Talk:Fibonacci sequence

Early discussion

The main page says:

A logarithmic spiral can be approximated as follows: start at the origin of the cartesian coordinate system, move F(1) units to the right, move F(2) units up, move F(3) units to the left, move F(4) units down, move F(5) units to the right etc.

Is it not the case that a logarithmic spiral can be approximated just as well by moving aⁿ units at step n for any fixed a whatsoever? For example, I might move 1 unit right, 2 units up, 4 units left, 8 units down, 16 units right, etc., and get an approximation to a logarithmic spiral also? In which case, this is not a property of the Fibonacci numbers except by virtue of the fact that they happen to approximate a geometric sequence.

The main page continues:

This is similar to the construction mentioned in the golden mean article. Fibonacci number s often occur in nature when logarithmic spirals are built from discrete units, such as in sunflowers or in pine cones.

The pine cone thing (phyllotaxis) is related to the Fibonacci numbers, or, more precisely, to the golden ratio. The arrangements of pine cone bracts and other plant parts into Fibonacci spirals can be explained by a number of arguments; see for example pp. 408-412 of A New Kind of Science.

I am going to remove the logarithmic spiral remark from the main page unless someone corrects me soon.

Dominus 09:03, 6 Nov 2003 (UTC)

Well, if you really want to get pernickety about it, it does say a logarithmic spiral, not the or all...I can't rememer exactly, but these logarithmic spirals involve the golden ratio, and are connected to the Fib numbers.

Just some thoughts :) Dysprosia

I removed the logarithmic spiral discussion. I was sad to remove the illustration of the approximation to the logarithmic spiral, because it was very attractive. I hope it can find a more apt home somewhere else.

I will add back discussion of the appearance of Fibonacci numbers in sunflowers, pine cones, pineapples, daisies, and so forth. I have something about that written up already that I think could be adapted for inclusion in this article without much trouble. -- Dominus 14:24, 11 Mar 2004 (UTC)

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Isn't the first fibonacci number 1 not 0? //Wellparp

Calculate F_2 by that definition in the article, and with F_0 = 1 in the way you are thinking of and you will see that the two sequences are basically the same. Yet, I will make clear of this in the article. Dysprosia 13:30, 10 Nov 2003 (UTC)

It's all a matter of convention. Having F(0)=0 makes several properties easier to write and more beautiful. For instance, if n divides m then F(n) divides F(m). --FvdP

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I removed this paragraph and link (sorry Alex):

Recursive computation of very long Fibonacci numbers in C++ with using STL features can be seen at http://groups.google.com/groups?selm=bnni5p%2412i47o%241%40ID-79865.news.uni-berlin.de. The algorithm was used to calculate Fibonacci [5,000,000] (see http://alexvn.freeservers.com/s1/fibonacci.html)

• the algorithm is uninteresting (it computes the numbers in sequence F1, F2, ...)
• the Bignum implementation is strictly limited to the needs of the above algorithm, i.e. not much;

Perhaps it may be interesting to someone who wants to compute Fibonacci numbers, who knows, but then a 5-line Python_programming_language program would do the same computations (at much the same speed)...

a = 0 # or 0L on earlier Python versions
b = 1 # or 1L on earlier Python versions
while 1:
    print a
    a,b = b,a+b

--FvdP 20:28, 19 Nov 2003 (UTC)

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Before we'll know it there will be a Fibonacci number program in 137 programming languages in this article. But this article is about math folks, not about you favorite programming language! So I've created a new Fibonacci number program article especially for you, and moved the two sample program from this article there. Lots of work to do guys, at least 135 programs to go!
—Herbee 02:41, 2004 Mar 11 (UTC)

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Phyllotaxis

I put a discussion of the appearance of Fibonacci numbers in Phyllotaxis (the distribution of leaves on a plant stem) at Talk:Fibonacci number/Phyllotaxis. The style isn't quite right for an encyclopedia article, but I think large chunks of it could be extracted and added to the Fibonacci article, or maybe even into a separate discussion of the Fibonacci number in phyllotaxis. -- Dominus 15:33, 11 Mar 2004 (UTC)

Looks good. Abigail 15:44, Mar 11, 2004 (UTC)

True or False??

True or false: It has been proven that 149 is the largest prime Tribonacci number.

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INCORRECT INFORMATION ON THE PAGE

Please review the page, the Fibonacci sequence begins as: 1, 1, 2, 3, 5, 8, ... NOT 0, 1, 1, 2, ...!!! This needs to be corrected everywhere in the page...

