Ma thuật Toán học với các số Fibonacci

The Mathematical Magic of the Fibonacci Numbers (Ma thuật Toán học với các số Fibônaci) – Dr Ron Knott

This page looks at some patterns in the Fibonacci numbers themselves, from the digits in the numbers to their factors and multiples and which are prime numbers. There is an unexpected pattern in the initial digits too. We also relate Fibonacci numbers to Pascal’s triangle via the original rabbit problem that Fibonacci used to introduce the series we now call by his name. We can also make the Fibonacci numbers appear in a decimal fraction, introduce you to an easily learned number magic trick that only works with Fibonacci-like series numbers, see how Pythagoras’ Theorem and right-angled triangles such as 3-4-5 have connections with the Fibonacci numbers and then give you lots of hints and suggestions for finding more number patterns of your own.

Take a look at the Fibonacci Numbers List or, better, see this list in another browser window, then you can refer to this page and the list together.

Patterns in the Fibonacci Numbers

The Final Digits

Here are some patterns people have already noticed in the final digits of the Fibonacci numbers:

• Look at the final digit in each Fibonacci number – the units digit:
  0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, …

  Is there a pattern in the final digits?

  0, 1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, 4, 3, 7, 0, 7, …

  Yes!
  It takes a while before it is noticeable. In fact, the series is just 60 numbers long and then it repeats the same sequence again and again all the way through the Fibonacci series – for ever. We say the series of final digits repeats with a cycle length of 60.

• Suppose we look at the final two digits in the Fibonacci numbers. Do they have a pattern?
  0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, …

  Yes, there is a pattern here too. After Fib(300) the last two digits repeat the same sequence again and again. The cycle length is 300 this time.

  So what about the last three digits?
  and the last four digits?
  and so on??

• For the last three digits, the cycle length is 1,500
• for the last four digits,the cycle length is 15,000 and
• for the last five digits the cycle length is 150,000
• and so on…

Digit Sums

Michael Semprevivo suggests this investigation for you to try.
If we add all the digits of a number we get its digit sum.

Find Fibonacci numbers for which the sum of the digits of Fib(n) is equal to its index number n:
For example:-

Fib(10)=55

the tenth Fibonacci number is Fib(10) = 55.
The sum of its digits is 5+5 or 10 and that is also the index number of 55 (10-th in the list of Fibonacci numbers). So the index number of Fib(10) is equal to its digit sum.

Fib(11)=89

This time the digit sum is 8+9 = 17.
But 89 is not the 17th Fibonacci number, it is the 11th (its index number is 11) so the digit sum of 89 is not equal to its index number.

Here are two more examples of the numbers we seek: Fib(1)=1 and Fib(5)=5.
There is also one more whose index number is less than 10 – what is it?

Can you find any more in this table of Fibonacci numbers up to Fib(300)?
As a check, you should be able to find TEN (including those above) up to Fib(200).
How many are there up to Fib(300)?
This makes a nice exercise in computer programming so the computer does the hard work.

A more difficult question is Does this series (of Fibonacci numbers which have a digit sum equal to their index number) go on for ever?

Robert Dawson of Saint Mary’s University, Nova Scotia, Canada summarizes a simple statistical argument (originally in the article referred to below by David Terr) that suggests there may be only a finite number (in fact, just 20 numbers) in this series:

“The number of decimal digits in Fib(N) can be shown to be about 0.2 N, and the average value of a decimal digit is (0+1+…+8+9)/10 = 4·5. Thus, unless the digits of Fibonacci numbers have some so-far undiscovered pattern, we would expect the digit sum to be about 0.9 N. This falls further behind N as N gets larger. Fib(2222) (with 465 digits) is the largest known Fibonacci number with this property. There are no others with N<5000, and it seems likely that Fib(2222) is actually the largest one. However, no proof exists!”

A new research question for you to try

If you want to try a brand new investigation, how about converting the Fibonacci numbers to a base other than 10 (binary is base 2 or undecimal is base 11, for example) and seeing what you get for the digit sums in different bases. Are there any bases where the Fibonacci numbers with a sum of their base B digits equal to their index numbers form an infinite series? In which bases is it a finite series?

On the Sums of Digits of Fibonacci Numbers David Terr, Fibonacci Quarterly, vol. 34, August 1996, pages 349-355.
Two series in Sloane’s Encyclopedia of Integer Sequences are relevant here: A020995 for the index numbers and A067515 for the Fibonacci numbers themselves.

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..

Factors of Fibonacci Numbers

There are some fascinating and simple patterns in the Fibonacci numbers when we consider their factors. You might like to click here to open a new browser window which shows the first 300 Fibonacci numbers and their factors. It will be helpful in the following investigations:

So every Fibonacci number is a factor of (an infinite number of) Fibonacci numbers, that is:

Fibonacci numbers as Factors of Fibonacci numbers

	i	3	4	5	6	7	8	9	10	11	12	…
	Fib(i)	2	3	5	8	13	21	34	55	89	144	…
F a c t o r s	2=Fib(3)											Every 3rd Fib number
	3=Fib(4)											Every 4th Fib number
	5=Fib(5)											Every 5th Fib number
	8=Fib(6)											Every 6th Fib number
	F(k)	…										F(all multiples of k)

Putting this into words we have:
Every 3-rd Fibonacci number is a multiple of 2 i.e. a multiple of F(3)
Every 4-th Fibonacci number is a multiple of 3 i.e. a multiple of F(4)
Every 5-th Fibonacci number is a multiple of 5 i.e. a multiple of F(5)
Every 6-th Fibonacci number is a multiple of 8 i.e. a multiple of F(6)

which suggests the general rule:

or, expressed mathematically,

A Primer For the Fibonacci Numbers: Part IX M Bicknell and V E Hoggatt Jr in The Fibonacci Quarterly Vol 9 (1971) pages 529 – 536 has several proofs that F(k) always divides exactly into F(nk): using the Binet Formula; by mathematical induction and using generating functions.

Every number is a factor of a Fibonacci number

But what about numbers that are not Fibonacci numbers?
Which other numbers exactly divide into (are factors of) Fibonacci numbers?
The surprising answer is that there is an infinite number of Fibonacci numbers with any chosen number as a factor!

For instance, here is a table of the first Fibonacci numbers that have each of the integers from 1 to 13 as a factor:

Here is a graph of the series, the index number j of the first Fibonacci number factor i for i from 1 to 200:

The series of index numbers j of the first Fib(j) for which each number is a factor:
1,3,4,6,5,12,8,6,12,15,10,12,7,24,20,12,9,12,18,30,8,30,… is Sloane’s A001177

The earliest reference for this result seems to be N N Vorob’ev in his Fibonacci Numbers booklet, published by Pergamon in 1961. It gives a proof that every whole number has a Fibonacci number for which it is a factor. Also, he showed that the first Fibonacci number with a factor i will be found within the first i² Fibonacci numbers and thereafter at double, treble etc that index number. The original version is in Russian, Chisla fibonachchi, 1951 and it is now again in print as Fibonacci Numbers, N N Vorob’ev, Birkhauser (Jan 2003).

