combinatorial proof of Zeckendorf’s theorem

Theorem 1.

Every nonnegative integer has a unique representation as a sum of nonconsecutive Fibonacci numbersMathworldPlanetmath, where we do not use F1 and instead use


(Ignoring F1 is necessary only to prevent the trivial counterexamples such as 6=F5+F2=F5+F1).


Given a sum i=2nZiFi where Zi{0,1}, we write that as a sequenceMathworldPlanetmath of binary digits; thus F4+F2 is written as 101, while F7+F5+F2 is 101001. Arrange such sequences with no two consecutive 1’s (i.e. those sequences corresponding to sums of nonconsecutive Fibonacci numbers) in increasing lexicographic orderMathworldPlanetmath starting from 0, and assign each one a rank corresponding to its order in the list:


Note that the rank is not the value of the string as a binary string. In fact, it is the sum of the string when interpreted as a sum of Fibonacci numbers, but we do not assume this; as far as we know up to this point, all the rank is is the row number in this table.

We say that such a sequence has length n if the leftmost 1 is in position n; thus 100 has length 3.

Claim that the number of representations with length n is Fn+2. We prove this by inductionMathworldPlanetmath on n; note that it is true for n=0 and for n=1. Now, given any representation of length n, it either has length n-1 or it has length exactly n. The number of representations of length n-1 is, by induction, Fn+1. To compute the number of representations of length exactly n, note that any such representation starts with 10 since we cannot have two consecutive 1’s. The digits after the 10 are arbitrary as long as there are not two consecutive 1’s, so that the number of such representations is precisely the number of representations of length n-2, or Fn. Thus the number of representations of length n is Fn+1+Fn=Fn+2. This proves the claim.

We will now prove the theorem by counting the number of rows preceding row n in two different ways. First, the number of such rows is obviously n (since we just numbered the row consecutively). We now count the number of predecessors by looking at the corresponding sequence. Given a sequence s, each of its predecessors p first differs from it in some position k; since s>p lexicographically, s must be 1 in that position while p is zero. Since p has a zero in position k, the remaining k-1 digits of p are arbitrary subject to the consecutivity condition, so the number of possibilities for p is the number of representations of length k-1, which is Fk+1. We can do this for each possible point of differencePlanetmathPlanetmath with s (which occur at the positions in which s has a 1, and thus must have nonconsecutive indices). This counts each predecessor of s once and only once, and gives a representation of the number of predecessors as a sum of nonconsecutive Fibonacci numbers. But there are n predecessors, so we have written n as the required sum.

As an example of this process, consider the row numbered 7, which is the sequence 1010. If a predecessor differs from it at the first 1, it is of the form 0*** where the *’s are arbitrary (so long as there are no two consecutive 1’s, so there are F5 of them. If a predecessor differs from it at the second 1, it is of the form 100* and again the *’s are arbitrary, so there are F3 of them. Thus 7=F5+F3.

This argument shows existence; uniqueness follows as well from this construction since every possible sum is represented in the list, and each one represents a different integer.

Title combinatorial proof of Zeckendorf’s theorem
Canonical name CombinatorialProofOfZeckendorfsTheorem
Date of creation 2013-03-22 19:05:53
Last modified on 2013-03-22 19:05:53
Owner rm50 (10146)
Last modified by rm50 (10146)
Numerical id 5
Author rm50 (10146)
Entry type Proof
Classification msc 11B39
Classification msc 11A63