cp's OEIS Frontend

Riordan array ((1+x)/(1-x)^2, x/(1-x)^2). Row sums are A002878. Diagonal sums are A003945. Inverse is A113187. An interesting factorization is (1/(1-x), x/(1-x))(1+2*x, x*(1+x)). - Paul Barry, Oct 17 2005

Central coefficients of rows with odd numbers of term are A052227.

From Wolfdieter Lang, Jun 26 2011: (Start)

T(k,s) appears as T_s(k) in the Knuth reference, p. 285.

This triangle is related to triangle A156308(n,m), appearing in this reference as U_m(n) on p. 285, by T(k,s) - T(k-1,s) = A156308(k,s), k>=s>=1 (identity on p. 286). T(k,s) = A156308(k+1,s+1) - A156308(k,s+1), k>=s>=0 (identity on p. 286).

(End)

A111125 is jointly generated with A208513 as an array of coefficients of polynomials v(n,x): initially, u(1,x)= v(1,x)= 1; for n>1, u(n,x)= u(n-1,x) +x*(x+1)*v(n-1) and v(n,x)= u(n-1,x) +x*v(n-1,x) +1. See the Mathematica section. The columns of A111125 are identical to those of A208508. Here, however, the alternating row sums are periodic (with period 1,2,1,-1,-2,-1). - Clark Kimberling, Feb 28 2012

This triangle T(k,s) (with signs and columns scaled with powers of 5) appears in the expansion of Fibonacci numbers F=A000045 with multiples of odd numbers as indices in terms of odd powers of F-numbers. See the Jennings reference, p. 108, Theorem 1. Quoted as Lemma 3 in the Ozeki reference given in A111418. The formula is: F_{(2*k+1)*n} = Sum_{s=0..k} ( T(k,s)*(-1)^((k+s)*n)*5^s*F_{n}^(2*s+1) ), k >= 0, n >= 0. - Wolfdieter Lang, Aug 24 2012

From Wolfdieter Lang, Oct 18 2012: (Start)

This triangle T(k,s) appears in the formula x^(2*k+1) - x^(-(2*k+1)) = Sum_{s=0..k} ( T(k,s)*(x-x^(-1))^(2*s+1) ), k>=0. Prove the inverse formula (due to the Riordan property this will suffice) with the binomial theorem. Motivated to look into this by the quoted paper of Wang and Zhang, eq. (1.4).

Alternating row sums are A057079.

The Z-sequence of this Riordan array is A217477, and the A-sequence is (-1)^n*A115141(n). For the notion of A- and Z-sequences for Riordan triangles see a W. Lang link under A006232. (End)

The signed triangle ((-1)^(k-s))*T(k,s) gives the coefficients of (x^2)^s of the polynomials C(2*k+1,x)/x, with C the monic integer Chebyshev T-polynomials whose coefficients are given in A127672 (C is there called R). See the odd numbered rows there. This signed triangle is the Riordan array ((1-x)/(1+x)^2, x/(1+x)^2). Proof by comparing the o.g.f. of the row polynomials where x is replaced by x^2 with the odd part of the bisection of the o.g.f. for C(n,x)/x. - Wolfdieter Lang, Oct 23 2012

From Wolfdieter Lang, Oct 04 2013: (Start)

The signed triangle S(k,s) := ((-1)^(k-s))*T(k,s) (see the preceding comment) is used to express in a (4*(k+1))-gon the length ratio s(4*(k+1)) = 2*sin(Pi/4*(k+1)) = 2*cos((2*k+1)*Pi/(4*(k+1))) of a side/radius as a polynomial in rho(4*(k+1)) = 2*cos(Pi/4*(k+1)), the length ratio (smallest diagonal)/side:

s(4*(k+1)) = Sum_{s=0..k} ( S(k,s)*rho(4*(k+1))^(2*s+1) ).

This is to be computed modulo C(4*(k+1), rho(4*(k+1)) = 0, the minimal polynomial (see A187360) in order to obtain s(4*(k+1)) as an integer in the algebraic number field Q(rho(4*(k+1))) of degree delta(4*(k+1)) (see A055034). Thanks go to Seppo Mustonen for asking me to look into the problem of the square of the total length in a regular n-gon, where this formula is used in the even n case. See A127677 for the formula in the (4*k+2)-gon. (End)

From Wolfdieter Lang, Aug 14 2014: (Start)

The row polynomials for the signed triangle (see the Oct 23 2012 comment above), call them todd(k,x) = Sum_{s=0..k} ( (-1)^(k-s)*T(k,s)*x^s ) = S(k, x-2) - S(k-1, x-2), k >= 0, with the Chebyshev S-polynomials (see their coefficient triangle (A049310) and S(-1, x) = 0), satisfy the recurrence todd(k, x) = (-1)^(k-1)*((x-4)/2)*todd(k-1, 4-x) + ((x-2)/2)*todd(k-1, x), k >= 1, todd(0, x) = 1. From the Aug 03 2014 comment on A130777.