The first Fibonacci number with a given prime as a factor

If we look at the prime numbers and ask when they first appear as a factor of a Fibonacci number, we find they do so within the first p+1 Fibonacci numbers. In this table, i is the index of the first Fibonacci number that has the prime as a factor:

The first-index numbers seem to be either equal to the prime (as for p=5) or one less (as p=11) or 1 more (p=7) in many cases. What about the others?
They are factors of p+1 (as in the case of p=17 where 9 is a factor of p+1=18) or of p–1 (e.g. p=29 where the index 14 is a factor of p-1=28).
This appears to be true in general.

This series on the lower row: 3,4,5,8,10,7,9,18,24,14,30,19,20,44,16,27,58,15,68,70,… is Sloane’s A001602. It is a subsequence of A001177 above, selecting the numbers at the prime positions.
Vajda (the full reference is at the foot of this page), on page 84 states:

Let F_u (u > 0) be the smallest Fibonacci number containing the prime p. The subscript u is called by Lucas the rank of apparition of p, and we know that it is a factor of, or equal to, p–1 or p+1.

Clark Kimberling has a brief biography of Lucas;
more information on Lucas at the St. Andrews MacTutor site.
On another page we look at a series of numbers similar to the Fibonacci numbers, called the Lucas numbers.
This section was suggested by an email from Allyn Shell.

Neighbouring Fibonacci Numbers have no common factors

You might already have noticed that no even Fibonacci number is next to another even Fibonacci number, or, no two neighbouring Fibonacci’s have a common factor of 2.
In the last section we saw that Fib(3)=2 so we would expect the even Fibonacci numbers (with a factor of 2) will appear every at every third place in the list of Fibonacci numbers. So perhaps we might have expected no even numbers to be neighbours in the Fibonacci series.

The same happens for a common factor of 3, since such Fibonacci’s are at every 4-th place (Fib(4) is 3).

In fact, there will not be a Fibonacci number as a common factor between two neighbouring Fibonacci’s for the same reason.

But what about other numbers as factors such as 6 or 7?

The answer is that no number is a factor of two neighbouring Fibonacci numbers. Well – that is not quite right, as 1 is always a common factor of course, but apart from this “obvious” case there are no common factors.

There is a proof of this that Tom E Ace wrote to me about — and it is so simple!

• 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
  for if B and A+B had a common factor, then their difference would too and the difference is just A.

Two numbers that have no common factors are called relatively prime (to each other).
So in any Fibonacci-type series which starts with A and B, if A and B are relatively prime then so are all pairs of consecutive numbers in te series.
Alternatively, if A and B have a common factor of k then so do B and A+B (the next pair in the series) and so on, so that this factor is a factor of all numbers in the series.

Since F(1)=1 and F(2)=1 have no common factor, then no neighbouring pairs in the Fibonacci series have a common factor.
We have just shown that

Now let’s look at Fibonacci numbers that have no factors at all, the prime Fibonacci numbers:

Fibonacci Numbers with Index number factor

Above we saw that F(nk) has both F(n) and F(k) as a factor, but what nk itself as a factor, that is:

For instance F(12)=144 which has 12 as a factor;
F(25)=75025 which has 25 as a factor;
but F(15)=610 which does not have 15 as a factor.

Here are the first few Fibonacci numbers F(n) with n as a factor:

The series of indices of such Fibonacci numbers is:

and is Sloane’s A023172.

More on Index numbers and Fibonacci Factors

If we look at the list of factors of Fibonacci numbers, some more patterns emerge:

If we try a simpler test, just choosing those Fibonacci numbers with a factor which is exactly (i+1) or (i–1) ignoring any other multiples of the index number, then we have two more series:

There are more patterns hidden here.

If so, let me know (click on Dr Ron Knott at the foot of this page for contact details) and I’ll try to include some of them here.

Fibonacci Primes

We have seen from investigations above that F(nk) is a multiple of F(k) for all values of n and k = 1,2,…
This means that if the subscript has factors (i.e. is composite, is not a prime) then so is that Fibonacci number — with one exception: can you find it?
So what about those Fibonacci numbers with no factors (apart from 1 and itself, of course)?
These are the Fibonacci numbers that are primes.
We can now deduce that

Any prime Fibonacci number must have a subscript which is prime
(again with one little exception – can you find it? Hint: you won’t have to search far for it . )

Unfortunately, the converse is not always true:
that is, it is not true that if a subscript is prime then so is that Fibonacci number.
The first case to show this is the 19th position (and 19 is prime) but
F(19)=4181 and F(19) is not prime because 4181=113×37.

In fact, a search using Maple finds that the list of index numbers, i, for which Fib(i) is prime begins as follows:

Now you should be able to spot the odd one out: that one number, i, which is not a prime in the list above, even though Fib(i) is.

The series continues (updated January 2007):

i	Fib(i)	Number of digits	Prime?
131	10663404174…72169	28
137	19134702400…23917	29
359	47542043773…76241	75
431	52989271100…62369	90
433	13872771278…68353	91
449	30617199924…65949	94
509	10597999265…29909	107
569	36684474316…65869	119
571	96041200618…74629	119
2971	35710356064…16229	621
4723	50019563612…91957	987
5387	29304412869…55833	1126
9311	34232086066…76289	1946
9677	10565977873…70357	2023
14431	35575535439…75869	3016
25561	38334290314…14961	5342
30757	30434499662…75737	6428
35999	99214776140…24001	7523
37511	96802910427…75089	7839	Jun 2005
50833	13159270824…02753	10624	Oct 2005
81839	97724940760…46561	17103
104911	5660323637…84189	21925	?
130021	2706998033…75321	27173	?
148091	6904738850…74809	30949	?
201107	3371962609…27913	42029	?
397379	8912712429…66921	83047	?
433781	3296782330…74981	90655	?
590041	8448035604…82641	123311	?
593689	2059052250…74289	124074	?
604711	5962634693…37389	126377	?

Those marked are definitely prime;
those marked? are probably prime but have not been proved so (explained here).

The largest known Fibonacci prime, F(81839) was reported in April 2001 by David Broadbent and Bouk de Water.
The series of index numbers i of the prime Fib(i): 3,4,5,7,11,13,17,23,29,43,47,83,… is Sloane’s A001605
The glossary entry on Chris Caldwell’s The Prime Pages under Fibonacci Prime has more information and references.

Fibonacci numbers and special prime factors

Every Fibonacci number is marked in a special way.
If we look at the prime factors of a Fibonacci number, there will be at least one of them that has never before appeared as a factor in any earlier Fibonacci number. This is known as Carmichael’s Theorem and applies to all Fibonacci numbers except 4 special cases:

1. Fib(1)=1 (has no prime factors),
2. Fib(2)=1 (has no prime factors),
3. Fib(6)=8 which only has prime factor 2 which is also Fib(3),
4. Fib(12)=144 which also only 2 and 3 as its prime factors and these have appeared earlier as Fib(3)=2 and Fib(4)=3.

Apart from these special cases, the theorem is true for all Fib(n).

Those prime factors that have never appeared earlier in the table are shown like this.

Here is the first part of a table of Fibonacci numbers and their prime factors:

[See the whole list up to Fib(300) on the next web page in a new window.]

