rkg> sum of the first n terms is [...] Good point, Macsyma could (c131) nusum(binomial(2*k,k)/4^k/(k+1),k,0,n) binomial(2 n + 1, n + 1) (d131) 2 - ------------------------ n 4 in 1977!-). Apropos Catalan numbers, a few weeks ago I conjectured the integerhood of 2 binomial(n, i) (n + 2 i + 3)! ------------------------------- . (i + 1)! (i + 2)! (n + 3)! This is the special case C(3,i,n-i) of k /===\ (n + k)! (n + m)! | | (j - 1)! (n + j m + m)! C(m, k, n) := ----------------- | | ----------------------- , n! (n + m + k)! | | (m + j - 1)! (n + j m)! j = 1 which I also conjecture integral. C(0, k, n) = 1, C(1, k, n) = binomial(n + k, k), and Catalan(n) = C(2, n, 0). n + 2 k + 2 (n + 2 k + 1)! C(2, k, n) = ----------------------- -------------- , (n + k + 1) (n + k + 2) k! (k + 1)! n! which may provide a stepping stone to the more general proof.