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%I #7 Dec 29 2023 10:54:22
%S 1,0,2,1,0,3,0,4,0,4,1,0,10,0,5,0,6,0,20,0,6,1,0,21,0,35,0,7,0,8,0,56,
%T 0,56,0,8,1,0,36,0,126,0,84,0,9
%N (1/2) * (A007318 - A007318^(-1)).
%C Row sums = (1, 2, 4, 8, ...). A131047 * (1,2,3, ...) = A087447 starting (1, 4, 10, 24, 56, ...). A generalized set of analogous triangles: (1/(Q+1)) * (P^Q - 1/P), Q an integer, generates triangles with row sums = powers of (Q+1). Cf. A131048, A131049, A131050, A131051 for triangles having Q = 2,3,4 and 5, respectively.
%C A007318, Pascal's triangle, = this triangle + A119467, since one triangle = the zeros or masks of the other. - _Gary W. Adamson_, Jun 12 2007
%F Let A007318 (Pascal's triangle) = P, then A131047 = (1/2) * (P - 1/P); deleting the right border of zeros.
%e First few rows of the triangle:
%e 1;
%e 0, 2;
%e 1, 0, 3;
%e 0, 4, 0, 4;
%e 1, 0, 10, 0, 5;
%e 0, 6, 0, 20, 0, 6;
%e 1, 0, 21, 0, 35, 0, 7;
%e ...
%Y Cf. A131048, A131049, A131050, A131051.
%Y Cf. A119467.
%K nonn,tabl
%O 1,3
%A _Gary W. Adamson_, Jun 12 2007
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