This discussion and my own (preliminary) benchmarking results have led me to rethink a few things about my
real problem.
For instance, when computing F(n), we can make use of the addition formula to break n into 2 pieces:
F(n) = F(i+j) = F(i-1)*F(j) + F(i)*F(j+1)
Then:
j = k**2 = int(sqrt(n))
i = n - j
F(n) = F(i+j) = F(i+k**2)
which we can compute using the first formula above and
F(k**2) = (F(k-1)+F(k+1))*F(k*(k-1))
- ((-1)**k)*F(k*(k-2))
It remains to be seen whether all of this extra work buys anything.
-QM
--
Quantum Mechanics: The dreams stuff is made of
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