Hmmm ... Fibonacci probably didn't have zero. But then, that means he would start with F(1) = F(2) = 1. So, if we start with F(0) = 0, wouldn't that be OK?

Charles Matthews 17:47, 30 Apr 2004 (UTC)

He did indeed have zero. The Fibonacci sequence is so-called because it was introduced by Fibonacci in his 1202 book Liber Abaci, which is about how to calculate with Hindu-Arabic numerals. In fact, it was this book that popularized the Hindu-Arabic numerals in Europe and introduced the zero to wide usage in Europe.

He started the sequence with F(1) = 1 and F(2)=2, contrary to your suggestion. Here's the relevant section of the book, taken from L. E. Sigler's recent translation:

A certain man had one pair of rabbits together in a certain enclosed place, and one wishes to know how many are created from the pair in one year when it is the nature of them in a single month to bear another pair, and in the second month those born to bear also.

Because the abovewritten pair in the first month bore, you will double it; there will be two pairs in one month. One of these, namely the first, bears in the second month, and thus there are in the second month 3 pairs; of these in on month two are pregnant, and in the third month 2 pairs of rabbits are born, and thus there are 5 pairs in the month...

You can indeed see in the margin how we operated, namely that we added the first number to the second, namely the 1 to the 2, and the second to the third, and the third to the fourth, and the fourth to the fifth, and thus one after another until we added the tenth to the eleventh, namely the 144 to the 233, and we had the abovewritten sum of rabbits, namely 377, and thus you can in order find it for an unending number of months.

There is also a tabulation in the book, but I don't remember whether it begins with "1, 1, 2, 3" or with "1, 2, 3"---I suspect the latter. Leonardo was almost certainly unaware of such identities as GCD(F(a), F(b)) = F(GCD(a, b)) for which the zero-indexing is important.

I hope this is helpful, or at least interesting. -- Dominus 15:49, 18 October 2005 (UTC)

F(0) = 0 Yes. [[1]]

The Fibonacci sequence can be continued towards negative infinity: F(−1) = 1, F(−2) = −1, F(−3) = 2, … So there is no natural "begin" to the sequence. If one wants the sequence to begin somewhere, the starting index is a matter of definition— it cannot be "correct" or "incorrect".
—Herbee 00:06, 2004 May 10 (UTC)

This discussion results from the anon's incorrect sentiment that Fibonacci's sequence has remained unexplored and should remain so, forever being presented as Fibonacci presented it 800 years ago. Not only is this bad math, science and history, it's just a plain bad idea. Tomer^talk 22:03, 27 March 2006 (UTC)

Miles to km conversion paragraph removed

I removed the paragraph about miles to kilometers conversion, I thought it of dubious value and potentially misleading. A5 02:05, 1 Sep 2004 (UTC)

Proven irrational??

Has the sum 1 + 1 + 1/2 + 1/3 + 1/5 + 1/8 + 1/13 + 1/21 + 1/34... (which is equal to 3.35988566624...) known to be irrational as of 2004?? 66.245.105.195 02:11, 9 Oct 2004 (UTC)

Yes, according to http://mathworld.wolfram.com/ReciprocalFibonacciConstant.html, this was proved irrational in 1989. Jwwalker 07:02, 13 February 2006 (UTC)

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H'm. "Fibonacci numbers can be obtained from the first term alone, by rounding to the nearest integer" yet "Computing Fibonacci numbers by computing powers of the golden mean is not very practical...since rounding errors will accrue and floating point numbers usually don't have enough precision." Needs elaboration, like, "if you need exact values beyond your machine's float precision, use the matrix method with bignums..." 'cause rounding errors aren't much of a problem if you don't even have to do the subtraction. Kwantus 04:27, 2004 Oct 10 (UTC)

Formula

While these nifty bignum formulas are nice and all, I think it would be very nice to have the actual formula for calculating them. The math isn't that hard to do by hand, because of cancelling pieces. I think that the article needs it because it is the non recursive form of it.

Gopala / Hemachandra

I've removed the clause from the introduction that says that the numbers are also called the "Gopala-Hemachandra numbers". The page already mentions that Fibonacci was anticipated by Gopala and Hemachandra, and I find no evidence that the numbers are actually called the "Gopala-Hemachandra numbers".

I'm also going to redirect the Gopala-Hemachandra numbers article to this one, since the two phrases mean the same thing and that article contains nothing that isn't already in this one.

