Fooling with more path invariant matrices last nite, the kids & I found HypergeometricPFQ[{1/2 - a, 1 - a, 1/2 + a, 1 + a, 23/14 + (2 a)/7 - 1/7 Sqrt[1 - 3 a - 3 a^2], 23/14 + (2 a)/7 + 1/7 Sqrt[1 - 3 a - 3 a^2]}, {5/4, 3/2, 7/4, 9/14 + (2 a)/7 - 1/7 Sqrt[1 - 3 a - 3 a^2], 9/14 + (2 a)/7 + 1/7 Sqrt[1 - 3 a - 3 a^2]}, -(1/48)] == (2 3^(3/2 + a) Sin[(a \[Pi])/3])/(a (11 + 12 a + 4 a^2)) I don't recall any F[-1/48]. Julian found 27 a which rationalize the parameters, e.g., HypergeometricPFQ[{46/43, 40/43, 49/86, 37/86, 883/602, 153/86}, {3/2, 5/4, 7/4, 281/602, 67/86}, -1/48] == 159014*3^(37/86)*Sin[Pi/43]/18827 There were three 3F2[-1/4] (which Macsyma could do), one 4F3, and no 5F4s. Along the way, he found the minor simplification puzzle 1 == 10*Sqrt[Pi]^3/(27*2^(2/3)*(1/3)!^2*(5/6)!*Sqrt[3]) and the unlikely looking 3*((4*Sqrt[3]*I + 1)/Sqrt[1 - I/4/Sqrt[3]] - (4*Sqrt[3]*I - 1)/Sqrt[I/4/Sqrt[3] + 1]) - 14*3^(1/4)*((47*Sqrt[3]*I + 24)/ Sqrt[4*Sqrt[3] - I]^5 - (47*Sqrt[3]*I - 24)/Sqrt[I + 4*Sqrt[3]]^5) == 121*Sqrt[Pi]^5/(288*2^(1/3)*(1/6)!*(1/4)!^2*(11/12)!^2*Sqrt[3]) We also accelerated Dixon's thm to get a three-parameter 7F6[-1/4]: HypergeometricPFQ[{a, b, c, 2*a - 2*b, -c - b + 2*a, (r + 3*c - b + 4*a)/10 + 1, (-r + 3*c - b + 4*a)/10 + 1}, {c - b + 1, (1 - b)/2 + a, -b/2 + a + 1, c + b - a + 1, (-r + 3*c - b + 4*a)/ 10, (r + 3*c - b + 4*a)/10}, -1/ 4] == (2*a - b)!*2^(2*b - 2*a)*(c - b)!*(c + b - a)!* Sqrt[Pi]/(a!*(-b + a - 1/2)!*c!*(c - a)!) r-> Sqrt[(3*c - b)^2 - 8*a*(2*c + b - 2*a)] a rational case of which is HypergeometricPFQ[{3/2 + a/2, 2 + a/2, 1 - c, 1 + a + 2 c, 1 + a/2 - c - (2 c)/a, 8/5 + a/5 - (2 c)/5 - (6 c)/(5 a), 1 + a/2 + 2 c + (2 c)/a}, {1 + a/2, 3/2 + a/2 + c/2, 2 + a/2 + c/2, 1 + a/2 - (2 c)/a, 3/2 - 2 c - (2 c)/a, 3/5 + a/5 - (2 c)/5 - (6 c)/(5 a)}, -(1/4)] == ( 2^(-1 - a - 2 c) Sqrt[\[Pi]] (2 + a + c)! (a/2 - (2 c)/a)! (1/2 - 2 c - (2 c)/ a)!)/(((3 + a)/ 2)! (a/2 + c)! (-(1/2) - c - (2 c)/a)! (1 + a/2 - c - (2 c)/a)!) --rwg