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%I #2 Mar 30 2012 17:34:33
%S 1,1,2,7,28,84,384,1824,6080,30400,304000,1064000,12768000,67488000,
%T 359936000,1934656000,30954496000,526226432000,2880397312000,
%U 15842185216000,316843704320000,1757042360320000,38654931927040000
%N a(n)=If[IntegerQ[((6*n - 4)/( n + 1))*a(n - 1)], ((6*n - 4)/(n + 1))* a(n - 1), If[IntegerQ[((4*n - 2)/(n + 1))*a( n - 1)], ((4*n - 2)/(n + 1))*a(n - 1), n*a(n - 1)]]
%C Catalan recursion is:
%C a[0] = 1; a[n_] := a[n] = ((4*n - 2)/(n + 1))*a[n - 1];
%C The object here is to get a sequence that is Catalan like, but lower ( bifurcates higher).
%F a(n)=If[IntegerQ[((6*n - 4)/( n + 1))*a(n - 1)], ((6*n - 4)/(n + 1))* a(n - 1),
%F If[IntegerQ[((4*n - 2)/(n + 1))*a( n - 1)], ((4*n - 2)/(n + 1))*a(n - 1), n*a(n - 1)]]
%t Clear [a, n]; a[0] = 1;
%t a[n_] := a[n] = If[IntegerQ[((6*n - 4)/(n + 1))*a[n - 1]], ((3*n - 2)/(n + 1))* a[n - 1],
%t If[IntegerQ[((6*n - 4)/(n + 1))*a[n - 1]], ((4*n - 2)/(n + 1))*a[n - 1], n*a[n - 1]]];
%t Table[a[n], {n, 0, 30}]
%Y A000108
%K nonn,uned
%O 0,3
%A _Roger L. Bagula_, Jan 24 2009
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