A Result about the Primes Dividing Fibonacci Numbers by M S Boase in Fibonacci Quarterly vol 39 (2001), pages 386-391 contains a proof of this result but does not refer to it as Carmichael’s Theorem. The problem is traced back to an analogous result proved by K Zsigmondy in 1892.
A Simple Proof of Carmichael’s Theorem on Primitive Divisors by M Yabuta in Fibonacci Quarterly vol 39 (2001), pages 439-443 also contains a proof and refers to the following article…
On the Numerical Factors of the Arithmetic Forms ⁿ± ⁿ by R D Carmichael in Annals of Maths vol 15 (1913) pages 30-70, where Carmichael refers to such first-occurrence-prime-factors as characteristic factors.

Fib(prime) and Carmichael’s Theorem

Shane Findley of Dover, USA, points out that all the factors of Fib(p) when p is a prime number are characteristic prime factors.
Let’s have a look at what this means in terms of our Table of Fibonacci Factors.

This relies on two properties of Fib(i) that we have already seen (on this page):

1. In the Factors of Fibonacci Numbers section we saw that
   if i itself has a factor k (so that we can write i as nk)
   then Fib(nk) has Fib(k) as a factor also.
2. Also, if Fib(i) is a prime number then i itself must be prime – see the Fibonacci Primes section.

So if i is prime – and let’s call it p here to remind ourselves we are considering the special case of Fib(i) for a prime number i – then p will have no factors and therefore Fib(p) also can have no earlier Fibonacci numbers as its factors.
Note that this does not mean Fib(p) itself must be prime, only that no smaller Fibonacci number can be a factor. We found an example of this in Fib(19) which is 4181 = 37 x 113, and, although 19 is a prime number, Fib(19) is not.
Carmichael’s Theorem says that there are special prime factors of Fib(p) that have not occurred earlier in our list of Fibonacci numbers.
So, if p is a prime number, then

• either Fib(p) is prime – in which case this is the first time we have seen this prime number in the list of Fibonacci factors
• or else, if Fib(p) has factors, at least one of them is new – a characteristic factor.

But Shane’s observation is that all the prime factors of Fib(p) are characteristic factors!

To put this more simply for a prime number p, Fib(p) is either

• a prime itself or
• is a product of prime factors that all appear to be characteristic (appear for the first time in our list of Fibonacci factors).

Here is a selection of lines from the Factors Table for those Fib(i) where i is a prime number. You will notice that either they are prime numbers or else their factors are all shown like this to show they are characteristic factors:

n :  F(n)=factorization
 
2 : 1 
3 : 2 PRIME
5 : 5 PRIME
7 : 13 PRIME
11 : 89 PRIME
13 : 233 PRIME
17 : 1597 PRIME
19 : 4181 = 37 x 113
23 : 28657 PRIME
29 : 514229 PRIME
31 : 1346269 = 557 x 2417
37 : 24157817 = 73 x 149 x 2221
41 : 165580141 = 2789 x 59369
43 : 433494437 PRIME
47 : 2971215073 PRIME
53 : 53316291173 = 953 x 55945741
59 : 956722026041 = 353 x 2710260697
61 : 2504730781961 = 4513 x 555003497
67 : 44945570212853 = 269 x 116849 x 1429913
71 : 308061521170129 = 6673 x 46165371073
73 : 806515533049393 = 9375829 x 86020717
79 : 14472334024676221 = 157 x 92180471494753
83 : 99194853094755497 PRIME
89 : 1779979416004714189 = 1069 x 1665088321800481
97 : 83621143489848422977 = 193 x 389 x 3084989 x 361040209

No primes next to Fibonacci’s!

Let’s look at the numbers next to each Fibonacci number…

We see there are a few small Fibonacci numbers which have a prime neighbour (shown like this) but then they seem to stop. (Yes, 143 is not a prime number because it is 11×13.)
Is this just a fluke or a feature of all but a few initial Fibonacci numbers – that their neighbours are never prime?
Toby Gee, a student at John of Gaunt’s School, Trowbridge proved this in the 1996/97 in…
Letter from Toby Gee in Mathematical Spectrum, vol 29 (1996/1997), page 68.

He gave formulae for the factors too:
For the Fibonacci numbers at with even index numbers, that is F(2n), we have:-

and for the odd index numbers, F(2n+1), we have similarly:

All of these are derived from Vajda-15a and Vajda-15b (see my Fibonacci, Phi and Lucas Numbers Formulae page).

Greatest Common Divisors in Altered Fibonacci Sequences U Dudley, B Tucker Fibonacci Quarterly 1971, pages 89-91 give these formulae too in an expanded form.

Almost no primes next to Fibonacci’s powers either!!

So having seen that the Fibonacci numbers influence their neighbours so that no neighbour is prime, what about neighbours of the squares of Fibonacci numbers?

Two formulae answer our question immediately:

These two formulae tell us that

So we could now investigate the neighbours of the cubes of Fibonacci Numbers and indeed I will leave you to discover the formulae that apply in those cases.
You will find that they too are never prime!!

The general result was found by Vernon Hoggatt Jr and Marjorie Bicknell-Johnson in 1977.

For example, all these neighbours on the “+1” side of a power of a Fibonacci number are prime:

F(9)⁴ + 1 = 1336337
F(198)⁴ + 1 (165 digits)
F(15)⁸ + 1 (23 digits)
F(48)⁸ + 1 (78 digits)
F(51)⁸ + 1 (83 digits)
F(21)³² + 1 (130 digits)

You will see that all the powers are themselves powers of 2 and all the indices are multiples of 3. It seems that such primes are quite rare though.

So Fibonacci numbers exert a powerful influence in that they (almost always) make any number next to them or their powers factorize!

Composites and Primes Among Powers of Fibonacci Numbers increased or decreased by one V E Hoggatt Jr and M Bicknell-Johnson, Fibonacci Quarterly vol 15 (1977), page 2.

A Prime Curio

G. L. Honaker Jr. pointed me to a little curio about the Fibonacci and the prime numbers: that the number of primes less than 144, which is a Fibonacci number, is 34, also a Fibonacci number. He asks:

I pass this question on to you – can it? The link to the Prime Curio page uses the notation that

(N) means the number of primes between 1 and N

and includes N too if N is prime. (See also a graph of this function.)
Since the prime numbers begin

then (8)=4 (there are 4 primes between 1 and 8, namely 2, 3, 5 and 7) and (11)=5.
Here are some smaller values that are also Fibonacci numbers:

(2) = 1
(3) = 2
(5) = 3
(21) = 8

Another Prime Curio

M J Zerger (Mathematics Teacher vol 89 (1996) page 26) noticed that the four consecutive Fibonacci numbers: F(6)=13, F(7)=21, F(8)=34 and F(9)=55 have a product of 13x3x7x17x2x5x11 or rearranging the factors into order: 2x3x5x7x11x13x17 and that is just the product of the first seven prime numbers!

More Links and References on Prime Numbers

There is a complete list of all Fibonacci numbers and their factors up to the 1000-th Fibonacci and 1000-th Lucas numbers and partial results beyond that on Blair Kelly’s Factorization pages
Chris Caldwell’s Prime Numbers site has a host of information.
For the real enthusiast, join the Yahoo group on the PrimeForm computer program and related matters to primes. Its Files folder has a section on Lucas and Fibonacci primes.
See Neil Sloane’s Online Encyclopedia of Integer Sequences where series number A001605 is the series of i’s for which Fib(i) is known to be prime: 3,4,5,7,11,13,17,23,29,… and contains fairly up-to-date information on the latest results.
Factorization of Fibonacci Numbers D E Daykin and L A G Dresel in The Fibonacci Quarterly, vol 7 (1969) pages 23 – 30 and 82 gives a method of factorising a Fib(n) for composite n using the “entry point” of a prime, that is, the index of the first Fibonacci number for which prime p is a factor.