-- Dominus 14:16, 11 Nov 2004 (UTC)

Addendum: even the external research paper linked to from the Gopala-Hemachandra numbers page does not refer to the numbers as the "Gopala-Hemachandra numbers". It says "The numbers in the sequence are called Fibonacci numbers." The phrase "Gopala-Hemachandra numbers" does not appear in that paper.

-- Dominus 14:18, 11 Nov 2004 (UTC)

Benford's law

The article has a section describing in details the conformance to Benford's law. But this conformance isn't surprising, (or, if it is surprising, the article doesn't say why) and since many collections of numbers conform to Benford's law, it's not clear what this is doing here. The detailed statistics about the distribution of initial digits in Fibonacci numbers does not seem to me to have any intrinsic interest, and smacks of original research. Am I missing something? -- Dominus 14:54, 10 Jan 2005 (UTC)

I was wondering that myself. I personally don't find Benford's law particularly relevant. And if it's relevant, couldn't the application of Benford's law to the Fibonacci sequence (or any other prominent integer sequence) be discussed in the article on Benford's law instead? --MarkSweep 15:55, 10 Jan 2005 (UTC)

My thoughts exactly. If there is no dissenting opinion in a day or two, I will remove the section. -- Dominus 16:09, 10 Jan 2005 (UTC)

I'd say if you want to remove the section and merge it with Benford's law, there's no loss of information and no need to wait. --MarkSweep 17:24, 10 Jan 2005 (UTC)

You can do what you want, but I'm going to wait. -- Dominus 18:44, 10 Jan 2005 (UTC)

F(n) or F_n?

I would prefer if this page used the latter format. Subscript is most common for integer sequences within Wikipedia and elsewhere, and certainly the most common for denoting the Fibonacci sequence. Fredrik | talk 29 June 2005 18:27 (UTC)

Both notations should be mentioned as both are common. Convenience and legibillity of formulae should dictate further use. --MarSch 11:12, 30 August 2005 (UTC)

I agree with MarSch. I find it strange that so many people seem to wish to enforce a nonexistent convention. -- Dominus 13:10, 31 August 2005 (UTC)

Tiling picture sloppy

The tiling picture of the Fibonacci series is rather sloppy. To do it right, each next box, should be adjacent to the previous two, to clearly show the construction. As it is now, the construction is not consistent per step. Box for nr 5 adjoins 1, 1 and 3. It should be on the upper side, along boxes 2 and 3. Then 8 can be two the right of 3 and 5 etc.

Or, alternatively, it should be consistently drawn on the opposite side of the previous two. That way the centers of the boxes follow show the Fibonacci spiral. −Woodstone 16:46, 24 September 2005 (UTC)

Relative Prime

The section below was moved from the main page, because it is incomprehensible as it stands. Can be re-inserted after thorough rework. −Woodstone 14:51, 18 October 2005 (UTC)

"All consecutive Fibonnaci numbers are relatively prime, this is to say that no two consecutive Fibonnaci numbers have a greatest common divisor greater than one. A proof is as follows:

Consider the following facts, if a and b have a common factor then it must also be a factor of a+b. If a and b have a common factor then it is also a factor of b-a if a and b have no common factor then neither do b and a+b. So let us say that the fibonnaci sequence starts with f(1)=a f(2)=b if a and b are relatively prime so are all consecutive numbers in the series. f(1)=1 and f(2)=1 have no common factor other than one therefore no neighbouring pairs in the Fibonnaci series have a common factor."

I see what the author is getting at, although I agree with you that it's a mess. I'll try to write something briefer and clearer and put it into the article. -- Dominus 15:31, 18 October 2005 (UTC)

Done. I hope you like this version better than what was there before. Thanks for pointing this out. -- Dominus 15:41, 18 October 2005 (UTC)

the sequence in music

1 Black 1 then 2 white are 3 all I see 5 in my infancy. 8 red and yellow then came to be, 5 reaching out to me. 3 lets me see. 13 As below, so above and beyond, I imagine 8 drawn beyond the lines of reason. 5 Push the envelope. 3 Watch it bend.

