Odd Fibbinary Numbers and the Golden Ratio

TL;DR

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Contribution

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Abstract

The fibbinary numbers are positive integers whose binary representation contains no consecutive ones. We prove the following result: If the $j$th odd fibbinary is the $n$th \emph{odd} fibbinary number, then $j = \lfloor n\phi^2 \rfloor - 1.$

Equations

$$F_n=F_{n-1}+F_{n-2}$$

$$z(4)=(3,1)=(F_4,F_2),$$

$$z(5)=(5)=(F_5),$$

$$z(100)=(89,8,3)=(F_{11},F_6,F_4).$$

$$\mathrm{fib}(n):=\sum _{j=1}^k{2}^{i_j-2}.$$

$$\mathrm{fib}(4)=\mathrm{fib}(F_4+F_2)=2^{4-2}+2^{2-2}=101_2=5.$$

$$\mathrm{fib}(5)=\mathrm{fib}(F_5)=2^{5-3}=1000_2=8.$$

$$\mathrm{fib}(100)=\mathrm{fib}(F_{11}+F_6+F_4)=2^{11-2}+2^{6-2}+2^{4-2}=1000010100_2=532,$$

$$\mathrm{odfib}(1)=1_2,\mathrm{odfib}(2)=101_2,\mathrm{odfib}(3)=1001_2,\dots .$$

$$Z(n):={\mathrm{fib}}^{-1}(\mathrm{odfib}(n)),$$

$$Z(1)=1,Z(2)=4,Z(3)=6,Z(4)=9,\dots .$$

$$1,4,6,9,12,14,17,\dots$$

$$\{\lfloor n{\varphi }^2\rfloor -1{\}}_{n=1}^{\infty }=\{\lfloor n\varphi \rfloor +n-1{\}}_{n=1}^{\infty }$$

$$\varphi =\frac{1+\sqrt{5}}{2}$$

$$\underset{n\to \infty }{\lim }\frac{F_{n+1}}{F_n}=\varphi .$$

$$j=Z(n)=\lfloor n{\varphi }^2\rfloor -1=\lfloor n\varphi \rfloor +n-1$$

$$\mathrm{odfib}(F_n+k)=2^n+\mathrm{odfib}(k).$$

$$\mathrm{odfib}(10)=\mathrm{odfib}(8+2)=\mathrm{odfib}(F_6+2)=1000101_2=2^6+101_2=2^6+\mathrm{odfib}(2).$$

$$Z(F_n+k)=F_{n+2}+Z(k).$$

$$Z(F_n+k)$$

$$={\mathrm{fib}}^{-1}(2^n+\mathrm{odfib}(k))$$

$$=F_{n+2}+{\mathrm{fib}}^{-1}(\mathrm{odfib}(k))$$

$$=F_{n+2}+Z(k).$$

$$Z(10)=Z(F_6+2)=25=21+4=F_8+Z(2).$$

$$Z(F_n+k)-Z(F_n+k-1)=Z(k)-Z(k-1)$$

$$Z(F_n+1)=F_{n+2}+1.$$

$$\lfloor (F_n+k)\varphi \rfloor -\lfloor (F_n+k-1)\varphi \rfloor$$

$$=\lfloor k{\varphi }^2\rfloor -\lfloor (k-1){\varphi }^2\rfloor$$

$$=\lfloor k\varphi \rfloor -\lfloor (k-1)\varphi \rfloor$$

$$\lfloor (F_n+1)\varphi \rfloor +F_n=\lfloor (F_n+1){\varphi }^2\rfloor -1=Z(F_n+1).$$

$$(-1{)}^n{\tau }^n=-F_n\tau +F_{n-1},$$

$$\{F_n\varphi \}=\{F_n\tau \}=\{-(-\tau {)}^n\}$$

$$\lfloor (F_n+k)\tau \rfloor -\lfloor (F_n+k-1)\tau \rfloor =\lfloor k\tau \rfloor -\lfloor (k-1)\tau \rfloor .$$

Full text

The work of the third author was partially supported by a grant from the Simons Foundation (#245307).

1. Background

The Fibonacci numbers $\{F_{n}\}_{n\geq 0}$ are given by $F_{0}=0$, $F_{1}=1$, and

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for $n\geq 2.$

Recall the following theorem of Zeckendorf [1]:

Zeckendorf’s theorem**.**

Every positive integer can be written uniquely as the sum of distinct, nonconsecutive Fibonacci numbers.

The Zeckendorf representation $z(n)$ of $n\in{\mathbb{N}}$ is the unique $k$-tuple of decreasing nonconsecutive Fibonacci numbers whose sum is $n$. Note that although $F_{2}=F_{1}=1$, we always associate $1$ with $F_{2}$ in the Zeckendorf representation.

