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%I #5 Jun 06 2016 23:43:34
%S 1,-1,1,-1,0,1,1,-1,-1,1,-1,0,0,0,1,1,-1,0,0,-1,1,1,0,-1,0,-1,0,1,-1,
%T 1,1,-1,1,-1,-1,1,-1,0,0,0,0,0,0,0,1,1,-1,0,0,0,0,0,0,-1,1,1,0,-1,0,0,
%U 0,0,0,-1,0,1,-1,1,1,-1,0,0,0,0,1,-1,-1,1,1,0,0,0,-1,0,0,0,-1,0,0,0,1
%N Matrix inverse of Sierpinski's triangle (A047999, Pascal's triangle mod 2).
%C In absolute values equal to A047999. - _M. F. Hasler_, Jun 06 2016
%e Triangle begins:
%e 1,
%e -1, 1,
%e -1, 0, 1,
%e 1,-1,-1, 1,
%e -1, 0, 0, 0, 1,
%e 1,-1, 0, 0,-1, 1,
%e 1, 0,-1, 0,-1, 0, 1,
%e -1, 1, 1,-1, 1,-1,-1, 1,
%e -1, 0, 0, 0, 0, 0, 0, 0, 1,
%e 1,-1, 0, 0, 0, 0, 0, 0,-1, 1,
%e 1, 0,-1, 0, 0, 0, 0, 0,-1, 0, 1,
%e -1, 1, 1,-1, 0, 0, 0, 0, 1,-1,-1, 1,
%e 1, 0, 0, 0,-1, 0, 0, 0,-1, 0, 0, 0, 1,
%e ...
%o (PARI) p=2; s=13; P=matpascal(s); PM=matrix(s+1,s+1,n,k,P[n,k]%p); IPM = 1/PM;
%o for(n=1,s,for(k=1,n,print1(IPM[n,k],","));print())
%Y Cf. A007318, A047999.
%K easy,sign,tabl
%O 0,1
%A _Gerald McGarvey_, Oct 10 2009