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A294519 Convolution triangle for Chebyshev S polynomials (rising powers).

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%I A294519 #7 Nov 11 2017 11:52:01
%S A294519 1,0,2,-2,0,3,0,-6,0,4,3,0,-12,0,5,0,12,0,-20,0,6,-4,0,30,0,-30,0,7,0,
%T A294519 -20,0,60,0,-42,0,8,5,0,-60,0,105,0,-56,0,9,0,30,0,-140,0,168,0,-72,0,
%U A294519 10,-6,0,105,0,-280,0,252,0,-90,0,11,0,-42,0,280,0,-504,0,360,0,-110,0,12,7,0,-168,0,630,0,-840,0,495,0,-132,0,13
%N A294519 Convolution triangle for Chebyshev S polynomials (rising powers).
%C A294519 See the array A128502 without zeros and falling powers. This is the main entry.
%C A294519 The coefficient triangle for Chebyshev S polynomials is given in A049310.
%C A294519 The self-convolution (or first convolution) of the S polynomials is S1(n, x) := Sum_{k=0..n} S(k, x)*S(n-k, x), n >= 0, and S1(n, x) = Sum_{m=0..n} T(n, m)*x^m.
%F A294519 T(n, m) = [x^m] S1(n, x), with the first convolution S1 of the Chebyshev S polynomials. See a comment above.
%F A294519 T(n, m) = 0 if n-m is odd and T(n, m) = (-1)^((n-m)/2)*((n-m)/2 + 1)*binomial(n - (n-m)/2 +1, (n-m)/2 +1) = (-1)^((n-m)/2)*(n - (n-m)/2 + 1)* binomial(n - (n-m)/2, (n-m)/2) if n-m is even.
%F A294519 O.g.f. of {S1(n, x)}_{n >= 0} is G1(z,x) = (1/(1 - x*z + z^2))^2.
%e A294519 The triangle T(n, m) begins:
%e A294519 n\m   0   1    2    3    4    5    6   7    8    9   10 11 12 ...
%e A294519 0:    1
%e A294519 1:    0   2
%e A294519 2:   -2   0    3
%e A294519 3:    0  -6    0    4
%e A294519 4:    3   0  -12    0    5
%e A294519 5:    0  12    0  -20    0    6
%e A294519 6:   -4   0   30    0  -30    0    7
%e A294519 7:    0 -20    0   60    0  -42    0   8
%e A294519 8:    5   0  -60    0  105    0  -56   0    9
%e A294519 9:    0  30    0 -140    0  168    0 -72    0   10
%e A294519 10:  -6   0  105    0 -280    0  252   0  -90    0   11
%e A294519 11:   0 -42    0  280    0 -504    0 360    0 -110    0 12
%e A294519 12:   7   0 -168    0  630    0 -840   0  495    0 -132  0 13
%e A294519 ...
%Y A294519 Cf. A049310, A128502.
%K A294519 sign,tabl,easy
%O A294519 0,3
%A A294519 _Wolfdieter Lang_, Nov 07 2017