A217479 -id:A217479 - cp's OEIS frontend

A217478 Triangle of coefficients of polynomials providing the second term of the numerator for the generating function for odd powers (2*m+1) of Chebyshev S-polynomials. The present polynomials are called P(m;1,x^2).

-2, 3, -4, -4, 10, -6, 5, -20, 21, -8, -6, 35, -56, 36, -10, 7, -56, 126, -120, 55, -12, -8, 84, -252, 330, -220, 78, -14, 9, -120, 462, -792, 715, -364, 105, -16, -10, 165, -792, 1716, -2002, 1365, -560, 136, -18, 11, -220, 1287, -3432, 5005, -4368, 2380, -816, 171, -20

Offset: 1

Wolfdieter Lang, Nov 14 2012

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).

			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,...

Cf. A049310, A217479 (P(m;2,x^2)).

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).

cp's OEIS Frontend

A217478 Triangle of coefficients of polynomials providing the second term of the numerator for the generating function for odd powers (2*m+1) of Chebyshev S-polynomials. The present polynomials are called P(m;1,x^2).

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