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..

Benford’s Law and initial digits

Having looked at the end digits of Fibonacci numbers, we might ask

What are the chances of a Fibonacci number beginning with “1”, say? or “5”? We might be forgiven for thinking that they probably are all the same – each digit is equally likely to start a randomly chosen Fibonacci number. You only need to look at the Table of the First 100 Fibonacci numbers or use Fibonacci Calculator to see that this is not so. Fibonacci numbers seem far more likely to start with “1” than any other number. The next most popular digit is “2” and “9” is the least probable!

This law is called Benford’s Law and appears in many tables of statistics. Other examples are a table of populations of countries, or lengths of rivers. About one-third of countries have a population size which begins with the digit “1” and very few have a population size beginning with “9”.

Here is a table of the initial digits as produced by the Fibonacci Calculator:

Initial digit frequencies of fib(i) for i from 1 to 100: 
    Digit:    1   2   3   4   5   6   7   8   9
Frequency:   30  18  13   9   8   6   5   7   4  100 values
  Percent:   30  18  13   9   8   6   5   7   4

What are the frequencies for the first 1000 Fibonacci numbers or the first 10,000? Are they settling down to fixed values (percentages)? Use the Fibonacci Calculator to collect the statistics. According to Benford’s Law, large numbers of items lead to the following statistics for starting figures for the Fibonacci numbers as well as some natural phenomena

When does Benford’s Law apply?

Random numbers are equally likely to begin with each of the digits 0 to 9. This applies to randomly chosen real numbers or randomly chosen integers.

Randomly chosen real numbers

If you stick a pin at random on a ruler which is 10cm long and it will fall in each of the 10 sections 0cm-1cm, 1cm-2cm, etc with the same probability. Also, if you look at the initial digits of the points chosen (so that the initial digit of 0.02cm is 2 even though the point is in the 0-1cm section) then each of the 9 values from 1 to 9 is as likely as any other value.

Randomly chosen integers

This also applies if we choose random integers.
Take a pack of playing cards and remove the jokers, tens, jacks and queens, leaving in all aces up to 9 and the kings. Each card will represent a different digit, with a king representing zero. Shuffle the pack and put the first 4 cards in a row to represent a 4 digit integer. Suppose we have King, Five, King, Nine. This will represent “0509” or the integer 509 whose first digit is 5. The integer is as likely to begin with 0 (a king) as 1 (an ace) or 2 or any other digit up to 9.
But if our “integer” began with a king (0), then we look at the next “digit”.
These have the same distribution as if we had chosen to put down just 3 cards in a row instead of 4. The first digits all have the same probability again. If our first two cards had been 0, then we look at the third digit, and the same applies again.
So if we ignore the integer 0, any randomly chosen (4 digit) integer begins with 1 to 9 with equal probability. (This is not quite true of a row of 5 or more cards if we use an ordinary pack of cards – why?)

So the question is, why does this all-digits-equally-likely property not apply to the first digits of each of the following:

• the Fibonacci numbers,
• the Lucas numbers,
• populations of countries or towns
• sizes of lakes
• prices of shares on the Stock Exchange

Whether we measure the size of a country or a lake in square kilometres or square miles (or square anything), does not matter – Benford’s Law will still apply.
So when is a number random? We often meant that we cannot predict the next value. If we toss a coin, we can never predict if it will be Heads or Tails if we give it a reasonably high flip in the air. Similarly, with throwing a dice – “1” is as likely as “6”. Physical methods such as tossing coins or throwing dice or picking numbered balls from a rotating drum as in Lottery games are always unpredictable.

The answer is that the Fibonacci and Lucas Numbers are governed by a Power Law.
We have seen that Fib(i) is round(Phiⁱ/5) and Lucas(i) is round(Phiⁱ). Dividing by sqrt(5) will merely adjust the scale – which does not matter. Similarly, rounding will not affect the overall distribution of the digits in a large sample.

Basically, Fibonacci and Lucas numbers are powers of Phi. Many natural statistics are also governed by a power law – the values are related to Bⁱ for some base value B. Such data would seem to include the sizes of lakes and populations of towns as well as non-natural data such as the collection of prices of stocks and shares at any one time. In terms of natural phenomena (like lake sizes or heights of mountains) the larger values are rare and smaller sizes are more common. So there are very few large lakes, quite a few medium sized lakes and very many little lakes. We can see this with the Fibonacci numbers too: there are 11 Fibonacci numbers in the range 1-100, but only one in the next 3 ranges of 100 (101-200, 201-300, 301-400) and they get increasingly rarer for large ranges of size 100. The same is true for any other size of range (1000 or 1000000 or whatever).

Benford’s Law for Fibonacci and Lucas Numbers, L. C. Washington, The Fibonacci Quarterly vol. 19, 1981, pages 175-177.
The original reference: The Law of Anomalous Numbers F Benford, (1938) Proceedings of the American Philosophical Society vol 78, pages 551-572.
The Math Forum’s archives of the History of Mathematics discussion group have an email from Ralph A. Raimi (July 2000) about his research into Benford’s Law. It seems that Simon Newcomb had written about it much earlier, in 1881, in American Journal of Mathematics volume 4, pages 39-40. The name Benford is, however, the one that is commonly used today for this law.
MathTrek by Ivars Peterson (author of The Mathematical Tourist and Islands of Truth) the editor of Science News Online has produced this very good, short and readable introduction to Benford’s Law.
M Schroeder Fractals, Chaos and Power Laws, Freeman, 1991, ISBN 0-7167-2357-3. This is an interesting book but some of the mathematics is at first year university level (mathematics or physics degrees), unfortunately, and the rest will need sixth form or college level mathematics beyond age 16. However, it is still good to browse through. It has only a passing reference to Benford’s Law: The Peculiar Distribution of the Leading Digit on page 116.

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..

The Fibonacci Numbers in Pascal’s Triangle

Each entry in the triangle on the left is the sum of the two numbers either side of it but in the row above. A blank space can be taken as “0” so that each row starts and ends with “1”.

Pascal’s Triangle has lots of uses including

Calculating probabilities.
If you throw n coins randomly onto a table then the chance of getting H heads among them is the entry in row N, col H divided by 2ⁿ:
for instance, for 3 coins, n=3 so we use row 3:
3 heads: H=3 is found in 1 way (HHH)
2 heads: H=2 can be got in 3 ways (HHT, HTH and THH)
1 head: H=1 is also found in 3 possible ways (HTT, THT, TTH)
0 heads: H=0 (ie all Tails) is also possible in just 1 way: TTT

Finding terms in a Binomial expansion: (a+b)ⁿ
EG. (a+b)³ = 1a³ + 3a²b + 3ab² + 1b³

Can you find the Fibonacci Numbers in Pascal’s Triangle?
Hints:

	The answer is in the formula on the right: where the big brackets with two numbers vertically inside them are a special mathematical notation for the entry in Pascal’s triangle on row n-k-1 and column k	Fib(n)= n–1 k=0 ( n–k–1 k )
	Or, an equivalent formula is:	Fib(n)= n k=1 ( n–k k–1 )
	If that still doesn’t help, then this animated diagram might:

Why do the Diagonals sum to Fibonacci numbers?

