Fast algorithm computing multiplications of 2^n number of decreasing natural numbers

[wij]

I just figured out the solution: n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)=
((n^2-7n+7)^2-21)^2+448n-784-64n^2 I.e. 2^3 number of decreasing natural
numbers can be computed using just 3 multiplications, though not exactly the
form in the beginning guess. So it is reasonable to move on the conjecture to
the next level: Could computing the product of 16 decreasing/increasing
natual numbers be done by using 4 multiplications?

https://math.stackexchange.com/questions/4099451/fast-algorithm-computing-multiplications-of-2n-number-of-decreasing-natural
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Computing 16 products is not applicable for me. Theoretical analysis of the
problem is also likely difficult. Can someone share ideas?
(note: the result is significantly different than the usual 1/2 mult. method)

[Keith Thompson]

How is this related to C?

--
Keith Thompson (The_Other_Keith) Keith.S.T...@gmail.com
Working, but not speaking, for Philips Healthcare
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