The Lucas Numbers
We have seen in earlier pages that there is another series quite similar to the
Fibonacci series that often occurs when working with the Fibonacci series.
Edouard
Lucas (18421891) (who gave the name
"Fibonacci Numbers" to the series written about by Leonardo of Pisa) studied
this second series of numbers: 2, 1, 3, 4, 7, 11, 18, .. called
the Lucas numbers in his honour.
On this page we examine
some of the interesting properties of the Lucas numbers themselves
as well as looking at its close relationship with the Fibonacci numbers. The following
page generalises further by taking any two starting values.
Contents
The
icon means there are Things To do
investigations at the end of that section.
2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843 ..More..
Other starting values for a "Fibonacci" series
The definition of the Fibonacci series is:
F_{n+1} = F_{n1} + F_{n} , if n>1
F_{0} = 0
F_{1} = 1
What if we have the same general rule: add the latest two values to get the next
but we started with different values instead of 0 and 1?
Things to do
 The Fibonacci series starts with 0 and 1.
What if we started a "Fibonacci" series with 1 and 2, using the same general rule is
for the Fibonacci series proper, so that F_{0} = 1 and
F_{1} = 2? What numbers follow?
 What if we started with 2 and 3 so that F_{0} = 2 and
F_{1} = 3?
 What other starting values give the same series as the previous two questions?
 The simplest values to start with are
0 and 1, or
1 and 1, or
1 and 2 or even
1 and 0 (in this order)
all of which we recognise as (part of) the Fibonacci series after a few terms.
The next two simplest numbers are 2 and 1.
What if we started with 2 and 1 so that F_{0} = 2 and
F_{1} = 1? Does this become part of the FIbonacci series too?
 Try some other starting values of your own.
 Investigate what happens to the ratio of successive terms in the
series of the earlier questions. We know that for the Fibonacci series,
the ratio gets closer and closer to Phi = (5+1)/2. Does it
look as (oh dear, I feel a pun coming on: Lucas
)
if all the series, no matter what starting values we choose, eventually
have successive terms whose ratio is Phi?
2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843 ..More..
The French mathematician,
Edouard Lucas
(18421891), who gave the
series of numbers 0, 1, 1, 2, 3, 5, 8, 13, .. the name the Fibonacci Numbers,
found a similar series occurs often when investigatng Fibonacci number patterns:
2, 1, 3, 4, 7, 11, 18, ...
The Fibonacci rule of adding the latest two to get the next is kept,
but here we start from 2 and 1 (in this order) instead of 0 and 1 for the
(ordinary) Fibonacci numbers.
The series, called the Lucas Numbers after him, is defined as follows:
where we write its members as L_{n}, for Lucas:
L_{n} = L_{n1} + L_{n2} for n>1
L_{0} = 2
L_{1} = 1
and here are some more values of L_{n} together with
the Fibonacci numbers for comparison:
n:  0  1  2  3  4  5  6  7  8  9  10  ... 
F_{n}:  0  1  1  2  3  5  8  13  21  34  55  ... 
L_{n}:  2  1  3  4  7  11  18  29  47  76  123  ... 
The Lucas numbers have lots of properties similar to those of Fibonacci numbers
and, uniquely among the series you examined in the Things To Do section above,
the Lucas numbers often occur in various formulae for the Fibonacci Numbers. Also,
if you look at many formulae for
the Lucas numbers, you will find the Fibonacci series is there too. The next section
introduces you to some of these equations. So of all the 'general
Fibonacci' series, these two seem to be the most important.
For instance, here is the graph of the ratios of successive Lucas numbers:
1 
= 0·5 

2 

3 
= 3 

1 

4 
= 1·333.. 

3 

7 
= 1·75 

4 

11 
= 1·5714.. 

7 

18 
= 1·6363.. 

11 

29 
= 1·6111.. 

18 

47 
= 1·6206.. 