It is easy to see that the diagonal sums really are the Fibonacci numbers if we remember that each number in Pascal’s triangle is the sum of two numbers in the row above it (blank spaces count as zero), so that 6 here is the sum of the two 3’s on the row above.

The numbers in any diagonal row are therefore formed from adding numbers in the previous two diagonal rows as we see here where all the blank spaces are zeroes and where we have introduced an extra column of zeros which we will use later:

Notice that the GREEN numbers are on one diagonal and the BLUE ones on the next. The sum of all the green numbers is 5 and all the blue numbers add up to 8.
Because all the numbers in Pascal’s Triangle are made the same way – by adding the two numbers above and to the left on the row above, then we can see that each red number is just the sum of a green number and a blue number and we use up all the blue and green numbers to make all the red ones.
The sum of all the red numbers is therefore the same as the sum of all the blues and all the greens: 5+8=13!
The general principle that we have just illustrated is:

The sum of the numbers on one diagonal is the sum of the numbers on the previous two diagonals.

If we let D(i) stand for the sum of the numbers on the Diagonal that starts with one of the extra zeros at the beginning of row i, then

are the two initial diagonals shown in the table above. The green diagonal sum is D(5)=5 (since its extra initial zero is in row 5) and the blue diagonal sum is D(6) which is 8. Our red diagonal is D(7) = 13 = D(6)+D(5).
We also have shown that this is always true: one diagonals sum id the sum of the previous two diagonal sums, or, in terms of our D series of numbers:

But…

is exactly the definition of the Fibonacci numbers! So D(i) is just F(i) and

Another arrangement of Pascal’s Triangle

By drawing Pascal’s Triangle with all the rows moved over by 1 place, we have a clearer arrangement which shows the Fibonacci numbers as sums of columns:

This table can be explained by referring to one of the (Easier) Fibonacci Puzzles – the one about Fibonacci for a Change. It asks how many ways you can pay n pence (in the UK) using only 1 pence and 2 pence coins. The order of the coins matters, so that 1p+2p will pay for a 3p item and 2p+1p is counted as a different answer. [We now have a new two pound coin that is increasing in circulation too!]
Here are the answers for paying up to 5p using only 1p and 2p coins:

Let’s look at this another way – arranging our answers according to the number of 1p and 2p coins we use. Columns will represent all the ways of paying the amount at the head of the column, as before, but now the rows represent the number of coins in the solutions:

If you count the number of solutions in each box, it will be exactly the form of Pascal’s triangle that we showed above!

Fibonacci’s Rabbit Generations and Pascal’s Triangle

Here’s another explanation of how the Pascal triangle numbers sum to give the Fibonacci numbers, this time explained in terms of our original rabbit problem.

Let’s return to Fibonacci’s rabbit problem and look at it another way. We shall be returning to it several more times yet in these pages – and each time we will discover something different!

We shall make a family tree of the rabbits but this time we shall be interested only in the females and ignore any males in the population. If you like, in the diagram of the rabbit pairs shown here, assume that the rabbit on the left of each pair is male (say) and so the other is female. Now ignore the rabbit on the left in each pair!
We will assume that each mating produces exactly one female and perhaps some males too but we only show the females in the diagram on the left. Also in the diagram on the left we see that each individual rabbit appears several times. For instance, the original brown female was mated with a while male and, since they never die, they both appear once on every line.
Now, in our new family tree diagram, each female rabbit will appear only once. As more rabbits are born, so the Family tree grows adding a new entry for each newly born female.
As in an ordinary human family tree, we shall show parents above a line of all their children.
Here is a fictitious human family tree with the names of the relatives shown for a person marked as ME:

The diagram shows that:

• Grandpa Abel and Grandma Mabel are the parents of my Dad;
• Grandma Freda and Grandpa Fred are the parents of my Mum.
• Bob is my Dad’s brother
• my Mum has two sisters, my aunts Hayley and Jane.
• Aunt Hayley became Hayley Weather when she married Clement Weather.
• They have two children, my cousins Sonny Weather and Gale Weather.
• Gale married Gustof Wind and so is now Gale Wind.
• My brother John and his wife Joan have two children, my nephew Dan and my niece Pam.

In this family tree of human relationships, the = joins people who are parents or signifies a marriage.
In our rabbit’s family tree, rabbits don’t marry of course, so we just have the vertical and horizontal lines:

The vertical line |

points from a mother (above) to the oldest daughter (below);

the horizontal line –

is drawn between sisters from the oldest on the left down to the youngest on the right;

the small letter r

represents a young female ( a little rabbit) and

the large letter R

shows a mature female (a big Rabbit) who can and does mate every month, producing one new daughter each time.

As in Fibonacci’s original problem (in its variant form that makes it a bit more realistic) we assume none die and that each month every mature female rabbit R always produces exactly one female rabbit r (we ignore males) each month.
So each month each r will change to R (each matures after one month) and each R will change to Rr, that is produce a new baby rabbit each month.
Here is the Rabbit Family tree as if grows month by month for the first 8 months:

   r   R   R   R     R       R          R                R
           |   |     |       |          |                |
           r   R_r   R_R_r   R___R_R_r  R_____R___R_R_r  R_________R_____R___R_R_r
                     |       |   |      |     |   |      |         |     |   |
                     r       R_r r      R_R_r R_r r      R___R_R_r R_R_r R_r r
                                        |                |   |     |
                                        r                R_r r     r

In month 2, our young female (r of month 1) becomes mature (R) and mates.
In month 3, she becomes a parent for the first time and produces her first daughter, shown on a line below – a new generation.
In month 4, the female born in month 3 becomes mature (R) and also her mother produces another daughter (r).
In month 5, the original female produces another female child added to the end of the line of the generation of her daughters, while the daughter born the previous month (the second in the line) becomes mature. Also the first daughter produces her own first daughter, so in month 5 the original female becomes a grand-mother and we have started a third line – the third generation.
The Family tree is shown for the first 8 months as more females are added to it. We can see that our original female becomes a great-grandmother in month 7 when a fourth line is added to the Family tree diagram – a fourth generation!

The numbers of rabbits in each generation, that is, along each level (line) of the tree, are the Pascal’s triangle numbers that add up to give each Fibonacci number – the total number of (female) rabbits in the Tree. In month n there are a total of F(n) rabbits, a number made up from the entry in row (n-k) and column (k-1) of Pascal’s triangle for each of the levels (generations) k from 1 to n. In other words, we are looking at this formula and explaining it in terms of generations, the original rabbit forming generation 1 and her daughters being generation 2 and so on:

Fib(n)=

n k=1

(

n–k k–1

)

Remember that the rows and columns of Pascal’s triangle in this formula begin at 0!
For example, in month 8, there are 4 levels and the number on each level is:
M o n t h 8:
Level 1: 1 rabbit which is Pascal’s triangle row 7=8-1 and column 0=1-1
Level 2: 6 rabbits which is Pascal’s triangle row 6=8-2 and column 1=2-1
Level 3: 10 rabbits which is Pascal’s triangle row 5=8-3 and column 2=3-1
Level 4: 4 rabbit which is Pascal’s triangle row 4=8-4 and column 3=4-1
When k is bigger than 4, the column number exceeds the row number in Pascal’s Triangle and all those entries are 0.

