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.