29 



In fact, for every series formed by adding the
latest two values to get the next, and no matter what two values we start with
we will always end up having terms whose ratio is Phi=1·6180339.. eventually!
2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843 ..More..
Suppose we add up alternate Fibonacci numbers, F_{n1} + F_{n+1};
that is, what do you notice about the two Fibonacci numbers either side of a Lucas
number in the table below: eg
n:  0  1  2  3  4  5  6  7  8  9  10  ... 
F_{n}:  0  1  1  2  3  5  8  13  21  34  55  ... 
L_{n}:  2  1  3  4  7  11  18  29  47  76  123  ... 
Now try your guess on some other Lucas numbers.
This gives our first equation connecting the Fibonacci numbers F(n)
to the Lucas numbers L(n):
L(n) = F(n1) + F(n+1) for all integers n
What about adding alternate Lucas numbers?
n:  0  1  2  3  4  5  6  7  8  9  10  ... 
F_{n}:  0  1  1  2  3  5  8  13  21  34  55  ... 
L_{n}:  2  1  3  4  7  11  18  29  47  76  123  ... 
The sum of L(2)=3 and L(4)=7 is not F(3)=2 However, try a few more additions in this pattern:
L(1)=1 and L(3)= 4 so their sum is 5 whereas F(2)=1;
L(2)=3 and L(4)= 7 so their sum is 10 whereas F(3)=2;
L(3)=4 and L(5)=11 so their sum is 15 whereas F(4)=3;
L(4)=7 and L(6)=18 so their sum is 25 whereas F(5)=5;
Have you spotted the pattern?
5 F(n) = L(n1) + L(n+1) for all integers n
Things to do

 What about the Fibonacci numbers that are TWO places away from Lucas(n)?
n:  0  1  2  3  4  5  6  7  8  9  10  ... 
F_{n}:  0  1  1  2  3  5  8  13  21  34  55  ... 
L_{n}:  2  1  3  4  7  11  18  29  47  76  123  ... 
What is the relationship between F(n2), and F(n+2) that will give L(n)?
 There is also a relationship between F(n3) and F(n+3) that gives L(n).
n:  0  1  2  3  4  5  6  7  8  9  10  ... 
F_{n}:  0  1  1  2  3  5  8  13  21  34  55  ... 
L_{n}:  2  1  3  4  7  11  18  29  47  76  123  ... 
What is it? Write it down as a mathematical formula.
 .. and between F(n4) and F(n+4) to give L(n)?
 Look back at the formula you have just found. Do they work if n is negative (n<0)?
 Can you write down a general expression that relates F(nk) and F(n+k) to give L(n)?
 How about the other way round now!
 We have already found the relationship between L(n1) and L(n+1) that gives F(n)  in fact 5F(n)  above.
What about L(n2) and L(n+2) to give F(n)?
 And now try using L(n3) and L(n+3) to get F(n).
 .. and how can you use L(n4) and L(n+4) to derive F(n)?
 Look back at the formula you have just found. Do they work if n is negative (n<0)?
 Can you write down a general expression that relates L(nk) and L(n+k) to give F(n)?
 Now  the really interesting part!
Have you spotted a pattern in these patterns?
If you have, can you write down a mathematical expression which covers ALL the
formula found in this Things To Do section?
2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843 ..More..
Here is Fibonacci and Lucas
Numbers Calculator to help with the investigations on this page. It opens the
calculator in a separate window.
When we began looking at properties of the Fibonacci numbers, we first examined
Factors of Fibonacci Numbers
and found that if an index number n is a factor of another number m,
then the Fibonacci numbers with n and m as index numbers
are also factors. For example, since 4 is a factor of 8
then Fib(4)=3 is a factor of Fib(8)=24.
If we look at the Fibonacci numbers in the even positions (even index numbers)
that is Fib(2n),
they will all be divisible by Fib(2). But this is 1, which is not very interesting,
so let's have a look at their Fib(n) factors (since n is a factor of 2n also).
Here's a table where F(2n)=kF(n) and we find k for the first few values of n:
n  Fib(n)  2n  Fib(2n)  k=Fib(2n)/Fib(n) 
1  1  2  1  1 
2  1  4  3  3 
3  2  6  8  4 
4  3  8  21  7 
5  5  10  55  
6  8  12  144  
7  13  14  377  
Fill in the gaps in the table above.
Do you recognise the numbers in the final column?
Yes... the Lucas Numbers!
So which Lucas number is a factor of Fib(2n)? Find the index numbers of the values in
the k column. Expressed mathematically, we seem to have found that
Fib(2n) = Fib(n) x Luc(n)
This result can be proved by Induction or by using Binet's formula for Fib(i) and a
similar formula that we will develop below for Lucas numbers.
Suppose we look at those Fibonacci numbers with an index number, n, whose only factor is 2,
that is, those at index numbers 2, 4=2^{2}, 8=2^{3}, 16=2^{4}, 32=2^{5},
64=2^{6}, and so on.
By the formula above:
Fib(4)=3 is a product of Fib(2)=1 and Luc(2)=3.
So Fib(4) = Luc(2)
The next case is Fib(8):
Fib(8) = Fib(4) x Luc(4). Using the result we have just found, we can write this as:
Fib(8) = Luc(2) x Luc(4)
The next case is Fib(16):
Fib(16) = Fib(8) x Luc(8). Again, using the result we just have for Fib(8), we can write this as:
Fib(16) = Luc(2) x Luc(4) x Luc(8)
Can you see the pattern developing here?
A Fibonacci number with an index number in the powersof2 series
2, 4, 8, 16, 32, 64, ...
is a product of all the Lucas numbers with index numbers before it in the same series
Mathematically:
Fib(2^{n}) = Luc(2) x Luc(4) x ... x Luc(2^{n1})
Binet's formula for the Fibonacci numbers in terms of Phi and phi is:
Fib(n) = 
Phi 
n 


