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.
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
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).
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
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)
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)
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
