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%I
%S 1,1,0,1,1,0,1,2,1,0,1,3,2,1,0,1,4,5,3,1,0,1,5,9,5,4,1,0,1,6,14,14,9,
%T 5,1,0,1,7,20,28,14,14,6,1,0,1,8,27,48,42,28,20,7,1,0,1,9,35,75,90,42,
%U 48,27,8,1,0,1,10,44,110,165,132,90,75,35,9,1,0,1,11,54,154,275,297,132
%N Pendular Catalan triangle, read by rows, where row n is formed from row n-1 by the recurrence: if n > 2k, T(n,k) = T(n,n-k) + T(n-1,k), else T(n,k) = T(n,n-1-k) - T(n-1,k) - T(n-1,k+1), for n>=k>=0, with T(n,0)=1 and T(n,n)=0^n.
%C See A118340 for definition of pendular triangles and pendular sums.
%F T(2*n+m,n) = [A000108^(m+1)](n), i.e., the m-th lower semi-diagonal forms the self-convolution (m+1)-power of A000108.
%e Row 6 equals the pendular sums of row 5:
%e [1, 4, 5, 3, 1, 0], where the sums proceed as follows:
%e [1,__,__,__,__,__]: T(6,0) = T(5,0) = 1;
%e [1,__,__,__,__, 1]: T(6,5) = T(6,0) - T(5,5) = 1 - 0 = 1;
%e [1, 5,__,__,__, 1]: T(6,1) = T(6,5) + T(5,1) = 1 + 4 = 5;
%e [1, 5,__,__, 4, 1]: T(6,4) = T(6,1) - T(5,4) - T(5,5) = 5-1-0 = 4;
%e [1, 5, 9,__, 4, 1]: T(6,2) = T(6,4) + T(5,2) = 4 + 5 = 9;
%e [1, 5, 9, 5, 4, 1]: T(6,3) = T(6,2) - T(5,3) - T(5,4) = 9-3-1 = 5;
%e [1, 5, 9, 5, 4, 1, 0] finally, append a zero to obtain row 6.
%e Triangle begins:
%e 1;
%e 1, 0;
%e 1, 1, 0;
%e 1, 2, 1, 0;
%e 1, 3, 2, 1, 0;
%e 1, 4, 5, 3, 1, 0;
%e 1, 5, 9, 5, 4, 1, 0;
%e 1, 6, 14, 14, 9, 5, 1, 0;
%e 1, 7, 20, 28, 14, 14, 6, 1, 0;
%e 1, 8, 27, 48, 42, 28, 20, 7, 1, 0;
%e 1, 9, 35, 75, 90, 42, 48, 27, 8, 1, 0;
%e 1, 10, 44, 110, 165, 132, 90, 75, 35, 9, 1, 0;
%e 1, 11, 54, 154, 275, 297, 132, 165, 110, 44, 10, 1, 0; ...
%e Central terms are Catalan numbers T(2*n,n) = A000108(n);
%e semi-diagonals form successive self-convolutions of the central terms:
%e T(2*n+1,n) = [A000108^2](n),
%e T(2*n+2,n) = [A000108^3](n).
%o (PARI) T(n,k)=if(n<k || k<0,0,if(k==0,1,if(n==k,0, if(n>2*k,T(n,n-k)+T(n-1,k),T(n,n-1-k)-T(n-1,k)-if(n-1>k,T(n-1,k+1)) ))))
%Y Cf. A000108, A118340, A033184.
%K nonn,tabl
%O 0,8
%A _Paul D. Hanna_, Apr 26 2006
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