Eli Bendersky's website Problem 104 was probably the most difficult I've tackled so far. At first I've attempted brute-force, but after several long minutes of runtime realized it just won't work. And well I did, because it turns out the number I'm looking for is over 68000 digits long. Such numbers are quite consuming to process.
So I had to resort to some numerical tricks to reach the solution.
Observation #1
To find numbers where the low 9 digits are pandigital, we don't really have to
consider the whole numbers, but only their lowest 9 digits. The higher digits
don't affect the low ones in addition, so once can generate the 9 lowest digits
of the Fibonacci numbers easily.
These are trivially checked for being pandigital, so it's simple to generate
"candidates" for the answer. This way I quickly find out the 
s for
which 
has
pandigital low digits.
Observation #2 Once we have the indices of candidates, we can directly find the numbers by using the following cool formula for Fibonacci numbers:
As you can see, the runtime of such a computation is logarithmic, and with some
simple caching it's a very efficient algorithm for generating the th
Fibonacci number.
So I could just test my candidates for upper-digit pandigital this way, and the
problem was solved in about a minute of runtime.
Observation #3 I'm pretty sure this can be done even more efficiently, using the golden ratio:
