fibonacci series what Iam is missing ?

[Rustom Mody]

On Tuesday, March 24, 2015 at 11:50:40 AM UTC+5:30, Rustom Mody wrote:
> On Tuesday, March 24, 2015 at 10:51:11 AM UTC+5:30, Ian wrote:
> > On Mon, Mar 23, 2015 at 4:53 PM, Terry Reedy wrote:
> > > Iteration with caching, using a mutable default arg to keep the cache
> > > private and the function self-contained.  This should be faster.
> > >
> > > def fib(n, _cache=[0,1]):
> > >     '''Return fibonacci(n).
> > >
> > >     _cache is initialized with base values and augmented as needed.
> > >     '''
> > >     for k in range(len(_cache), n+1):
> > >         _cache.append(_cache[k-2] + _cache[k-1])
> > >     return _cache[n]
> > >
> > > print(fib(1), fib(3), fib(6), fib(5))
> > > # 1 2 8 5
> > 
> > Iteration in log space. On my desktop, this calculates fib(1000) in
> > about 9 us, fib(100000) in about 5 ms, and fib(10000000) in about 7
> > seconds.
> > 
> > def fib(n):
> >     assert n >= 0
> >     if n == 0:
> >         return 0
> > 
> >     a = b = x = 1
> >     c = y = 0
> >     n -= 1
> > 
> >     while True:
> >         n, r = divmod(n, 2)
> >         if r == 1:
> >             x, y = x*a + y*b, x*b + y*c
> >         if n == 0:
> >             return x
> >         b, c = a*b + b*c, b*b + c*c
> >         a = b + c
> 
> This is rather arcane!
> What are the identities used above?

Seems to be close to these (with some spices added!)
f₂ₙ = (fₙ)² + 2fₙ₋₁fₙ
f₂ₙ₊₁ = (fₙ)² + (fₙ₊₁)²