cp's OEIS Frontend

According to Barequet-Barequet-Rote, p. 261, the value DX(7, 6) = 134209 given by W. F. Lunnon is incorrect; it should be 153664. - Alexander Knapp, May 13 2013

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

a(n+1) is the number of labeled incomplete ternary trees on n vertices in which each left child has a larger label than its parent. - Brian Drake, Jul 28 2008

Put a(0) = 1. For n > 0, let x(n,k) = 2*cos((2*k-1)*Pi/(2*n+1)), k=1..n. Define the recurrences S(n;0,x(n,k)) = 1, S(n;1,x(n,k)) = x(n,k), S(n;r,x(n,k)) = x(n,k)*S(n;r-1,x(n,k)) - S(n;r-2,x(n,k)), r > 1 an integer, k=1..n. CONJECTURE: For n > 0, a(n) = Product_{k=1..n} (Sum_{m=0..n-1} S(n;2*m,x(n,k))^2). - L. Edson Jeffery, Sep 11 2013

From Wolfdieter Lang, Dec 16 2013: (Start)

Discriminants of the first difference of Chebyshev S-polynomials.

The coefficient table for the first difference polynomials P(n, x) = S(n, x) - S(n-1, x), n >= 0, S(-1, x) = 0, with the Chebyshev S polynomials (see A049310), is given in A130777.

For the discriminant of a polynomial in terms of the square of a determinant of a Vandermonde matrix build from the zeros of the polynomial see, e.g., A127670.

For the proof that D(n) := discriminant(P(n,x)) = (2*n + 1)^(n - 1), n >= 1, use the formula given e.g., in the Rivlin reference, p. 218, Theorem 5.13, eq. (5.3), namely D(n) = (-1)^(n*(n-1)/2)*Product_{j=1..n} P'(n, x(n,j)), with the zeros x(n,j) = -2*cos(2*Pi*j/(2*n+1)) of P(n, x) (see A130777). P'(n, x(n,j)) = (2*n+1)*P(n-1, x(n,j))/(2*sin(Pi*j/(2*n+1))*2*cos(Pi*j/(2*n+1)))^2. P(n-1, x(n,j)) = (-1)^(n+j)*2*cos(Pi*j/(2*n+1)). Product_{j=1..n} 2*sin(Pi*j/(2*n+1)) = 2*n+1 (see the Oct 10 2013 formula in A005408. Product_{j=1..n} 2*cos(Pi*j/(2*n+1)) = 1, because S(2*n, 0) = (-1)^n.

(End)

a(n) is the number of labeled 2-trees with n+2 vertices, rooted at a given edge. - Nikos Apostolakis, Nov 30 2018

a(n) is also the number of 2-trees with n labeled triangles and with a distinguished oriented edge. - Nikos Apostolakis, Dec 14 2018

According to Barequet-Barequet-Rote, p. 261, the value DX(7, 6) = 134209 given by W. F. Lunnon is incorrect; it should be 153664, see A127670. - Alexander Knapp, May 13 2013

The array of coefficients of the (monic) Chebyshev C-polynomials is found under A127672 (where they are called, in analogy to the S-polynomials, R-polynomials).

See A127670 for the formula in terms of the square of a Vandermonde determinant, where now the zeros are xn[j]:=2*cos(Pi*(2*j+1)/(2*n)), j=0,..,n-1.

One could add a(0)=0 for the discriminant of C(0,x)=2.

Except for sign, a(n) is the field discriminant of 2^(1/n); see the Mathematica program. - Clark Kimberling, Aug 03 2015

Discriminant of Chebyshev polynomial T_n (x) of first kind.