– ( –phi ) 
n 





_{}5

Some alternative forms for this equation are:
On the Phi's Fascinating Figures page
the Things To Do in the
Numerical Relationships between Phi and its Powers
section asked you to investigate what happens
when, instead of subtracting the powers of Phi and (phi) as in the formula for
Fib(n) above, we added them:
n:  Phi^{n}  (phi)^{n}  Phi^{n}+(phi)^{n} 
0  1·000000000  1·000000000  2·000000000 
1  1·618033989  0·618033989  1·000000000 
2  2·618033989  0·318196601  3·000000000 
3  4·236067978  0·236067978  4·000000000 
Extend this table by a few more rows. Do the values look like they are integers
always? What integers do they Lucas if they are (hint!)?
Yes! They are the Lucas numbers again:
Lucas(n) = Phi^{n} + ( –phi )^{n}
A.W.W.J.M. van Loon noticed that, since Phiphi=1 and Phi+phi=5
there is a particularly nice way of writing
the Lucas numbers formula that shows a closer relationship
with the Fibonacci numbers formula:
F(n)  =  Phi^{n} – (–phi)^{n}   Phi– (–phi) 


L(n)  =  Phi^{n} + (–phi)^{n}   Phi + (–phi) 

Things to do
 Make a table of the first few powers of Phi=(5+1)/2=1·618033..
starting at the second power (Phi squared).
Round each value to the nearest whole number.
What do you notice?
This is an easier method than the formula given above.
 Take a Fibonacci number, double it and add it to its neighbour on its right. What
do you notice?
Can you prove that your observation is always true?
[Hint: Use the formula for the Lucas numbers given in terms of the Fibonacci numbers.]
 Take F_{2} and multiply it by the Fibonacci number after it:
F_{2}=1 and F_{3}=2 and 1x2=2.
Do this with F_{4},
with F_{6},
with F_{8} and so on.
There is a relationship between the new numbers you have found and the Lucas series.
What is it?
[Hint: multiply your number by 5 and see if it is near a number in the Lucas series.]
Now write the relationship as a mathematical formula.
[You should be able to prove this one if you keep applying the
basic definition of that a Fibonacci number is the sum of the
two previous ones and do this several times!]
Optional extra!
Can you prove that your formula is always true?
This result may help:
F_{n+m} = F_{n1}F_{m} + F_{n}F_{m+1}
 If we sum the first k Fibonacci numbers, the answer is almost another
Fibonacci number. First find the exact formula by continuing the
pattern below for a few more rows, filling in the gaps marked ? and ! so that the
! values are as small as possible and ! is the same value on each line:
F_{1} = F_{?}  !
F_{1} + F_{2} = F_{?}  !
F_{1} + F_{2} + F_{3} = F_{?}  !
...
Now fill in this sentence replacing ? and ! symbols with something more precise:
The sum of the first K Fibonacci numbers is ! less than the ?th Fibonacci number.
 Now try the same pattern as in the previous question, but using L instead of F:
and again @ is to be the same value on each line:
L_{1} = L_{?}  @
L_{1} + L_{2} = L_{?}  @
L_{1} + L_{2} + L_{3} = L_{?}  @
...
and so fill in this sentence:
The sum of the first K Lucas numbers is @ less than the ?th Lucas number.
 Compare F_{2} with F_{1}.
Compare F_{4} with F_{2}.
Compare F_{6} with F_{3}.
Compare F_{8} with F_{4}.
What pattern is emerging?
[Hint: does one divide exactly into the other?]
How is this pattern related to the Lucas numbers?
Now express the pattern as a mathematical equation.
 We have seen that Lucas Number L(n) is also just F(n1)+F(n+1).
So we can ask:
Is there anything special about F(n2)+F(n+2)?
Yes!
They are all multiples of 3 but can you spot which
multiples they are, that is, can you fill in this equation:
F(n2)+F(n+2) = 3 × ?
Try the same thing for
 F(n3)+F(n+3) = 2 × ?
 F(n4)+F(n+4) = 7 × ?
 F(n5)+F(n+5)
 ...
Can you put all these results into one formula:
F(nk) + F(n+k) = ?? x ??
Hint: consider even values of k then look at the odd values of k.
 Surprisingly, there is a similar formula for the Lucas numbers
L(nk)+L(n+k).
Repeat the above investigation for this new expression,
spotting the patterns for k=1, then k=2, k=3,
k=4, and so on, until you can spot the general pattern.
2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843 ..More..
A number trick based on Phi, Lucas and Fibonacci numbers!
Here is a trick that you can use to amaze your friends with your (supposed)
stupendous calculating powers. All you need to remember is a few
Lucas and Fibonacci numbers and you can write down a complicated expression like this:
You can ask them to verify these formulas on their calculators and they will always work out!
The ^{4} by the sign means the fourthroot.
So if
2^{4}  =  16 
"2 to the fourth is 16"
 then 
2  =  ^{4}16
 "2 is the fourthroot of 16" 
You will often find a button on your calculator which
extracts roots (perhaps marked
^{y}x) near the
button which computes the power of a number (marked x^{y}).
If there is no ^{y}x button on your calculator, you can compute ^{4}16
for instance by computing 1/4 first and using this as the y power with x as 16. This is
because
^{y}x = x^{1/y}
What's the secret?
You will need to learn a few of the early Lucas and Fibonacci numbers and their positions
in the sequences:
n:  0  1  2  3  4  5  6  7  8  9  10  ... 
F_{n}:  0  1  1  2  3  5  8  13  21  34  55  ... 
L_{n}:  2  1  3  4  7  11  18  29  47  76  123  ... 
For the example at the head of this section,
I randomly picked the index (column) 4 numbers, F(4)=3 and L(4)=7.
We will use
these three numbers, 4, 3 and 7 in both expressions. Notice that the first expression
has a plus inside its 4throotsign whereas the second has a minus.
Since the position number, 4, is EVEN, I will use a MINUS sign BETWEEN the two expressions.
Now just substitute your values into this formula:

