OFFSET
1,1
COMMENTS
The o.g.f. for S(n,x)^(2*m+1), m >= 0, with Chebyshev's S-polynomials (see A049310), is
G(m;z,x) := sum(S(n,x)^(2*m+1)*z^n, n=0..infinity)= sum(T(m,k)*S(2*k,x)/(1-z*x*tau(k,x)+z^2), k=0..m)/(x^2-4)^m, with the signed Riordan triangle
T(m,k) = binomial(2*m+1,m-k)*(-1)^(m-k) = A113187(m,k),
and tau(k,x):= R(2*k+1,x)/x with R the monic integer Chebyshev T-polynomials (see A127672). The proof uses the de Moivre-Binet formula: S(n,x) = (q^(n+1) - 1/q^(n+1))/(q-1/q), with q:=(x+sqrt(x^2-4))/2, and the one for tau(n,x) = (q^(2*n+1) + 1/q^(2*n+1))/(q+1/q). This can be written as G(m;z,x) = Z(m;z,x)/N(m;z,x) with N(m;z,x) = product((1+z^2) - z*x*tau(k,x),k=0..m), and Z(m;z,x) = sum((1+z^2)^(m-l)*(-z*x)^l*P(m;l,x^2),l=0..m), where P(m;l,x^2) = sum(T(m,k)*S(2*k,x)*sigma(m;k,l,x^2), k=0..m)/(x^2-4)^m, with sigma(m;k,l,x^2) the elementary symmetric function of a product of l factors from tau(j,x), for j=0..m, with tau(k,x) missing. E.g., sigma(3;1,2,x^2) = tau(0,x)*tau(2,x) + tau(0,x)*tau(3,x) + tau(2,x)*tau(3,x), (tau(1,x) is missing). P(m;0,x^2) = 1 due to the identity Id(0;m,x^2) := sum(T(m,k)*S(2*k,x), k=0..m) = (x^2-4)^m (proof by using the de Moivre-Binet formula and the formula mentioned in a comment on A113187). Also the other P(m;l,x^2) turn out to be polynomials in x^2.
The present triangle a(m,k) provides the P(m;1,x^2) coefficients: P(m;1,x^2) = sum(a(m,k)*(x^2)^k, k=0..m-1), m>=1.
Using inclusion-exclusion one can write (x^2-4)^m*P(m;1,x^2) = sum(T(m,k)*S(2*k,x)*(sum(tau(k,x),k=0..m) - tau(k,x)), k=0..m) = sum(tau(k,x),k=0..m)*(x^2-4)^m - sum(T(m,k)*S(2*k,x)* tau(k,x),k=0..m), using the mentioned identity Id(0;m,x^2). In the second term S(2*k,x)*tau(k,x) = S(4*k+1,x)/x (de Moivre-Binet formulas for S and tau). This leads to the l=1 identity Id(1;m,x^2) := sum(T(m,k)*S(4*k+1,x)/x,k=0..m) =
((x^2-4)*x^2)^m, using again de Moivre-Binet and the identity
given in a comment on A113187. Therefore, after dividing by (x^2-4)^m, P(m;1,x^2) = sum(tau(k,x),k=0..m) - x^(2*m).
FORMULA
a(m,k) = [x^(2*k)] P(m;1,x^2) = [x^(2*k)](sum(tau(k,x),k=0..m) - x^(2*m)) (see the comment above), m>=1, k = 0..m-1.
a(m,k) = (-1)^(m-k)*binomial(m+k+1,2*k+1). For the proof one uses the identity sum(tau(j,x),j=0..m) = S(m,x^2-2) which holds by comparing the o.g.f.s of both sides (see a Nov 13 2012 comment on Riordan A053122 where tau is called r).
EXAMPLE
The triangle a(m,k) begins:
m\k 0 1 2 3 4 5 6 7 8 9 ...
1: -2
2: 3 -4
3: -4 10 -6
4: 5 -20 21 -8
5: -6 35 -56 36 -10
6: 7 -56 126 -120 55 -12
7: 8 84 -252 330 -220 78 -14
8: 9 -120 462 -792 715 -364 105 -16
9: -10 165 -792 1716 -2002 1365 -560 136 -18
10: 11 -220 1287 -3432 5005 -4368 2380 -816 171 -20
...
P(2;1,x^2) = 3 - 4*x^2, appears in the second term of the numerator of the o.g.f. for S(n,x)^5 which is Z(2;z,x) = (1+z^2)^2 + (1+z^2)*(-x*z)*(3-4*x^2) + ((-x*z)^2)*2*(-4 +3*x^2). The last term is taken from row m=2 of A217479. The denominator is N(2;z,x) = product((1+z^2)-z*x*tau(k,x), k=0..2). This checks with [1,x^5,-1+5*x^2-10*x^4+10*x^6-5*x^8
+x^10,-32*x^5+80*x^7-80*x^9+40*x^11-10* x^13+x^15,...] for S(n,x)^5, n=0,1,2,3,...
CROSSREFS
KEYWORD
AUTHOR
Wolfdieter Lang, Nov 14 2012
STATUS
approved