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%I #11 Aug 29 2019 16:26:56
%S 1,2,3,-2,4,-6,5,-12,3,6,-20,12,7,-30,30,-4,8,-42,60,-20,9,-56,105,
%T -60,5,10,-72,168,-140,30,11,-90,252,-280,105,-6,12,-110,360,-504,280,
%U -42,13,-132,495,-840,630,-168,7,14,-156,660,-1320,1260,-504,56,15,-182,858,-1980,2310,-1260,252,-8,16,-210,1092
%N Convolution array for Chebyshev's S(n,x)=U(n,x/2) polynomials.
%C S1(n,x):=sum(S(n-k,x)*S(k,x),k=0..n)= sum(a(n,m)*x^(n-2*m),m=0..floor(n/2)).
%C The unsigned column sequences, m>=0, divided by (m+1) give Pascal triangle column sequences for m+1.
%C G.f. for column m sequence: ((-1)^m)*(m+1)*(x^(2*m))/(1-x)^(m+2), m>=0.
%C Row polynomials P1(n,x):= sum(a(n,m)*x^m,m=0..floor(n/2)) (increasing powers of x).
%C Written as a triangle with increasing powers of x this is A294519. - _Wolfdieter Lang_, Nov 12 2017
%H W. Lang, <a href="/A128502/a128502.txt">First 15 rows and more.</a>
%F a(n,m)=binomial(n-m,m)*(n+1-m)*(-1)^m, m=0..floor(n/2), n>=0.
%F a(n,m)=binomial(n+1-m,m+1)*(m+1)*(-1)^m, m=0..floor(n/2), n>=0.
%F G.f. for S1(n,x): 1/(1-x*z+z^2)^2.
%F G.f. for P1(n,x): 1/(1-z+x*z^2)^2.
%e [1];[2];[3,-2],[4,-6];[5,-12,3];[6,-20,12];[7,-30,30,-4];[8,-42,60,-20];...
%e n=4: [5,-12,3] stands for the polynomial S1(4,x) = 5*x^4-12*x^2+3 = 2*(S(4,x)*1+S(3,x)*S(1,x))+S(2,x)*S(2,x).
%e n=4: [5,-12,3] stands also for the row polynomial P1(4,x) = 5-12*x+3*x^2.
%Y Row sums (signed array) give A099254. Unsigned row sums are A001629(n+2).
%Y Cf. A115139 (with offset n>=0 is S(n, x) array, decreasing powers of x).
%Y Cf. A294519 (as triangle).
%K sign,tabf,easy
%O 0,2
%A _Wolfdieter Lang_ Apr 04 2007
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