± 

= 1 
The SIGN in the middle is + if n is ODD and – if n is EVEN 
Here is a Fibonacci and Lucas
Numbers Calculator which also generates these expressions for you.
Click on the "Amaze me!" button and see a new example every time.
An even more complicatedlooking variation!
If you want to make it look even more complicated, choose TWO columns in the table,
one for the first expression and one for the second. Here's an example where I use the
fifth and ninth columns:
The sign in the middle (between the two rootexpressions)
will depend on the SECOND POSITION (in the example it was 9):
if it is ODD (and 9 here is odd),
then use PLUS and if it is EVEN put a MINUS sign.
In the new example above, I chose two different positions: 5 for the first
expression and 9 for the second.
For the first expression with position=5, I will then use Fib(5)=5 and Lucas(5)=11.
For the second, with position 9, I will use Fib(9)=34 and Lucas(9)=76.
Since 9, the second choice, is ODD, I will put a PLUS sign between
the two expressions.
Just substitute your two sets of values: N, Lucas(N) and Fib(N); K, Lucas(K) and Fib(K)
in each expression like this, taking care not to mix up your two sets of numbers:
REMEMBER that the first expression always has a plus(+) inside the root sign and
the second always has a minus () inside its rootsign but the
sign inbetween depends on the second (K) value.
Why does it work?
Follow through the suggestions in the following Investigation section and the secret will
be revealed!
Things to do