-Lateralus -TOOL

he sings the sylabols.

http://toolshed.down.net/lyrics/latmaster.html cite

Christa rim 06:30, 22 October 2005 (UTC)

Nice. It should also be noted that BT used the Fibonacci sequence in his song--wait for it-- "Fibonacci Sequence". Note that this was not referenced on the main page at all, which is immensly stupid. //user:DJRaveN4x 12:11 EST, 28 December 2005

Bee generations

The statement about the bee generations is incorrect. It assumes that no parentage can be shared. However it is quite likely that two grandmothers of a bee are sisters, so no new parents are added going backwards. −Woodstone 12:59, 24 October 2005 (UTC)

Formula of finding fibonacci number

lets say a = sqrt(5), then: F(n) = ((a+1)^(n+1)-(a-1))/(2^(n+1)*a)

Roulette

The roulette page mentions that the fibonacci numbers can be used as a method of gambling. No chance of a mention on this page? (And an extrapolation - including why it is flawed?) --JohnO You found the secret writing! 08:24, 21 February 2006 (UTC)

spirals

Why is the conch picture here? Is the ratio in size between successive turns known to be τ? If not, let's avoid encouraging the popular misconception that all logarithmic spirals have something to do with Fibonacci numbers. —Tamfang 19:30, 21 February 2006 (UTC)

Pseudocode Computer Programs

The current examples of pseudocode are sloppy.

They both will give invalid results for negative input.

That's right, but a check for negative input is not common in CS textbooks. Just assume the input type is natural/unsigned. Qwertyus 14:12, 18 March 2006 (UTC)

The second example shows a line

a,b := a+b,a

I would say that should be

a,b := b,a+b

And it is not elegant either because it calculates one number too many (b is the next one). −Woodstone 21:51, 17 March 2006 (UTC)

It now follows SICP. I'll add a reference. Qwertyus 14:12, 18 March 2006 (UTC)

I see you've changed it; but you've put the inelegance that you complained about back in :) Qwertyus 13:05, 10 April 2006 (UTC)

I did not put it back, because it was still there. There is no difference in cycles though the loop. In the previous form "a" was the excessively computed value, better hidden, but still present. But I'm glad you seem to agree it's not elegant. How about:

var a,b = 0,1
if n=0 return a
while n>1 
  a,b = a,a+b
  n=n-1
return b

Does that strike you as more elegant? −Woodstone 19:00, 10 April 2006 (UTC)

No, because it's not as short as it could be. Elegance lies not in saving a small constant number of processor cycles. Qwertyus 19:07, 10 April 2006 (UTC)

I fully agree. Elegance is much more than that. That's why I didn't put it in. However I still maintain that the show program is not elegant because it computes non-required data. My improvement was just making it slightly more logical, the values a,b being successive results in the right sequence. −Woodstone 22:03, 10 April 2006 (UTC)

I've found a more elegant version at Data Structures and Algorithms with Object-Oriented Design Patterns in C++ and inserted a pseudocode version. I think this is as elegant as it gets. It uses the idiosyncratic loop construct repeat n times because otherwise we'd have to use a counting var, as the referenced book/site does. Using while n≥0, as I did at first, makes the counter stop at -1, thus requiring it to be in signed representation. Qwertyus 23:16, 10 April 2006 (UTC)

The stated version was incorrect because it shifts the result by 1 index number. Now corrected, but it brings back the inelegance of more work then needed (now shifted from the end to the beginning). But surely the "repeat" idea makes it more elegant than the while and decrement. However the trickery for start-up, makes it less obviously correct, so I'm not convinced it's more elegant altogether. I'm inclined to go back to

function fib(n)
var a,b = 0,1
repeat n times
  a,b = a,a+b
return a

−Woodstone 10:22, 11 April 2006 (UTC)

Another question: what is the complexity of algorithms derived from the closed form? When using exponentiation by squaring, I'd say I would be O($\log n$), and thus not asymptotically better than the algorithm derived from the matrix form. Qwertyus 20:33, 12 April 2006 (UTC)

Can you explain the math?

I don't understand the last step of the derivation in the relaytion to goldn ratio section. Can someone explain it to me?

How does

$1+{\frac {1}{\displaystyle \lim _{n\to \infty }{\frac {F(n)}{F(n-1)}}}}$

equal

$1+{\frac {1}{x}}$

when

$x={\lim _{n\to \infty }{\frac {F(n+1)}{F(n)}}}$

?

Thank you. Loom91 12:26, 28 March 2006 (UTC)

It amounts to a change of variable: $\lim _{n\to \infty }{\frac {F(n)}{F(n-1)}}$ is the same as $\lim _{m\to \infty }{\frac {F(m+1)}{F(m)}}$ where $m=n-1$.

