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%I #2 Mar 30 2012 17:25:32
%S 1,0,1,2,0,2,4,2,0,3,-4,4,2,0,9,0,-4,4,6,0,11,-8,0,-4,12,18,0,17,-16,
%T -8,0,-12,36,22,0,35,16,-16,-8,0,-36,44,34,0,57,0,16,-16,-24,0,-44,68,
%U 70,0,91,32,0,16,-48,-72,0,-68,140,114,0,161
%N Eigentriangle, row sums = A144181
%C Row sums = A144181: (1, 1, 3, 9, 11, 17, 35,...).
%C Left border = A118434: (1, 0, 2, 4, -4, 0, -8,...); (i.e. row sums of the self-inverse triangle A118433).
%C Triangle A144183 = partial sums starting from the right of A144182.
%C Sum of n-th row terms = rightmost term of next row.
%F Triangle read by rows, T(n,k) = A118434(n-k)*A144181(k-1); where A144181(k-1) = A144181 shifted to (1, 1, 1, 3, 9, 11, 17, 35, 57, 91, 161,...).
%e First few rows of the triangle are:
%e 1;
%e 0, 1;
%e 2, 0, 1;
%e 4, 2, 0, 3;
%e -4, 4, 2, 0, 9;
%e 0, -4, 4, 6, 0, 11;
%e -8, 0, -4, 12, 18, 0, 17;
%e -16, -8, 0, -12, 36, 22, 0, 35;
%e ...
%e row 3 = (4, 2, 0, 3) = termwise products of (4, 2, 0, 1) and (1, 1, 1, 3) = (4*1, 2*1, 0*1, 1*3).
%Y A144181, Cf. A118434, A144183
%K tabl,sign
%O 0,4
%A _Gary W. Adamson_, Sep 13 2008
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