(a) See what happens in the first nthroot expression if we let n=2.
The first expression is just:
Use your calculator and find its value.
(b) Now try the second expression with n (or k) =2:
Use your calculator and find this value.
(c) Adding the numbers in (a) and (b) should give 1. Does it?
 Repeat the above for n=3 finding the two values:
Check that combining them really does give 1, remembering that since n is ODD, you must
subtract the second from the first, not add it.
 You can try n=4, if you like:
 What do you notice about the values of the separate square, cube and fourthroots
in all the questions above?
 Look at the
Table of relationships between Phi, phi and
5 and see if you can spot the two expressions
in each of questions. So when we take the squareroots in question (1) and the cuberoots
in question (2), and the fourthroots in question (3), what are the results for each
expression?
 Finally, does it matter if we use different columns of figures for the two
expressions in the trick?
Now you know the secret behind this trick!
With thanks to R S (Chuck) Tiberio of Wellesley, MA, USA
for pointing out to me the basic relationships that this trick depends upon.
He was one of the solvers of the original problem which you can find in:
Problem 402 in The College Mathematics Journal, vol. 21, No. 4,
September 1990, page 339.
For a similar unlikelylooking collection of identities see:
Incredible Identities by D Shanks in Fibonacci Quarterly vol 12 (1974)
pages 271 and 280.
2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843 ..More..
The Lucas Numbers in Pascal's Triangle
We found the Fibonacci numbers appearing as
sums of "diagonals" in Pascal's Triangle
on the Mathematical Patterns in the Fibonacci Numbers
page. We can also find the Lucas numbers there too.
Here is the alternative form of Pascal's triangle that we referred to above,
with the diagonals
realigned as columns and the sums of the new columns are the Fibonacci numbers:
 0  1  2  3  4  5  6  7  8  9 
0  1  .  .  .  .  .  .  .  .  . 
1  .  1  1  .  .  .  .  .  .  . 
2  .  .  1  2  1  .  .  .  .  . 
3  .  .  .  1  3  3  1  .  .  . 
4  .  .  .  .  1  4  6  4  1  . 
5  .  .  .  .  .  1  5  10  10  5 
6  .  .  .  .  .  .  1  6  15  20 
7  .  .  .  .  .  .  .  1  7  21 
8  .  .  .  .  .  .  .  .  1  8 
9  .  .  .  .  .  .  .  .  .  1 
 1  1  2  3  5  8  13  21  34  55 
To derive the Lucas numbers we still add the columns, but
to each number in the column we first multiply by its column number and
divide by its row number! Here's an example:
Let's take the third column which, when after the appropriate
multiplications and divisions should sum to L(3) which is 4.
The lowest number in column 3 is 1 and it is on row 3, so we need:
1 x column / row = 1 x 3 / 3 = 1
which, in this case, doesn't alter the number by much!
The other number in column 3 is 2 on row 2, so this time we have:
2 x column / row = 2 x 3 / 2 = 3
Note that for all the numbers in the same column, we will always multiply by the same
number  the column number is the same for all of them  but the divisors will alter
each time.
Adding the numbers we have derived for this column we have 1+3=4
which is the third Lucas number L(3).
Here is what happens in column 4, starting from the bottom again:
1 x 4 / 4 = 1
3 x 4 / 3 = 4
1 x 4 / 2 = 2
SUM = 7
Here's our revised Pascal's triangle from above showing some of the fractions
that we use to derive the Lucas numbers  it shows the pattern in the
multipliers and divisors more easily:

0 
1 
2 
3 
4 
5 
6 
7 
8 
9 
0 
1 
   
   

1 

1x1/1=1 
1x2/1=2 
   
  
2 
 
1x2/2=1 
2x3/2=3 
1x4/2=2 
  
 
3 
  
1x3/3=1 
3x4/3=4 
3x5/3=5 
1x6/3=2 

 
4 
   
1x4/4=1 
4x5/4=5 
6x6/4=9 
4x7/4= 7 
1x8/4= 2 

5 
    
1x5/5=1 
5x6/5=6 
10x7/5=14 
10x8/5=16 
... 
6 
     
1x6/6=1 
6x7/6= 7 
15x8/6=20 
... 
7 
      
1x7/7= 1 
7x8/7= 8 
... 
8 
       
1x8/8= 1 
... 


1 
3 
4 
7 
11 
18 
29 
47 
... 

2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843 ..More..
References
Lucas and Primality Testing Hugh C Williams, Wiley, 1998,
ISBN: 0471 14852 0
is a new book on how to test if a number is prime
without factoring it using a technique developed by Edouard Lucas, with
modern extensions to his work.
Primality testing has become a focus of
modern numbertheory and algorithmics research. Our present inability to find
prime factors of a number
in a fast and efficient way is relied upon in encryption systems  systems which
encode information to send over phone lines.
Such encryption systems
are now built into computer chips in
 cashcard machines which communicate with your bank's central computing service
to check your PIN and to verify the transaction;
 electronic cash transfer over the WWW where your browser encodes the message
 credit card transactions when your card is swiped through a machine at the till
Each of these systems must send the information in a secure way, free
from tampering by fraudsters.
© 19962007 Dr Ron Knott
updated 26 March 2007