If that explanation is insufficient you can expand the limit statement in terms of the $\delta$ and $\epsilon$ definition of a limit. Roger Hui 15:36, 28 March 2006 (UTC)

so you are saying $(F(n))^{2}=F(n-1)F(n+1)$? Doesn't this require an explanation? Loom91 10:33, 29 March 2006 (UTC)

- No, that is not what I am saying. The two limits are both on the ratio of consecutive elements of a sequence, whether it's the Fibonacci sequence or some other sequence. There is no difference between $\lim _{n\to \infty }a_{n}$ and $\lim _{n\to \infty }a_{n+1}$ and $\lim _{m\to \infty }a_{m}$ . Roger Hui 16:19, 29 March 2006 (UTC)

Thanks for the explanation, now I get it. But I think someone should make this clearer in the article. Loom91 09:28, 30 March 2006 (UTC)

indexing

68.155.181.96 says "a set cannot start with an index of zero". Must be a Fortran programmer! —Tamfang 06:58, 31 March 2006 (UTC)

Fibonacci primes

I have removed the following contrbution from an anonymous editor from the main article, as I think it is very unclear, and the final sentence sounds like original research:

However, since Fib(p) can share no common divisors exceeding one, with F(n<p).

Using the identity:

http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibmaths.html#fibprimecarm

GCD(Fn, Fm) = FGCD(m,n)

Proof:

Since if we are saying that m is a prime p, and n is less than p, then the

GCD(Fp, Fn ) = F(GCD(p,n)) = F(1) = 1

Which means that all F(p), will be a product of all new prime factor/s, also called characteristic factor/s (Charmichael's theorem), or primitive factor/s. It is reasonable so assume that, F(p) is more likely to be a single prime factor (itself) because the field of potential factors at random has been reduced.

If anyone can sort this tangle out, please feel free to copyedit it and re-insert it. Gandalf61 09:47, 4 April 2006 (UTC)

The section was re-inserted without change, so I have removed it again. My specific objections to this section, as written, are:

1. It is ungrammatical. For example, the first sentence has no main clause.
2. It uses inconsistent notation - Fib(n), F(n) and Fn all seem to be used for the Fibonacci sequence.
3. It does not give adequate references - what is "Charmichael's theorem", for example ?
4. Its final sentence sounds as if it is an unverified claim, or at best original research.

If all these errors can be fixed, then of course the section could be re-inserted. Gandalf61 13:18, 5 April 2006 (UTC)

What errors? It's rude to have removed this just because you don't understand. This is well known Gandalf61!

A. All prime Fibonacci numbers have a prime index, "p". The converse is not true, because not all Fibonacci with a prime index are prime.

B. Fibonacci numbers with a prime index "p" do not share any common divisors exceeding one, with preceeding Fibonacci numbers, due to the well known identity:

GCD(F(n), F(m)) = F(GCD((m),(n)))

The greatest common divisor between any two Fibonacci numbers, is equal to the greatest common divisor between their index values.

Simple Proof: If m is a prime p from the identity above, and n is less than p, then it is clear that F(p), cannot share any common divisors with preceeding Fibonacci numbers.

GCD(F(p), F(n)) = F(GCD((p),(n))) = F(1) = 1

C. Charmichaels' theorem states, that for every Fibonacci number there will always be at least one unique prime factor, that has not been a factor of a preceeding Fibonacci number.

By combining B&C, a Fibonacci number with a prime index, will be a product of all new prime factor/s called characteristic factor/s(Charmichael's theorem). There is nothing to rule out one single prime factor, as stated by C.

1. The sentences now have a main clause with a subject, and a verb.
2. Notation is fixed, and could be by anyone, rather than deleted.
3. Charmichael's theorem is stated, and there had been a link, but I guess you could not follow it as a reference.
4. The final sentence just words it correctly, if you call that original.

• Do you mean Carmichael's theorem rather than Charmichaels' theorem? Roger Hui 15:47, 9 April 2006 (UTC)

Well, I don't think the original contributor means this Carmichael's theorem, which agrees with the definition of Carmichael's Theorem in this mathworld entry. Maybe he means a different Carmichael's Theorem ? Maybe this is where he gets the statement GCD(F(n), F(m)) = F(GCD((m),(n))) from ?? This is exactly why this tangle need more clarity. Gandalf61 10:01, 10 April 2006 (UTC)

Fixed this. Carmichael's theorem now describes the theorem that the original contributor was refering to (it was previously a redierct to Carmichael function). Gandalf61 10:44, 11 April 2006 (UTC)

There is a spelling error. That identity is not a made up statement, but is well known. You can find a link and post it, but dont delete this again. http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci/fibmaths.html#primefactor Oh you and your toil and tangle, you british....errrgh As you can tell I don't like you personally now. A am Shane Findley, and this was published in the Fibonacci quarterly. You can also do a google search and find Shane On the Mizar Encoding of the Fibonacci numbers.

