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%I #4 Mar 30 2012 18:36:44
%S 1,1,2,3,6,15,29,63,160,333,749,1914,4135,9490,24335,53791,125104,
%T 321521,721887,1694914,4362855,9907851,23429158,60379623,138320021,
%U 328917615,848432824,1957091277,4674847097,12067450014,27992976565
%N Antidiagonal sums of the slanted Catalan convolution table A100247.
%F a(n) = Sum_{k=0..[2n/3]} C(n+k-[k/2], k)*(n-k-[k/2])/(n+k-[k/2]), with a(0)=1. G.f. A(x) satisfies: A(x^2) = ((1+x)/(2*x-(1-sqrt(1-4*x^3)))-(1-x)/(2*x+(1-sqrt(1+4*x^3)))).
%o (PARI) {a(n)=if(n==0,1,sum(k=0,(2*n)\3,binomial(n+k-(k\2),k)*(n-k-(k\2))/(n+k-(k\2))))}
%Y Cf. A100247.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Nov 09 2004