A datum, probably long known, apropos the computational complexity of the Gamma fcn: 3 1 6 9 5 1 6 8 1 6 10 hyper_2f1 (-, -, - ) hyper_2f1 (-, -, - ) hyper_2f1(-, -, -- ) 7 7 7 7 7 7 7 7 7 7 1 %pi 588 (-)! cos(---) 7 14 = ------------------, %pi so (n/d)! is probably easy, but with effort proportional to d. (Or maybe d^2). ------- My POWERSERIES function got a double sum instead of Clausen's 4F3[z] for not knowing the well-poised, 3-balanced (sesquiSaalschutzian?), terminating 4F3[1] in the inner sum. "SesquiPfaffian" is more just, but also more confusing. I got ambitious and derived the whole 4F3[a,b,c,d;e,f,g] (and 3F2[z]) matrix system, and extracted the general, noterminating case: 1 hyper_f ([a, b, c, c - b - a + -], 4, 3 2 1 [c - a + 1, c - b + 1, b + a + -], 1) = 2 1 (b + a - -)! (c - a)! (c - b)! (sqrt(%pi) (c - 2 b - 2 a)! 2 1 1 /((a - -)! (b - -)! (c - 2 a)! (c - 2 b)! (c - b - a)!) 2 2 - hyper_f ([1, c - 2 a + 1, c - 2 b + 1, c - b - a + 1], 4, 3 3 [c + 1, c - 2 b - 2 a + 2, c - b - a + -], 1) 2 c + 1 1 /((a - 1)! (b - 1)! (----- - b - a) c! (c - b - a + -)!)) 2 2 Notice the rhs 4F3 getting divided by infinity exactly when an upper parameter on the left is a nonpositive integer. Also note the weird oo - oo when (c+1)/2 = a+b. As usual, the matrices provide the complete system of identities contiguous to the above. The algebra gets very heavy in transforming the general 4x4 system (which has eight dimensions), but there's a way to exploit the symmetry in a,b,c,d and e,f,g. You can actually substitute a=b=c=d=S and later recover the original generality with an extension of the following trick: Supposing there were only two variables, a and b: Replace every f(S)^3 with f(a) f(b) f((a+b)/2), then f(S)^2 with f(a) f(b) and finally f(S) with f((a+b)/2). This isn't rigorous, but elementary theology seems to find the right symmetric functions. A rigorous way to fight malignant algebroma is to get even more ambitious and derive the q-extension of the 4F3 system. Normally, the addition of a new variable makes matters nonlinearly worse, but q-extending (n + d + c + b + a) (n + e + c + b + a) (n + e + d + b + a) (n + e + d + c + a) (n + e + d + c + b) to n + d + c + b + a n + e + c + b + a (1 - q ) (1 - q ) n + e + d + b + a n + e + d + c + a (1 - q ) (1 - q ) n + e + d + c + b (1 - q ) actually *reduces* the expanded form from 247 terms to 31 terms, with five 5-term exponents and 26 sixes. All the nonlinear terms (n^5, a^4, ...) disappear, but are recoverable with l'Hospital's rule. Unfortunately, 4F3 is just shy of the breakeven point. But at least I now have the 4phi3 system too. --rwg (-: Mesotonic economist comes-into emoticons :-) PS, I'm dabbling in sudoku as a programming exercise. Our program (brute-force backtrack) typically finds one or two "redundant" clues in published puzzles, but of course removing them renders the puzzle both unsymmetrical and insoluble without (perhaps humanly infeasible) lookahead.