I have now included an external link, to the greatest common divisor identity.

Divineprime, I suggest we continue this discussion on your talk page. Gandalf61 08:39, 11 April 2006 (UTC)

Common factors

Any two consecutive Fibonacci numbers are relatively prime. Suppose that Fn and Fn+1 have a common factor g. Then Fn+1 − Fn = Fn−1 similarly. Thus by induction, all preceding Fibonacci number must be a multiple of g. But F1 = 1, so g = 1.

Hello what's this? What does the induction mean? None of this is clear at all, I think I'll delete it then. This needs to be fixed or reworded. — Preceding unsigned comment added by 4.233.134.240 (talk • contribs) 12:29, 11 April 2006 (UTC)

What is unclear about it? It seems to me a perfectly correct proof, by descending induction, albeit exceedingly simple. −Woodstone 21:07, 11 April 2006 (UTC)

Perhaps noname doesn't know what mathematical induction is. —Tamfang 23:13, 12 April 2006 (UTC)

Perhaps Gandolf61 does not know what mathematical induction is.

Noname — Preceding unsigned comment added by 4.233.149.161 (talk • contribs) 13 apr 2006 01:40 (UTC)

Please people, this is no way of arguing. Here's my attempt at clarifying the passage by making it more explicit:

Any two consecutive Fibonacci numbers are relatively prime. Let $g$ be a common factor of $F_{n}$ and $F_{n+1}$, i.e. there exist positive integers $a$ and $b$ such that $F_{n}=ag$ and $F_{n+1}=bg$. Then

$F_{n-1}=F_{n+1}-F_{n}=bg-ag=(b-a)g$,

i.e. $F_{n-1}$ is a multiple of $g$. By descending induction all lower Fibonacci numbers must be multiples of $g$. But $F_{0}=0$ and $F_{1}=1$, and 0 and 1 are relatively prime, so $g=1$.

But maybe this is too long and explicit? Qwertyus 12:22, 13 April 2006 (UTC)

Why bring zero into it? g must be a factor of 1, that's sufficient. Speaking of the (relative) primeness of zero makes me nervous. —Tamfang 17:04, 14 April 2006 (UTC)

"g must be a factor of 1" (equiv. g=1) is what I'm trying to prove there.

What makes you nervous about it? gcd(0,1) = 1 because 1 is the greatest n such that 1|0 and 1|1. Also, try calculating it using Euclid's algorithm. Qwertyus 18:31, 14 April 2006 (UTC)

I also think it is perfectly clear as written, and that the proposed replacement is considerably less so. Making it "more explicit", and so longer and more convoluted and turgid, does not make it clearer, in my opinion. -- Dominus 02:30, 15 April 2006 (UTC)

Personally, I don't think it's "convoluted and turgid", it clearly and precisely states the steps of the proof. Qwertyus 02:34, 15 April 2006 (UTC)

I think the version above is not very precise. The statement "But F0 = 0 and F1 = 1, and 0 and 1 are relatively prime, so g = 1." is not the right way to close the proof. As Tamfang stated above, the relative primeness of 0 and 1 is not directly relevant. The proof should continue: so g is a factor of 1 and consequently g=1 (ignoring the sign). So the only common factor between consecutive Fibonacci numbers is 1 and they are relative prime. Also, talking about "lower" Fibonacci numbers is unnecessarily assuming they are ascending (true, but why use it?). Earlier versions used the more accurate "preceding" (in the sequence). −Woodstone 10:32, 15 April 2006 (UTC)

Broccoli?

"Romanesco Broccoli clearly illustrating the fractal repeating spiral pattern found within nature"

I've removed the image to the right from the article, as it makes no claim to have anything to do with Fibonacci numbers. In fact, the whole "Fibonacci numbers in nature" section needs citations. Melchoir 22:31, 23 April 2006 (UTC)

HI!! THIS WAS CONNOR!!!! XD IM SPAMMING!!!