Simon>hello everybody, about that formula for 1/pi, SP>it is nice , the coefficient a(k) is a binomial term SP>The sequence a(k) is : 1, 1, 33, 97, 1729, 8001, 105441, 627873, 6989697, 48363649, 488206753, 3701949153, 35289342529, 283146701761, 2610495177057, 21695983405857, 196218339243777, 1667338615773441, 14917038493453089, 128562758660255073, 1143482133220664769, 9946278255903268929, 88205310329762729697, 771946983805271894433, 6837125121111415598721,... SP>a(k] has this g.f. 1 -------------------- = 2 1/2 (1 - 2 x - 63 x ) How did you get this? SP>Now, this has a closed binomial expression, of course Julian used the trinomial theorem to get a(k)==Hypergeometric2F1[1/2 - k/2, -(k/2), 1, 64] Did you mean something different? Nicer?? Julian's yields numerous triangular double sums for 24/pi, e.g. Sum[((Sum[((Binomial[2 * (2 * i - k), 2 * i - k])^2 * (2 * i - k)! * ( - 30 * k + 60 * i + 7) * ( - 1)^k * 2^(4 * k)/(((i - k)!)^2 * k!)), List[k, 0, i]])/(2^(12 * i))), List[i, 0, Infinity]], Sum[(((30 * k + 7) * k! * (Binomial[2 * k, k])^2 * Sum[((16^i)/((i!)^2 * (k - 2 * i)!)), List[i, 0, Floor[k/2]]])/(( - 256)^k)), List[k, 0, Infinity]], Sum[Sum[((16^i * (30 * (k + i) + 7) * (k + i)! * (Binomial[2 * (k + i), k + i])^2)/((i!)^2 * ( - 256)^(k + i) * (k - i)!)), List[i, 0, k]], List[k, 0, Infinity]], and this wild thing: 160/(7*Pi) + (28*EllipticK[(1/2)*(1 - (3*Sqrt[7])/8)]^2)/Pi^2 - (25*Gamma[1/7]^2*Gamma[2/7]^2*Gamma[4/7]^2)/(14*Sqrt[7]*Pi^4) - 30*Sum[(k*Binomial[2*k, k]^2* HypergeometricPFQ[{1 - k, 1/2 + k, 1/2 + k}, {1, 1 + k}, 1])/(-256)^k, {k, 1, Infinity}] (We're trying to get 24/pi and it gives us something plus 160/(7 pi).) But unless one of these can be written with the inner summand free of the outer index, these have an asymptotic convergence rate of *zero* bits/term. The >0 estimates cheat by assuming a(k) is free, or rather constant(k). It remains, however an intriguing conjecture. --rwg