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%I #39 Dec 06 2016 02:57:38
%S 1,1,1,1,0,-1,1,1,-1,-1,1,0,-2,0,1,1,1,-2,-2,1,1,1,0,-3,0,3,0,-1,1,1,
%T -3,-3,3,3,-1,-1,1,0,-4,0,6,0,-4,0,1,1,1,-4,-4,6,6,-4,-4,1,1,1,0,-5,0,
%U 10,0,-10,0,5,0,-1,1,1,-5,-5,10,10,-10,-10,5,5,-1,-1,1,0,-6,0,15,0,-20
%N Coefficients in expansion of (1-x)^floor(n/2)(1+x)^ceiling(n/2).
%C Triangle T(n,k), 0<=k<=n, read by rows given by: [1,0,-1,0,0,0,0,0,...]DELTA[1,-2,1,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - _Philippe Deléham_, Sep 22 2008
%C The production matrix for this array has bivariate e.g.f. equal to exp(-t*x)*(1-t). - _Paul Barry_, Nov 22 2008
%C The elements of the matrix inverse are apparently T^(-1)(n,k) = (-1)^(n+k)*T(n,k). - _R. J. Mathar_, Apr 08 2013
%C Row sums give A130706. - _Philippe Deléham_, Oct 21 2013
%H E. Burlachenko, <a href="https://arxiv.org/abs/1612.00970">Fractal generalized Pascal matrices</a>, arXiv:1612.00970 [math.NT], 2016. See p. 3.
%F T(n, k) = -T(n-2, k-2) + T(n-2, k). T(0, 0) = T(1, 0) = T(1, 1) = 1.
%F T(n,k) = T(n-1,k) + (-1)^(n-1)*T(n-1,k-1), T(0,0)=1. - _Jose Ramon Real_, Nov 10 2007
%F G.f.: (1+x+x*y)/(1-x^2+x^2*y^2). - _Philippe Deléham_, Oct 21 2013
%e Triangle begins:
%e 1;
%e 1, 1;
%e 1, 0, -1;
%e 1, 1, -1, -1;
%e 1, 0, -2, 0, 1;
%e 1, 1, -2, -2, 1, 1;
%e ...
%p A051160 := proc(n,k)
%p (1-x)^floor(n/2)*(1+x)^ceil(n/2) ;
%p coeftayl(%,x=0,k) ;
%p end proc: # _R. J. Mathar_, Apr 08 2013
%t t[n_, k_] := Coefficient[(1-x)^Floor[n/2]*(1+x)^Ceiling[n/2], x, k]; Table[t[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Jan 09 2014 *)
%o (PARI) {T(n, k) = polcoeff( (1 - x)^(n\2) * (1 + x)^ceil(n/2), k)}
%Y Cf. A007318, A051159(n, k) = (-1)^[ k/2 ]*T(n, k).
%K sign,tabl,easy
%O 0,13
%A _Michael Somos_, Oct 14 1999
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