The general pattern for month n and level (generation) k is
Level k: is Pascal’s triangle row n-k and column k-1 For month n we sum all the generations as k goes from 1 to n (but half of these will be zeros).

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..

The Fibonacci Series as a Decimal Fraction

Have a look at this decimal fraction:

           0·0112359550561...

It looks like it begins with the Fibonacci numbers, 0, 1, 1, 2, 3 and 5 and indeed it does if we express it as:

           0·0                +
              1               +
               1              +
                2
                 3
                  5
                   8
                   13
                    21
                     34
                      55
                       89
                       144
                         ...
           --------------------
           0·011235955056179...

What is the value of this decimal fraction?

It can be expressed as

or, using powers of 10 and replacing the Fibonacci numbers by F(i):

or, if we use the negative powers of 10 to indicate the decimal fractions:

To find the value of the decimal fraction we look at a generalization, replacing 10 by x.

Let P(x) be the polynomial in x whose coefficients are the Fibonacci numbers:

or

To avoid confusion between the variable x and the multiplication sign x, we will represent multiplication by *: The decimal fraction 0.011235955… above is just

which is just P(x) with x taking the value (1/10), which we write as P(1/10).

Now here is the interesting part of the technique!
We now write down xP(x) and x²P(x) because these will “move the Fibonacci coefficients along”:

We can align these terms up so that all the same powers of x are in the same column (as we would do when doing ordinary decimal arithmetic on numbers) as follows:

We have done this so that each Fibonacci number in P(x) is aligned with the two previous Fibonacci numbers. Since the sum of the two previous numbers always equals the next in the Fibonacci series, then, when we take them away, the result will be zero – the terms will vanish!
So, if we take away the last two expressions (for xP(x) and x²P(x)) from the first equation for P(x), the right-hand side will simplify since all but the first few terms vanish, as shown here:

P(x)=	F(0)x +	F(1)x2 +	F(2)x3 + F(3)x4 + … +F(n-1)xn + …
xP(x)=		F(0)x2 +	F(1)x3 + F(2)x4 + … +F(n-2)xn + …
x2P(x)=			F(0)x3 + F(1)x4 + … +F(n-3)xn + …
(1-x-x2)P(x)=	F(0)x +	(F(1)-F(0))x2 +	(F(2)-F(1)-F(0))x3+…

Apart from the first two terms, the general term, which is just the coefficient of xⁿ, becomes F(n)-F(n-1)-F(n-2) and, since F(n)=F(n-1)+F(n-2) all but the first two terms become zero which is why we wrote down xP(x) and x²P(x):
(1-x-x²)P(x) = x²

P(x) =

x2 1 – x – x2

=

1 x–2 – x–1 – 1

So now our fraction is just P(1/10), and the right hand side tells us its exact value:
1 / (100-10-1) = 1/89 = 0·0112358…
From our expression for P(x) we can also deduce the following:

            10/89 = 0·112359550561...

If x=1/100, we have

      P(1/100) = 0·00 01 01 02 03 05 08 13 21 34 55 ... = 1/(10000-100-1) = 1/9899 
  and
         100/9899 = 0·01010203050813213455...

and so on.

References

The Decimal Expansion of 1/89 and related Results, Fibonacci Quarterly, Vol 19, (1981), pages 53-55.

Calvin Long solves the general problem for all Fibonacci-type sequences i.e. G(0)=c, G(1)=d are the two starting terms and G(i) = a G(i-1) + b G(i-1) defines all other values for integers a and b. For our “ordinary” Fibonacci sequence, a=b=1 and c=d=1. He gives the exact fractions for any base B (here B=10 for decimal fractions) and gives conditions when the fraction exists (i.e. when the series does not get too large too quickly so that we do have a genuine fraction).

A Complete Characterization of the Decimal Fractions that can be Represented as SUM(10^-k(i+1)F(ai)) where F(ai) is the ai^th Fibonacci number Richard H Hudson and C F Winans, Fibonacci Quarterly, 1981, Vol 19, pp 414 – 421.

This article examines all the decimal fractions where the terms are F(a), F(2a), F(3a) taken k digits at a time in the decimal fraction.

A Primer For the Fibonacci Numbers: Part VI, V E Hoggatt Jr, D A Lind in Fibonacci Quarterly, vol 5 (1967) pages 445 – 460

is a nice introduction to Generating Functions (a polynomial in x where the coefficients of the powers of x are the members of a particular series). The whole collection of articles is now available in book form by mail order from The Fibonacci Association. It is readable and not too technical. There is also a list of formulae for all kinds of generating functions, which, if we substitute a power of 10 for x, will give a large collection of fractions whose decimal expansion is , for example:

• the Lucas Numbers (see this page at this site) e.g. 1999/998999
• the squares of the Fibonacci numbers e.g. 999000/997998001
• the product of two neighbouring Fibonacci numbers e.g. 1000/997998001
• the cubes of the Fibonacci numbers e.g. 997999000/996994003001
• the product of three neighbouring Fibonacci numbers e.g. 2000000000/996994003001
• every k^th Fibonacci number e.g. 1000/997001 or 999000/997001
• etc

Scott’s Fibonacci Scrapbook, Allan Scott in Fibonacci Quarterly vol 6 number 2, (April 1968), page 176

is a follow-up article to the one above, extending the generating functions to Lucas cubes and Fibonacci fourth and fifth powers.
Note there are several corrections to these equations on page 70 of vol 6 number 3 (June 1968).

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..

A Fibonacci Number Trick

Here is a little trick you can perform on friends which seems to show that you have amazing mathematical powers. We explain how it works after showing you the trick.

Alice and Bill

Here is Alice performing the trick on Bill:

Alice: Choose any two numbers you like, Bill, but not too big as you’re going to have to do some adding yourself. Write them as if you are going to add them up and I’ll, of course, be looking the other way!

Bill: OK, I’ve done that.

Bill chooses 16 and 21 and writes them one under the other:	16
	21

Alice: Now add the first to the second and write the sum underneath to make the third entry in the column.

Bill: I don’t think I’ll need my calculator just yet…. Ok, I’ve done that.

Bill writes down 37 (=16+21) under the other two:	16
	21
	37

Alice: Right, now add up the second and your new number and again write their sum underneath. Keep on doing this, adding the number you have just written to the number before it and putting the new sum underneath. Stop when you have 10 numbers written down and draw a line under the tenth.

There is a sound of lots of buttons being tapped on Bill’s calculator!

Bill: OK, the ten numbers are ready.

Bills column now looks like this:

16 21 37 58 95 153 248 401 649 1050

Alice: Now I’ll turn round and look at your numbers and write the sum of all ten numbers straight away!

She turns round and almost immediately writes underneath: 2728.

Bill taps away again on his calculator and is amazed that Alice got it right in so short a time [gasp!]

So how did Alice do it?

