  Zeckendorf proved that every positive integer has a unique representation as
a sum of non-consecutive Fibonacci numbers. Once this has been shown, it's
natural to ask how many summands are needed. Using a continued fraction
approach, Lekkerkerker proved that the average number of such summands needed
for integers in $[F_n, F_{n+1})$ is $n / (\varphi^2 + 1) + O(1)$, where
$\varphi = \frac{1+\sqrt{5}}2$ is the golden mean. Surprisingly, no one appears
to have investigated the distribution of the number of summands; our main
result is that this converges to a Gaussian as $n\to\infty$. Moreover, such a
result holds not just for the Fibonacci numbers but many other problems, such
as linear recurrence relation with non-negative integer coefficients (which is
a generalization of base $B$ expansions of numbers) and far-difference
representations.
  In general the proofs involve adopting a combinatorial viewpoint and
analyzing the resulting generating functions through partial fraction
expansions and differentiating identities. The resulting arguments become quite
technical; the purpose of this paper is to concentrate on the special and most
interesting case of the Fibonacci numbers, where the obstructions vanish and
the proofs follow from some combinatorics and Stirling's formula; see [MW] for
proofs in the general case.
