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From Petros Hadjicostas, Jun 10 2020: (Start)

Swamy's (1966) inequality states that F[n]^2 <= G[n] for all real x and all integers n >= 3. Because F[n] >= 1 for all real x >= 1, we get F[n] <= G[n] for all integers n >= 3 and all real x >= 1.

Row n >= 3 of this irregular table gives the coefficients of the polynomial G[n] - F[n] (with exponents in increasing order). The degree of G[n] - F[n] is 2*n - 2, so row n >= 3 contains 2*n - 1 terms.

Guilfoyle (1967) notes that F[n] = det(A_n), where A_n is the (n-1) X (n-1) matrix [[x, -1, 0, 0, ..., 0, 0, 0], [1, x, -1, 0, ..., 0, 0, 0], [0, 1, x, -1, ..., 0, 0, 0], ..., [0, 0, 0, 0, ..., 1, x, -1], [0, 0, 0, 0, ..., 0, 1, x]], and Swamy's original inequality follows from Hadamard's inequality.

Koshy (2019) writes Swamy's original inequality in the form x^(n-3)*F[n]^2 <= F[3]^2*F[4]^(n-3) for x >= 1, and gives a counterpart inequality for Lucas polynomials. Notice, however, that the original form of Swamy's inequality is true for all real x. (End)

Swamy's (1966) inequality states that F[n]^2 <= G[n] for all real x and all integers n >= 3.

Row n >= 3 of this irregular table gives the coefficients of the even powers of the polynomial G[n] - F[n]^2 (with exponents in increasing order). The coefficients of the odd powers are zero, and they are thus omitted. The degree of G[n] - F[n]^2 is 2*n - 6, so row n >= 3 contains n - 2 terms.

To prove that the degree of G[n] - F[n]^2 is 2*n - 6, note that the first few terms of G[n] are x^(2*n-2) + 2*(n-2)*x^(2*n-4) + (2*n^2 - 10*n + 13)*x^(2*n-6) + ... while the first few terms of F[n]^2 are x^(2*n-2) + 2*(n-2)*x^(2*n-4) + (2*n^2 - 11*n + 16)*x^(2*n-6) + ..., so the leading term of the polynomial G[n] - F[n]^2 is (n-3)*x^(2*n-6).

Guilfoyle (1967) notes that F[n] = det(A_n), where A_n is the (n-1) X (n-1) matrix [[x, -1, 0, 0, ..., 0, 0, 0], [1, x, -1, 0, ..., 0, 0, 0], [0, 1, x, -1, ..., 0, 0, 0], ..., [0, 0, 0, 0, ..., 1, x, -1], [0, 0, 0, 0, ..., 0, 1, x]], and Swamy's original inequality follows from Hadamard's inequality.

Koshy (2019) writes Swamy's original inequality in the form x^(n-3)*F[n]^2 <= F[3]^2*F[4]^(n-3) for x >= 1, and gives a counterpart inequality for Lucas polynomials. Notice, however, that the original form of Swamy's inequality is true for all real x.

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).

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

Alternating row sums A049347(n) = S(n,-1) = periodic(1,-1,0). - Wolfdieter Lang, Nov 04 2011

S(n,x) is the characteristic polynomial of the adjacency matrix of the n-path. - Michael Somos, Jun 24 2002

S(n,x) is also the matching polynomial of the n-path. - Eric W. Weisstein, Apr 10 2017

|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

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

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

From Wolfdieter Lang, Jul 12 2011: (Start)

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.

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

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.

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:

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

(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]

(End)

The discriminants of the S(n,x) polynomials are found in A127670. - Wolfdieter Lang, Aug 03 2011

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

The determinant of the N x N matrix S(N,[x[1], ..., 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[1], ..., 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[1], ...., x[N]]) if and only if the x[j], j = 1..N, are pairwise distinct. - Wolfdieter Lang, Aug 26 2013

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

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

From M. Sinan Kul, Jan 30 2016; edited by Wolfdieter Lang, Jan 31 2016 and Feb 01 2016: (Start)

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.

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.

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

Then we have with the present T triangle

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

A051286(n) = Sum_{i=0..n} T(n,i)^2 (see the Philippe Deléham, Nov 21 2005 formula),

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

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

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

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.

(End)

For more on the Diophantine equation presented by Kul, see the Ismail paper. - Tom Copeland, Jan 31 2016

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

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 (Instances of the bivariate o.g.f. for this entry are on pp. 5 and 18 of Sunada. - Tom Copeland, Jan 18 2021)

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

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

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

From Wolfdieter Lang, May 08 2017: (Start)

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, 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.

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)

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

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

From Wolfdieter Lang, Apr 12 2018: (Start)

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:

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).

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).

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

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).

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

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

Also the convolution triangle of A101455. - Peter Luschny, Oct 06 2022

From Wolfdieter Lang, Apr 26 2023: (Start)

Multi-section of S-polynomials: S(m*n+k, x) = S(m+k, x)*S(n-1, R(m, x)) - S(k, x)*S(n-2, R(m, x)), with R(n, x) = S(n, x) - S(n-2, x) (see A127672), S(-2, x) = -1, and S(-1, x) = 0, for n >= 0, m >= 1, and k = 0, 1, ..., m-1.

O.g.f. of {S(m*n+k, y)}_{n>=0}: G(m,k,y,x) = (S(k, y) - (S(k, y)*R(m, y) - S(m+k, y))*x)/(1 - R(m,y)*x + x^2).

See eqs. (40) and (49), with r = x or y and s =-1, of the G. Detlefs and W. Lang link at A034807. (End)

S(n, x) for complex n and complex x: S(n, x) = ((-i/2)/sqrt(1 - (x/2)^2))*(q(x/2)*exp(+n*log(q(x/2))) - (1/q(x/2))*exp(-n*log(q(x/2)))), with q(x) = x + sqrt(1 - x^2)*i. Here log(z) = |z| + Arg(z)*i, with Arg(z) from [-Pi,+Pi) (principal branch). This satisfies the recurrence relation for S because it is derived from the Binet - de Moivre formula for S. Examples: S(n/m, 0) = cos((n/m)*Pi/4), for n >= 0 and m >= 1. S(n*i, 0) = (1/2)*(1 + exp(n*Pi))*exp(-(n/2)*Pi), for n >= 0. S(1+i, 2+i) = 0.6397424847... + 1.0355669490...*i. Thanks to Roberto Alfano for asking a question leading to this formula. - Wolfdieter Lang, Jun 05 2023

Lim_{n->oo} S(n, x)/S(n-1, x) = r(x) = (x - sqrt(x^2 -4))/2, for |x| >= 2. For x = +-2, this limit is +-1. - Wolfdieter Lang, Nov 15 2023

Take a(0)=-2 instead of 1. The recurrence begins immediately (at the third instead of the fourth polynomial). Companion: A192011(n). - Paul Curtz, Sep 20 2011

Row sums 0,1,1,2,3,5,... are the Fibonacci numbers, A000045.

Note that the coefficients are given in decreasing order. - M. F. Hasler, Dec 07 2011

Essentially a mirror image of A168561. - Philippe Deléham, Dec 08 2013

There are several versions of Lucas and Fibonacci polynomials in this database. Our naming follows the convention of calling polynomials after the values of the polynomials at x = 1. This assumes a regular sequence of polynomials, that is, a sequence of polynomials where degree(p(n)) = n. This view makes the coefficients of the polynomials (the terms of a row) a refinement of the values at the unity.

A remarkable property of the polynomials under consideration is that they are dual in this respect. This means they give the Lucas numbers at x = 1 and the Fibonacci numbers at x = -1 (except for the sign). See the example section.

The Pell numbers and the dual Pell numbers are also values of the polynomials, at the points x = -1/2 and x = 1/2 (up to the normalization factor 2^n). This suggests a harmonized terminology: To call 2^n*P(n, -1/2) = 1, 0, 1, 2, 5, ... the Pell numbers (A000129) and 2^n*P(n, 1/2) = 1, 4, 9, 22, ... the dual Pell numbers (A048654).

Based on our naming convention one could call A162515 (without the prepended 0) the Fibonacci polynomials. In the definition above only the initial values would change to: T(n, k) = k + 1 for k < 1. To extend this line of thought we introduce A374438 as the third triangle of this family.

The triangle is closely related to the qStirling2 numbers at q = -1. For the definition of these numbers see A333143. This relates the triangle to A065941 and A103631.

Member of the family of Fibonacci polynomials (A011973, A162515, ...) and Chebyshev polynomials (A053119).

a(n)=

-1, :-1,

3, 0, :3*x,

3, 0, 1, :3*x^2+1,

3, 0, -2, 0, :3*x^3-2*x,

3, 0, -5, 0, -1,

3, 0, -8, 0, 1, 0

3, 0, -11, 0, 6, 0, 1.

Row sum=period 6:repeat -1, 3, 4, 1, -3, 4=-A117378(n)=A117378(n+3).

In the notation of A195673, this defines polynomials P(n,x,p=-3,q=2), where p and q are the values of the constant and linear order for n=0 and 1.

Row sums -- the value P(n,1,-3,2) of the polynomial -- are A130848(n+5).

For general seed values in the two top rows of the triangle, the recurrence T(n,k) = T(n-1,k) - T(n-2,k-2) defines the triangle

p;

q, 0;

q, 0, -p;

q, 0, -p-q, 0;

q, 0, -p-2q, 0, p;

q, 0, -p-3q, 0, 2p+q, 0;

and a companion triangle by adding 1 to both seed values:

p+1;

q+1, 0;

q+1, 0, -p-1;

q+1, 0, -p-q-2, 0;

q+1, 0, -p-2q-3, 0, p+1;

q+1, 0, -p-3q-4, 0, 2p+q+3, 0;

The point-by-point difference between two companions is P(n,x,p+1,q+1) - P(n,x,p,q) = S(n,x) as represented (with increasing exponents) by A053119.

Examples of such triangles are A053119 (p=q=1), A192575 (p=1, q=2),

A162514 (p=2, q=1, up to a sign factor), A192011 (p=-1, q=2), A135929 (p=-2, q=1, apart from a irregular leading T(0,0)).

Obviously T(n,k) = 0 for all odd k.

Conjecture: The polynomials p(n,x) = sum_{k=0..n} T(n,k)*x^(n-k) based on this simple recurrence for other initial constant values of T(0,0)=p and T(1,0)=q are related to the S-polynomials of A053119: p(n,x,p+1,q+1)-p(n,x,p,q) = S(n,x).