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From: Ian Kelly <ian.g.kelly@gmail.com>
Date: Fri, 20 May 2011 01:32:52 -0600
Subject: Re: Faster Recursive Fibonacci Numbers
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Xref: x330-a1.tempe.blueboxinc.net comp.lang.python:5836

2011/5/20 Ian Kelly <ian.g.kelly@gmail.com>:
> def fib_decimal(n):
> =A0 =A0with localcontext() as ctx:
> =A0 =A0 =A0 =A0ctx.prec =3D n // 4 + 1
> =A0 =A0 =A0 =A0sqrt_5 =3D Decimal('5').sqrt()
> =A0 =A0 =A0 =A0phi =3D (1 + sqrt_5) / 2
> =A0 =A0 =A0 =A0numerator =3D (phi ** n) - (1 - phi) ** n
> =A0 =A0 =A0 =A0return int((numerator / sqrt_5).to_integral_value())

Note that I believe this is O(n) since the necessary precision grows
linearly with n.