combinatorial proof of Zeckendorf’s theorem

combinatorial proof of Zeckendorf’s theorem

Theorem 1

.

Every

nonnegative integer

has a unique representation as a sum of nonconsecutive

Fibonacci numbers

, where we do not use

F

1

and instead use

F

2

=

1

,

F

3

=

2

,

F

4

=

3

,

F

5

=

5

,

F

6

=

8

,

…

(Ignoring

F

1

is necessary only to prevent the

trivial

counterexamples such as

6

=

F

5

+

F

2

=

F

5

+

F

1

).

Proof.

Given a sum

∑

i

=

2

n

Z

i

⁢

F

i

where

Z

i

∈

{

0

,

1

}

, we write that as a

sequence

of binary digits; thus

F

4

+

F

2

is written as

101

, while

F

7

+

F

5

+

F

2

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 order

starting from

0

, and assign each one a rank corresponding to its order in the list:

5

4

3

2

1

r

⁢

a

⁢

n

⁢

k

0

0

1

1

1

0

2

1

0

0

3

1

0

1

4

1

0

0

0

5

1

0

0

1

6

1

0

1

0

7

1

0

0

0

0

8

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

F

n

+

2

. We prove this by

induction

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,

F

n

+

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

F

n

. Thus the number of representations of length

≤

n

is

F

n

+

1

+

F

n

=

F

n

+

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

F

k

+

1

. We can do this for each possible point of

difference

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

F

5

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

F

3

of them. Thus

7

=

F

5

+

F

3

.

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

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