a(n+1) = sum{j = 0,...n}[C(2n,2j)a(j)a(n-j)] with a(0) = 1.

A063902

a(n+1) = sum{j = 0,...n}[C(2n,2j)a(j)a(n-j)] with a(0) = 1.

Terms

    a(0) =1a(1) =1a(2) =2a(3) =10a(4) =80a(5) =1000a(6) =17600a(7) =418000a(8) =12848000a(9) =496672000a(10) =23576960000

External references