This leads to a recurrence for the signed triangle, call it S(k,s) as in the Oct 04 2013 comment: S(k,s) = (1/2)*(1 + (-1)^(k-s))*S(k-1,s-1) + (2*(s+1)*(-1)^(k-s) - 1)*S(k-1,s) + (1/2)*(-1)^(k-s)*Sum_{j=0..k-s-2} ( binomial(j+s+2,s)*4^(j+2)* S(k-1, s+1+j) ) for k >= s >= 1, and S(k,s) = 0 if k < s and S(k,0) = (-1)^k*(2*k+1). Note that the recurrence derived from the Riordan A-sequence A115141 is similar but has simpler coefficients: S(k,s) = sum(A115141(j)*S(k-1,s-1+j), j=0..k-s), k >= s >=1.

(End)

From Tom Copeland, Nov 07 2015: (Start)

Rephrasing notes here: Append an initial column of zeros, except for a 1 at the top, to A111125 here. Then the partial sums of the columns of this modified entry are contained in A208513. Append an initial row of zeros to A208513. Then the difference of consecutive pairs of rows of the modified A208513 generates the modified A111125. Cf. A034807 and A127677.

For relations among the characteristic polynomials of Cartan matrices of the Coxeter root groups, Chebyshev polynomials, cyclotomic polynomials, and the polynomials of this entry, see Damianou (p. 20 and 21) and Damianou and Evripidou (p. 7).

As suggested by the equations on p. 7 of Damianou and Evripidou, the signed row polynomials of this entry are given by (p(n,x))^2 = (A(2*n+1, x) + 2)/x = (F(2*n+1, (2-x), 1, 0, 0, ... ) + 2)/x = F(2*n+1, -x, 2*x, -3*x, ..., (-1)^n n*x)/x = -F(2*n+1, x, 2*x, 3*x, ..., n*x)/x, where A(n,x) are the polynomials of A127677 and F(n, ...) are the Faber polynomials of A263196. Cf. A127672 and A127677.

(End)

The row polynomials P(k, x) of the signed triangle S(k, s) = ((-1)^(k-s))*T(k, s) are given from the row polynomials R(2*k+1, x) of triangle A127672 by

P(k, x) = R(2*k+1, sqrt(x))/sqrt(x). - Wolfdieter Lang, May 02 2021

Let b(n,x) = Sum_{k = 0..n} binomial(n+k,2*k)*x^k denote the Morgan-Voyce polynomials of A085478. This triangle lists the coefficients (in ascending powers of x) of the related polynomial sequence R(n,x) := (1/2)*b(n,2*x) + 1/2. Several sequences already in the database are of the form (R(n,x))n>=0 for a fixed value of x. These include A101265 (x = 1), A011900 (x = 2), A182432 (x = 3), A054318 (x = 4) as well as signed versions of A133872 (x = -1), A109613(x = -2), A146983 (x = -3) and A084159 (x = -4).

The polynomials R(n,x) factorize in the ring Z[x] as R(n,x) = P(n,x)*P(n+1,x) for n >= 1: explicitly, P(2*n,x) = 1/2*(b(2*n,2*x) + 1)/b(n,2*x) and P(2*n+1,x) = b(n,2*x). The coefficients of P(n,x) occur in several tables in the database, although without the connection to the Morgan-Voyce polynomials being noted - see A211956 for more details. In terms of T(n,x), the Chebyshev polynomials of the first kind, we have P(2*n,x) = T(2*n,u) and P(2*n+1,x) = 1/u * T(2*n+1,u), where u = sqrt((x+2)/2). Hence R(n,x) = 1/u * T(n,u) * T(n+1,u).

Let p(n)=p(n,x) be the characteristic polynomial of the n-th principal submatrix. The zeros of p(n) are real, and they interlace the zeros of p(n+1). See A202605 and A204016 for guides to related sequences.

a(0)=1 by convention. - Philippe Deléham, Nov 17 2013

The n roots of the n-th polynomial are 1/(1+cos((2*k-1)*Pi/(2*n))) for k = 1..n. See my pdf in the link section for the proof. - Jianing Song, Dec 01 2023

The row generating polynomials R(n,x) of A211955 factorize in the ring Z[x] as R(n,x) = P(n,x)*P(n+1,x) for n >= 1: explicitly, P(2*n,x) = 1/2*(b(2*n,2*x) + 1)/b(n,2*x) and P(2*n+1,x) = b(n,2*x), where b(n,x) := Sum_{k = 0..n} binomial(n+k,2*k)*x^k are the Morgan-Voyce polynomials of A085478. This triangle lists the coefficients in ascending powers of x of the polynomials P(n,x).

The odd numbered rows of the present triangle produce triangle A123519; the even numbered row entries are recorded separately in A211957 and appear to equal the unsigned and row reversed form of A204021. The even numbered rows with a factor of 2^(k-1) removed from the k-th column entries produce triangle A208513.