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A244422 Quasi-Riordan triangle ((2-z)/(1-z), -z^2/(1-z)). Row reversed monic Chebyshev T-polynomials without vanishing columns.

%I A244422 #16 Feb 26 2025 02:31:51
%S A244422 2,1,0,1,-2,0,1,-3,0,0,1,-4,2,0,0,1,-5,5,0,0,0,1,-6,9,-2,0,0,0,1,-7,
%T A244422 14,-7,0,0,0,0,1,-8,20,-16,2,0,0,0,0,1,-9,27,-30,9,0,0,0,0,0,1,-10,35,
%U A244422 -50,25,-2,0,0,0,0,0,1,-11,44,-77,55,-11,0,0,0,0,0,0,1,-12,54,-112,105,-36,2,0,0,0,0,0,0
%N A244422 Quasi-Riordan triangle ((2-z)/(1-z), -z^2/(1-z)). Row reversed monic Chebyshev T-polynomials without vanishing columns.
%C A244422 This is a signed version of the triangle A061896.
%C A244422 The coefficient table for the monic Chebyshev polynomials of the first kind R(n, x) = 2*T(n, x/2) is given in A127672. For the T-polynomials see A053120. The present table is obtained from the row reversed coefficient table A127672 by deleting all odd numbered columns which have only zeros, and appending in the rows numbered n >= 1 zeros in order to obtain a triangle. This becomes the quasi-Riordan triangle T = ((2-z)/(1-z), -z^2/(1-z)). This means that the o.g.f. of the row polynomials Rrev(n, x) := sqrt(x)^n*R(n, 1/sqrt(x)) = Sum_{k=0..n} T(n, k)*x^k have o.g.f. (2-z)/(1 - z + x*z^2) like for ordinary Riordan triangles. However this is not a Riordan triangle (or lower triangular infinite dimensional matrix) in the usual sense because it is not invertible. Therefore, this lower triangular matrix is not a member of the Riordan group.
%C A244422 The row sums give repeat(2,1,-1,-2,-1) which is A057079(n+1), n >= 0. The alternating row sums give the Lucas numbers A000032.
%H A244422 Wolfdieter Lang, <a href="/A244422/a244422.pdf">First rows of the triangle.</a>
%F A244422 T(n,k) = [x^k] Rrev(n, x), k=0, 1, ..., n, with the row polynomials Rrev(n, x) = sqrt(x)^n*R(n,1/sqrt(x)), with R(n, x) given in A127672 (monic Chebyshev polynomials of the first kind).
%F A244422 O.g.f. row polynomials Rrev(n,x) = Sum_{k=0..n} T(n,k)*x^k: (2-z)/(1 - z + x*z^2) (quasi-Riordan).
%F A244422 O.g.f. for column number k entries with leading zeros: ((2-x)/(1-x))*(-x^2/(1-x))^k, k > = 0. See A054977, -A000027, A000096, -A005581, A005582, -A005583, A005584.
%F A244422 Recurrence: T(n,k) = T(n-1, k) - T(n-2, k-1), n >= k >= 1, T(n,k) = 0 if n < k, T(0,0) = 2, T(n,0) = 1 if n>=1, (Compare with A061896).
%F A244422 For n >= 1 the entries without trailing zeros are given by T(n,k) = (-1)^k*(n/(n-k))*binomial(n-k,k) where k=0..floor(n/2).
%e A244422 The triangle T(n,k) begins:
%e A244422   n\k  0   1   2     3    4     5  6   7  8  9 10 11
%e A244422   0:   2
%e A244422   1:   1   0
%e A244422   2:   1  -2   0
%e A244422   3:   1  -3   0     0
%e A244422   4:   1  -4   2     0    0
%e A244422   5:   1  -5   5     0    0     0
%e A244422   6:   1  -6   9    -2    0     0  0
%e A244422   7:   1  -7  14    -7    0     0  0   0
%e A244422   8:   1  -8  20   -16    2     0  0   0  0
%e A244422   9:   1  -9  27   -30    9     0  0   0  0  0
%e A244422   10:  1 -10  35   -50   25    -2  0   0  0  0  0
%e A244422   11:  1 -11  44   -77   55   -11  0   0  0  0  0  0
%e A244422   ...
%e A244422 Rrev(3, x) = 1 - 3*x = sqrt(x)^3*R(3,1/sqrt(x)) = sqrt(x)^3*(-3/sqrt(x) + 1/sqrt(x)^3 ) = -3*x + 1.
%e A244422 Rrev(4, x) = 1 - 4*x + 2*x^2 = sqrt(x)^4*(2 - 4/sqrt(x)^2 + 1/sqrt(x)^4) = 2*x^2 - 4*x + 1.
%e A244422 Recurrence: T(4,1) = T(3, 1) - T(2, 0) = -3 -1 = -4.
%Y A244422 Cf. A053120, A127672, A054977, -A000027, A000096, -A005581, A005582, -A005583, A005584, A057079, A000032.
%K A244422 sign,easy,tabl
%O A244422 0,1
%A A244422 _Wolfdieter Lang_, Aug 08 2014