For example,

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and

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The sequence $\{\mathrm{fib}(n)\}_{n\geq 1}$ of fibbinary numbers111According to [3], the name “fibbinary” is due to Marc LeBrun: “…integers whose binary representation contains no consecutive ones and noticed that the number of such numbers with $n$ bits was $F_{n}$.”—posting to sci.math by Bob Jenkins, July 17, 2002. is given as follows: For $n>0$, if $z(n)=(F_{i_{1}},F_{i_{2}},\dots F_{i_{k}})$ is the Zeckendorf representation of $n$, then

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For example,

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where the $2$ subscript indicates the usual binary (base two) representation.

2. Statement of main result relating odd fibbinaries to the golden ratio

It is easy to generate the odd fibbinary numbers from the binary representations and find the corresponding value of the original integer, e.g.,

• •

$1_{2}=2^{0}=\mathrm{fib}(F_{2})=\mathrm{fib}(1)$;

• •

$101_{2}=2^{2}+2^{0}=\mathrm{fib}(F_{4}+F_{2})=\mathrm{fib}(4)$;

• •

$1001_{2}=2^{3}+2^{0}=\mathrm{fib}(F_{5}+F_{2})=\mathrm{fib}(6)$;

• •

$\ldots$.

Let $\mathrm{odfib}(n)$ denote the $n$th odd fibbinary number, i.e.,

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Let

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so that

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In other words, if the $n$th odd fibbinary number is the $j$th fibbinary number, then $j=Z(n)$.

This sequence

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appears to be “A003622” in OEIS [2], defined as

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where

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is the golden ratio, which satisfies $\phi^{2}=\phi+1$, and of course arises as

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The correspondence displayed above is naturally conjectured to be true in general.

Theorem 2.1**.**

Let $j$ be a positive integer such that the $j$th fibbinary number is odd. Suppose that this $j$th fibbinary number is the $n$th odd fibbinary number. Then

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for any $n\geq 1$.

3. Proof of Theorem 2.1

It is easy to check that (2) is true for small values of $n$. In the rest of this note we provide a general proof for $n\geq 3$.

Next, we record the following observation on $\mathrm{odfib}(n)$ as a lemma.

Lemma 3.1**.**

For any $1\leq k\leq F_{n-1}$ and $n\geq 2$, we have

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Proof.

Immediate as each digit in the binary representation of $\mathrm{odfib}(n)$ corresponds to a specific Fibonacci number. ∎

For example,

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Lemma 3.2**.**

For any $1\leq k\leq F_{n-1}$ and $n\geq 2$, we have

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Proof.

Since each digit in the binary representation of $\mathrm{odfib}(n)$ corresponds to a distinct Fibonacci number in the sum of $Z(n)$ under the Zeckendorf representation, we can claim the same for $Z(n)$. That is,

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∎

For example,

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From Lemma 3.2, we have

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for $2\leq k\leq F_{n-1}$ and

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Now to show (2) for any $n$, we only need to show analogues of (3) and (4) for $\left\lfloor n\phi^{2}\right\rfloor-1$, i.e.,

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for $2\leq k\leq F_{n-1}$, and

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Remark**.**

Intuitively, this can be considered as using $Z(F_{n}+1)$ as the “stepping stone” to prove (2) for $Z(F_{n}+k)$ for $k=2,3,\ldots,F_{n-1}$ using induction.

In order to establish (5), it is essentially sufficient to show that $\{F_{n}\phi\}$ is never large enough to affect the difference $\left\lfloor k\phi\right\rfloor-\left\lfloor(k-1)\phi\right\rfloor$, where $\{x\}:=x-\lfloor x\rfloor$, the fractional part of the real number $x$.

We make use of the following fact, which is easily established by induction in $n$:

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where $\tau=\frac{\sqrt{5}-1}{2}=\phi-1$ satisfying $\tau^{2}=-\tau+1$. Consequently,

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for any $n$.

Making use of the fact that $\tau=\phi-1$, it suffices to show

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To show (8), simply consider

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for any $1\leq k\leq F_{n-1}$. We will show that this value never reaches 1 or goes below zero and hence $\{F_{n}\tau\}$ will not affect $\left\lfloor k\tau\right\rfloor-\left\lfloor(k-1)\tau\right\rfloor$.

(i) To show $\{k\tau\}+\{F_{n}\tau\}<1$, consider the Zeckendorf representation of $k$ as the sum of non-consecutive Fibonacci numbers.

If $k<F_{n-1}$, then

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where

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Then

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If $k=F_{n-1}$, we have

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for any $n\geq 3$.

(ii) To show $\{k\tau\}-\{F_{n}\tau\}>0$, simply note that

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if $k<F_{n-1}$ and

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if $k=F_{n-1}$.

Cases (i) and (ii) imply that

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Thus (8) and (5) are proved.

By (4), (6) is equivalent to

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Fact (7) implies that

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for $n\geq 3$. Thus (9) and hence (6) is proved. ∎

Acknowledgements

We thank the anonymous referee for carefully reading the manuscript and providing helpful suggestions.