I asked Scooby Dougall's thm Dougall's thm: a hyper_f ([a, - + 1, b, c, d, n - d - c - b + 2 a + 1, - n], 7, 6 2 a [-, - b + a + 1, - c + a + 1, - d + a + 1, - n + d + c + b - a, n + a + 1]) 2 (a + 1) (- c - b + a + 1) (- d - b + a + 1) (- d - c + a + 1) n n n n = -------------------------------------------------------------------. (- b + a + 1) (- c + a + 1) (- d + a + 1) (- d - c - b + a + 1) n n n n I found an early 70s (pre matrix-calculus) printout from a secret Xerox device (a LASER printer!) implying a hyper_f ([a, b, c, d, - n, n - d - c - b + 2 a, - + 1], 7, 6 2 a [- b + a + 1, - c + a + 1, - d + a + 1, - n + d + c + b - a + 1, n + a + 1, -], 1) 2 = - (a + 1) (- c - b + a + 1) (- d - b + a + 1) (- d - c + a + 1) n n - 1 n - 1 n - 1 ((d + c + b - a) n (n - d - c - b + 2 a) + (c + b - a) (d + b - a) (d + c - a)) /((- b + a + 1) (- c + a + 1) (- d + a + 1) (- d - c - b + a) ). n n n n (Tested through n=5). The latter cannot be a variable-rename of the former, because the lower parameters minus the upper parameters total 2 in the former and 4 in the latter. Presumably, there's a q-extension. Can this be "new"? George Andrews and Dick Askey kindly inform me that they've known this for decades, and sketched a fancy derivation. I just mucked with my usual matrices, but there is an embarrassingly more obvious way: Call Dougall's lhs above D(n). Then ScoobyDougall(n) is just n (n + a) D(n) - ----------------------------------------- D(n - 1) (n - d - c - b + a) (n - d - c - b + 2 a) combined termwise. Oddly, this trick seems to fail for the q-extension ("ScoobyJackson"), and my matrices bog down in algebra. If there is a ScoobyJackson, why isn't it in BHS? --rwg TRANSSEXUAL EX-SATURNALS