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%I #32 Mar 17 2019 03:47:19
%S -1,1,0,1,0,1,1,0,0,0,1,0,-1,0,-1,1,0,-2,0,-1,0,1,0,-3,0,0,0,1,1,0,-4,
%T 0,2,0,2,0,1,0,-5,0,5,0,2,0,-1,1,0,-6,0,9,0,0,0,-3,0,1,0,-7,0,14,0,-5,
%U 0,-5,0,1,1,0,-8,0,20,0,-14,0,-5,0,4,0
%N Triangle T(n,k) of the coefficients [x^(n-k)] of the polynomial p(0,x)=-1, p(1,x)=x and p(n,x) = x*p(n-1,x) - p(n-2,x) in row n, column k.
%C Consider the Catalan triangle A009766 antisymmetrically extended by a mirror along the diagonal (see also A176239):
%C 0, -1, -1, -1, -1, -1, -1, -1,
%C 1, 0, -1, -2, -3, -4, -5, -6,
%C 1, 1, 0, -2, -5, -9, -14, -20,
%C 1, 2, 2, 0, -5, -14, -28, -48,
%C 1, 3, 5, 5, 0, -14, -42, -90,
%C 1, 4, 9, 14, 14, 0, -42, -132,
%C 1, 5, 14, 28, 42, 42, 0, -132,
%C 1, 6, 20, 48, 90, 132, 132, 0.
%C The rows in this array are essentially the columns of T(n,k).
%F Sum_{k=0..n} T(n,k) = A057079(n-1).
%F Apparently T(3s,2s-2) = (-1)^(s+1)*A000245(s), s >= 1.
%e Triangle begins
%e -1; # -1
%e 1, 0; # x
%e 1, 0, 1; # x^2+1
%e 1, 0, 0, 0; # x^3
%e 1, 0, -1, 0, -1; # x^4-x^2-1
%e 1, 0, -2, 0, -1, 0;
%e 1, 0, -3, 0, 0, 0, 1;
%e 1, 0, -4, 0, 2, 0, 2, 0;
%e 1, 0, -5, 0, 5, 0, 2, 0, -1;
%e 1, 0, -6, 0, 9, 0, 0, 0, -3, 0;
%e 1, 0, -7, 0, 14, 0, -5, 0, -5, 0, 1;
%e 1, 0, -8, 0, 20, 0,-14, 0, -5, 0, 4, 0;
%e 1, 0, -9, 0, 27, 0,-28, 0, 0, 0, 9, 0, -1;
%p p:= proc(n,x) option remember: if n=0 then -1 elif n=1 then x elif n>=2 then x*procname(n-1,x)-procname(n-2,x) fi: end: A192174 := proc(n,k): coeff(p(n,x),x,n-k): end: seq(seq(A192174(n,k),k=0..n), n=0..11); # _Johannes W. Meijer_, Aug 21 2011
%Y Cf. A023443, A080956 and A000096, A129936 and A005586, A005587.
%Y Cf. A194084. - _Paul Curtz_, Aug 16 2011
%K sign,tabl,easy
%O 0,18
%A _Paul Curtz_, Jun 24 2011
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