389.

%F The non-vanishing antidiagonals are signed rows of the Pascal triangle A007318.

%F For a reversed, unsigned version with the zeros removed, see A011973. (End)

%F The Boas-Buck recurrence (see a comment above) for the sequence of column k is: S(n, k) = ((k+1)/(n-k))*Sum_{p=0..n-1-k} (1 - (-1)^p)*(-1)^((p+1)/2) * S(n-1-p, k), for n > k >= 0 and input S(k, k) = 1. - _Wolfdieter Lang_, Aug 11 2017

%F The m-th row consecutive nonzero entries in order are (-1)^c*(c+b)!/c!b! with c = m/2, m/2-1, ..., 0 and b = m-2c if m is even and with c = (m-1)/2, (m-1)/2-1, ..., 0 with b = m-2c if m is odd. For the 8th row starting at a(36) the 5 consecutive nonzero entries in order are 1,-10,15,-7,1 given by c = 4,3,2,1,0 and b = 0,2,4,6,8. - _Richard Turk_, Aug 20 2017

%e The triangle T(n, k) begins

%e n\k 0 1 2 3 4 5 6 7 8 9 10 11

%e 0: 1

%e 1: 0 1

%e 2: -1 0 1

%e 3: 0 -2 0 1

%e 4: 1 0 -3 0 1

%e 5: 0 3 0 -4 0 1

%e 6: -1 0 6 0 -5 0 1

%e 7: 0 -4 0 10 0 -6 0 1

%e 8: 1 0 -10 0 15 0 -7 0 1

%e 9: 0 5 0 -20 0 21 0 -8 0 1

%e 10: -1 0 15 0 -35 0 28 0 -9 0 1

%e 11: 0 -6 0 35 0 -56 0 36 0 -10 0 1

%e ... Reformatted and extended by _Wolfdieter Lang_, Oct 24 2012

%e For more rows see the link.

%e E.g., fourth row {0,-2,0,1} corresponds to polynomial S(3,x)= -2*x + x^3.

%e From _Wolfdieter Lang_, Jul 12 2011: (Start)

%e Zeros of S(3,x) with rho(4)= 2*cos(Pi/4) = sqrt(2):

%e +- t(1,sqrt(2)) = +- sqrt(2) and

%e +- t(2,sqrt(2)) = +- 0.

%e Factorization of S(3,x) in terms of Psi polynomials:

%e S(3,x) = (2^3)*Psi(4,x/2)*Psi(8,x/2) = x*(x^2-2).

%e (End)

%e From _Wolfdieter Lang_, Nov 04 2011: (Start)

%e A- and Z- sequence recurrence:

%e T(4,0) = - (C(0)*T(3,1) + C(1)*T(3,3)) = -(-2 + 1) = +1,

%e T(5,3) = -3 - 1*1 = -4.

%e (End)

%e Boas-Buck recurrence for column k = 2, n = 6: S(6, 2) = (3/4)*(0 - 2* S(4 ,2) + 0 + 2*S(2, 2)) = (3/4)*(-2*(-3) + 2) = 6. - _Wolfdieter Lang_, Aug 11 2017

%e From _Wolfdieter Lang_, Apr 12 2018: (Start)

%e Factorization into C polynomials (see the Apr 12 2018 comment):

%e S(4, x) = 1 - 3*x^2 + x^4 = (-1 + x + x^2)*(-1 - x + x^2) = (-C(5, -x)) * C(5, x); the number of factors is 2 = 2*A095374(2).

%e S(5, x) = 3*x - 4*x^3 + x^5 = x*(-1 + x)*(1 + x)*(-3 + x^2) = C(2, x)*C(3, x)*(-C(3, -x))*C(6, x); the number of factors is 4 = A302707(2). (End)

%p A049310 := proc(n,k): binomial((n+k)/2,(n-k)/2)*cos(Pi*(n-k)/2)*(1+(-1)^(n-k))/2 end: seq(seq(A049310(n,k), k=0..n),n=0..11); # _Johannes W. Meijer_, Aug 08 2011

%t t[n_, k_] /; EvenQ[n+k] = ((-1)^((n+k)/2+k))*Binomial[(n+k)/2, k]; t[n_, k_] /; OddQ[n+k] = 0; Flatten[Table[t[n, k], {n, 0, 12}, {k, 0, n}]][[;; 86]] (* _Jean-François Alcover_, Jul 05 2011 *)

%t Table[Coefficient[(-I)^n Fibonacci[n + 1, - I x], x, k], {n, 0, 10}, {k, 0, n}] //Flatten (* _Clark Kimberling_, Aug 02 2011; corrected by _Eric W. Weisstein_, Apr 06 2017 *)

%t CoefficientList[ChebyshevU[Range[0, 10], -x/2], x] // Flatten (* _Eric W. Weisstein_, Apr 06 2017 *)

%t CoefficientList[Table[(-I)^n Fibonacci[n + 1, -I x], {n, 0, 10}], x] // Flatten (* _Eric W. Weisstein_, Apr 06 2017 *)

%o (PARI) {T(n, k) = if( k<0 || k>n || (n + k)%2, 0, (-1)^((n + k)/2 + k) * binomial((n + k)/2, k))} /* _Michael Somos_, Jun 24 2002 */

%o (Sage)

%o @CachedFunction

%o def A049310(n,k):

%o if n< 0: return 0

%o if n==0: return 1 if k == 0 else 0

%o return A049310(n-1,k-1) - A049310(n-2,k)

%o for n in (0..9): [A049310(n,k) for k in (0..n)] # _Peter Luschny_, Nov 20 2012

%Y Cf. A000005, A000217, A000292, A000332, A000389, A001227, A007318, A008611, A008615, A010892, A011973, A053112 (without zeros), A053117, A053119 (reflection), A053121 (inverse triangle), A055034, A097610, A099774, A099777, A100258, A112552 (first column clipped), A168561 (absolute values), A187360. A194960, A232624, A255237.

%Y Triangles of coefficients of Chebyshev's S(n,x+k) for k = 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5: A207824, A207823, A125662, A078812, A101950, A049310, A104562, A053122, A207815, A159764, A123967.

%K easy,nice,sign,tabl,core

%O 0,8

%A _Wolfdieter Lang_

%E _M. Sinan Kul_'s Jan 30 2016 comment edited by _Wolfdieter Lang_, Jan 31 2016 and Feb 01 2016

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Last modified November 12 14:43 EST 2019. Contains 329058 sequences. (Running on oeis4.)