The sum of all ten numbers is just eleven times the fourth number from the bottom. Also, Alice knows the quick method of multiplying a number by eleven. The fourth number from the bottom is 248, and there is the quick and easy method of multiplying numbers by 11 that you can easily do in your head:

Starting at the right, just copy the last digit of the number as the last digit of your product. Here the last digit of 248 is 8 so the product also ends with 8 which Alice writes down:	... 248 401 649 1050 8
Now, continuing in 248, keep adding up from the right each number and its neighbour, in pairs, writing down their sum as you go. If ever you get a sum bigger than 10, then write down the units digit of the sum and remember to carry anything over into your next pair to add. Here the pairs of 248 are (from the right) 4+8 and then 2+4. So, next to the 8 Alice thinks “4+8=12” so she writes 2 and remembers there is an extra one to add on to the next pair:	... 248 401 649 1050 28
Then 2+4 is 6, adding the one carried makes 7, so she writes 7 on the left of those digits already written down:	... 248 401 649 1050 728
Finally copy down the left hand digit (plus any carry). Alice sees that the left digit is 2 which, because there is nothing being carried from the previous pair, becomes the left-hand digit of the sum.The final sum is therefore 2728 = 11 x 248 .	... 248 401 649 1050 2728

Why does it work?

You can see how it works using algebra and by starting with A and B as the two numbers that Bill chooses.
What does he write next? Just A+B in algebraic form.
The next sum is B added to A+B which is A+2B.
The other numbers in the column are 2A+3B, 3A+5B, … up to 21A+34B.

  A
       B
  A +  B
  A + 2B
 2A + 3B
 3A + 5B
 5A + 8B
 8A +13B
13A +21B
21A +34B
--------
55A +88B

If you add these up you find the total sum of all ten is 55A+88B.
Now look at the fourth number up from the bottom. What is it?
How is it related to the final sum of 55A+88B?

So the trick works by a special property of adding up exactly ten numbers from a Fibonacci-like sequence and will work for any two starting values A and B!

Perhaps you noticed that the multiples of A and B were the Fibonacci numbers? This is part of a more general pattern which is the first investigation of several to spot new patterns in the Fibonacci sequence in the next section.

On a Fibonacci Arithmetical Trick C T Long, Fibonacci Quarterly vol 23 (1985), pages 221-231. This article introduces the above trick and generalizes it.

Practice here with “Bill”

Another Number Pattern

Dave Wood has found another number pattern that we can prove using the same method.

He notices that

       f(10)-f(5)  is   55 -   5 which is  50  or   5 tens and  0;

       f(11)-f(6)  is   89 -   8 which is  81  or   8 tens and  1;

       f(12)-f(7)  is  144 -  13 which is  131 or  13 tens and  1.

It looks like the differences seem to be ‘copying’ the Fibonacci series in the tens and in the units columns.
If we continue the investigation we have:

       f(13)-f(8)  is  233 -  21 which is  212 or  21 tens and  2;

       f(14)-f(9)  is  377 -  34 which is  343 or  34 tens and  3;

       f(15)-f(10) is  610 -  55 which is  555 or  55 tens and  5;

       f(16)-f(11) is  987 -  89 which is  898 or  89 tens and  8;

       f(17)-f(12) is 1597 - 144 which is 1453 or 144 tens and 13;

From this point on, we have to borrow a ten in order to make the ‘units’ have the 2 digits needed for the next Fibonacci number. Later we shall have to ‘borrow’ more, but the pattern still seems to hold.

In words we have:

In mathematical terms, we can write this as:

A Proof
That the pattern always holds is found by extending the table we used in the Why does it work section of the Number Trick above:

        A
             B
        A +  B
        A + 2B
       2A + 3B
       3A + 5B
       5A + 8B
       8A +13B
      13A +21B
      21A +34B
      34A +55B

We can always write any Fibonacci number Fib(n) as 34A+55B because, since the Fibonacci series extends backwards infinitely far, we just pick A and B as the two numbers that are 10 and 9 places before the one we want.

Now let’s look at that last line: 34A +55B.
It is almost 11 times the number 5 rows before it:

      11 x (3A+5B) = 33A+55B,

and it is equal to it if we add on an extra A, i.e. the number ten rows before the last one:

      34A + 55B = 11 (3A+5B) + A

Putting this in terms of the Fibonacci numbers, where the 34A+55B is F(n) and 3A+5B is “the Fibonacci number 5 before it”, or Fib(n-5) and A is “the Fibonacci number 10 before it” or Fib(n-10), we have:

      34A + 55B = 11 (3A+5B) +  A
or
      Fib(n)    = 11 Fib(n-5) + Fib(n-10)

We rearrange this now by taking Fib(n-5) from both sides and we have:

      Fib(n) - Fib(n-5) = 10 Fib(n-5) + Fib(n-10)

which is just what Dave Wood observed!

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..

Fibonacci Numbers and Pythagorean Triangles

A Pythagorean Triangle is a right-angled triangle with sides which are whole numbers.

In any right-angled triangle with sides s and t and longest side (hypotenuse) h, the Pythagoras Theorem applies:

However, for a Pythagorean triangle, we also want the sides to be integers (whole numbers) too. A common example is a triangle with sides s=3, t=4 and h=5:
We can check Pythagoras theorem as follows:

s² + t²
= 3² + 4²
= 9 + 16
= 25 = 5² = h²

Here is a list of some of the smaller Pythagorean Triangles:

Here is another longer list of Triples generated using Autograph from Oundle School, Peterborough, UK.

You will see that some are just magnifications of smaller ones where all the sides have been doubled, or trebled for example. The others are “new” and are usually called primitive Pythagorean triangles.
Any Pythagorean triangle is either primitive or a multiple of a primitive and this is shown in the table above. Primitive Pythagorean triangles are a bit like prime numbers in that every integer is either prime or a multiple of a prime.

Using the Fibonacci Numbers to make Pythagorean Triangles

There is an easy way to generate Pythagorean triangles using 4 Fibonacci numbers. Take, for example, the 4 Fibonacci numbers:

Let’s call the first two a and b. Since they are from the Fibonacci series, the next is the sum of the previous two: a+b and the following one is b+(a+b) or a+2b:-

You can now make a Pythagorean triangle as follows:

1. Multiply the two middle or inner numbers (here 2 and 3 giving 6);
2. Double the result (here twice 6 gives 12). This is one side, s, of the Pythagorean Triangle.
3. Multiply together the two outer numbers (here 1 and 5 giving 5). This is the second side, t, of the Pythagorean triangle.
4. The third side, the longest, is found by adding together the squares of the inner two numbers (here 2²=4 and 3²=9 and their sum is 4+9=13). This is the third side, h, of the Pythagorean triangle.

We have generated the 12, 5,13 Pythagorean triangle, or, putting the sides in order, the 5, 12, 13 triangle this time.

Try it with 2, 3, 5 and 8 and check that you get the Pythagorean triangle: 30, 16, 34.
Is this one primitive?

In fact, this process works for any two numbers a and b, not just Fibonacci numbers. The third and fourth numbers are found using the Fibonacci rule: add the latest two values to get the next.
Four such numbers are part of a generalised Fibonacci series which we could continue for as long as we liked, just as we did for the (real) Fibonacci series.

All Pythagorean triangles can be generated in this way by choosing suitable starting numbers a and b! Try it with this Calculator:

Connections in Mathematics: An Introduction to Fibonacci via Pythagoras E A Marchisotto in Fibonacci Quarterly, vol 31, 1993, pages 21 – 27.
This article explores many ways of introducing the Fibonacci numbers in class starting from the Pythagorean triples, with an extensive Appendix of references useful for the teacher and comparing different approaches. Highly recommended!
Pythagorean Triangles from the Fibonacci Series C W Raine in Scripta Mathematica vol 14 (1948) page 164.

Fibonacci Numbers as the sides of Pythagorean Triangles

No, because of the following condition that must be true for any and all triangles:

This is called the Triangle Inequality.
Since three consecutive Fibonacci numbers already have the third number equal to the sum of the other two, then the Triangle Inequality fails. Or, if you prefer, the two shorter sides collapse onto the third to form a straight line when you try to construct a triangle from these numbers.
If the smallest side is smaller, that makes it worse, as it does if the longer side gets longer!
So we have

Yes: 3,3,5 will do. The longest side, 5, is now less than the sum of the other two, 3+3. But this triangle is not right-angled: 3² + 3² is 18 whereas 5² is 25.

No Pythagorean triangle has two equal sides. If we ask for just two sides which are Fibonacci numbers, the third being any whole number, then there are at least two Pythagorean triangles with Fibonacci numbers on two sides:

It is still an unsolved problem as to whether there are any more right-angled (Pythagorean) triangles with just two Fibonacci numbers as sides.

We can make every odd-indexed Fibonacci number the hypotenuse of a Pythagorean triangle using the technique of the section above.
We take 4 consecutive Fibonacci numbers:

and get the two sides of a Pythagorean Triangle:

The hypotenuse is the sum of the squares of the middle two numbers: F(n)² + F(n+1)² and Lucas showed in 1876 that this is just F(2n+1).
So we have the following Pythagorean Triangle with an odd-indexed Fibonacci number as hypotenuse:

Here are some examples:

Pythagorean Triples A F Horadam Fibonacci Quarterly vol 20 (1982) pages 121-122.

Square Fibonacci Numbers

Let’s have a further look at that formula of Lucas from 1876:

We can make this into a Pythagorean triangle whenever F(2n+1) is a square number. So we now have the question

We only have to look at the first few Fibonacci numbers to spot these square numbers:

But a longer look does not reveal any more squares among the Fibonacci’s. Are these the only Fibonacci squares? Yes, as was proved by Cohn in the article below. Therefore…

Square Fibonacci Numbers Etc J H E Cohn in Fibonacci Quarterly vol 2 (1964) pages 109-113.

Other right-angled triangles and the Fibonacci Numbers

Even if we don’t insist that all three sides of a right-angled triangle are integers, Fibonacci numbers still have some interesting applications.
For instance, if we choose two consecutive Fibonacci numbers as the sides next to the right angle, then the third side squared is also a Fibonacci number.
For instance, if the sides are 3 and 5, by Pythagoras’ Theorem we have that the hypotenuse, h, is given by:
3² + 5² = h²
and 9 + 25 = 34, another Fibonacci number. So h is √34. If we look at the indices of the Fibonacci numbers, we can directly predict which Fibonacci number will be the square of the hypotenuse.
For our example here, F(4)=3 and F(5)=5 and the hypotenuse-squared is 34=F(9). 9, the index of the h² Fibonacci number is the sum of the other two indices, 4 and 5. This is also true in general, provided the two Fibonacci sides are consecutive Fibonacci numbers, say F(n) and F(n+1):

F(n)² + F(n+1)² = F(2 n + 1) ………. Lucas (1876)

This rule was known (and proved) by E Lucas in 1876.
I am grateful to Richard Van De Plasch for pointing out this application of Lucas’s formula to right-angled triangles.

The investigations above will lead you to Catalan’s Identity of 1879 which is on this site’s Formula page. We can rearrange it as follows:

For the investigations above we use a special case where r is an odd number (2k+1) more than n; i.e. (n-r) is (2k+1) for any integer k.
This means that:

• the -(-1)^n-r on the left-hand side becomes just a + sign since n-r is odd;
• the factor F(n-r) on the right is just F(2k+1) which is why the factors above: 2, 5, 13, 34,… are alternate Fibonacci numbers – the ones with an odd index.
• the Fibonacci number F(n+r) in the hypotenuse has an index (n+r) which is the sum of the indices of the Fibonacci numbers on the other two sides of the triangle (n and r).

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..

Maths from the Fibonacci Spiral diagram

Let’s look again at the Fibonacci squares and spiral that we saw in the Fibonacci Spiral section of the Fibonacci in Nature page.
Wherever we stop, we will always get a rectangle, since the next square to add is determined by the longest edge on the current rectangle. Also, those longest edges are just the sum of the latest two sides-of-squares to be added. Since we start with squares of sides 1 and 1, this tells us why the squares sides are the Fibonacci numbers (the next is the sum of the previous 2).

Also, we see that each rectangle is a jigsaw puzzle made up of all the earlier squares to form a rectangle. All the squares and all the rectangles have sides which are Fibonacci numbers in length. What is the mathematical relationship that is shown by this pattern of squares and rectangles? We express each rectangle’s area as a sum of its component square areas:
The diagram shows that

and also, the smaller rectangles show:

This picture actually is a convincing proof that the pattern will work for any number of squares of Fibonacci numbers that we wish to sum. They always total to the largest Fibonacci number used in the squares multiplied by the next Fibonacci number.
That is a bit of a mouthful to say – and to understand – so it is better to express the relationship in the language of mathematics:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987 ..More..

..and now it’s your turn!

Here are some series that use the Fibonacci numbers. Compute a few terms and see if you can spot the pattern, ie guess the formula for the general term and write it down mathematically:

So try them for yourself. This is where Mathematics becomes more of an Art than a Science, since you are relying on your intuition to “spot” the pattern. No one is quite sure where this ability in humans comes from. It is not easy to get a computer to do this (although Maple is quite good at it) – and it may be something specifically human that a computing machine can never really copy, but no one is sure at present. Here are two references if you want to explore further the arguments and ideas of why an electronic computer may or may not be able to mimic a human brain:

Prof Roger Penrose’s book Shadows of the Mind published in 1994 by Oxford Press makes interesting reading on this subject.
An on-line Journal, Psyche has many articles and reviews of this book in Volume 2.

Dr. Math has some interesting replies to questions about the Fibonacci series and the Golden section together with a few more formulae for you to check out.

S. Vajda, Fibonacci and Lucas numbers, and the Golden Section: Theory and Applications, Halsted Press (1989).
This is a wonderful book – now out of print – which is full of formulae on the Fibonacci numbers and Phi. Do try and find it in your local college or university library. It is well worth dipping in to if you are studying maths at age 16 or beyond! Most of Vajda’s formulae are available on my Fibonacci, Phi and Lucas Numbers Formulae page too, with some corrections of Vajda’s (rare) errors.
Mathematical Mystery Tour by Mark Wahl, 1989, is full of many mathematical investigations, illustrations, diagrams, tricks, facts, notes as well as guides for teachers using the material. It is a great resource for your own investigations.

The Puzzling World of the Fibonacci Numbers

Fibonacci Home PageThis is the first page on this Topic. WHERE TO NOW? The first 500 Fibonacci Numbers A Formula for Fibonacci Numbers

The next Topic is… The Golden Section – the Number and Its Geometry

© 1996-2007Dr Ron Knott
updated 29 January 2007