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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 (1842-1891) (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.
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?
The Fibonacci numbers!
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?
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?
Yes! No matter what values we start with, positive or negative,
the ratio of two neighbouring terms will either be Phi or else -phi
The French mathematician,
Edouard Lucas
(1842-1891), 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 he was investigating 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:
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 positive values we start with
we will always end up having terms whose ratio is Phi=1·6180339.. eventually!
Suppose we add up alternate Fibonacci numbers, F_{n-1} + F_{n+1};
that is, what do you notice about the two Fibonacci numbers either side of a Lucas
number in the table below? Tap (click) on a grayed Lucas number to highlight the neighbouring Fibonacci 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
...
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(n-1) + 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(n-1) + L(n+1) for all integers n
Things to do
What about the Fibonacci numbers that are TWO places away from Lucas(n)?
Click on a Lucas number to highlight the Fibonacci's to add.
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(n-2), and F(n+2)? You should be able to find a simple formula
that does not involve any Lucas number.
F(n-2) + F(n+2) = 3 F(n)
There is also a relationship between F(n-3) and F(n+3) that does involve 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.
F(n-3) + F(n+3) = 2 L(n) for all integers n
.. and between F(n-4) and F(n+4)?
F(n-4) + F(n+4) = 7 F(n) for all integers 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(n-k) and F(n+k) which covers all the
formula above and applies to a general k? It is easier if you consider first the even values of k:
F(n-k) + F(n+k) = F(n)L(k) for all integers n if k is even
and then the odd values of k:
F(n-k) + F(n+k) = F(k)L(n) for all integers n if k is odd
How about adding Lucas numbers in the same way as we did with Fibonacci's above?
We have already found the relationship between L(n-1) and L(n+1) that gives F(n) - in fact 5F(n) - above.
What about L(n-2) and L(n+2)?
L(n-2) + L(n+2) = 3 L(k) for all integers n
And now try using L(n-3) and L(n+3) to get F(n).
L(n-3) + L(n+3) = 10 F(k) for all integers n
.. and of what is L(n-4) and L(n+4) a simple multiple?
L(n-4) + L(n+4) = 11 F(n) for all integers 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(n-k) and L(n+k) first
for even k and then for odd k?
L(n-k) + L(n+k) = L(n)L(k) for all integers n if k is even
L(n-k) + L(n+k) = 5F(k)F(n) for all integers n if k is odd
Now repeat this Things To Do but for F(n+k) - F(n-k) and L(n+k) - L(n-k).
What are the formulas this time?
F(n+k) – F(n–k) = F(n)L(k), k odd
F(n+k) – F(n–k) = L(n)F(k), k even
L(n+k) – L(n–k) = L(n)L(k), k odd
L(n+k) – L(n–k) = 5F(n)F(k), k even
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. Can you write this mathematically?
F(2n) = F(n) L(n)
This result can be proved by Induction or by using Binet's formula for F(n) and a
similar formula that we will develop below for Lucas numbers.
Suppose we look at those Fibonacci numbers with an index number, n, which is a power of 2,
that is, those Fibonacci numbers 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:-
F(4)=3 is a product of F(2)=1 and L(2)=3.
So F(4) = L(2)
The next case is F(8):-
F(8) = F(4) x L(4). Using the result we have just found, we can write this as: F(8) = L(2) × L(4)
The next case is F(16):-
F(16) = F(8) x L(8). Again, using the result we just have for F(8), we can write this as: F(16) = L(2) × L(4) × L(8)
Can you see the pattern developing here?
A Fibonacci number with an index number in the powers-of-2 series
2, 4, 8, 16, 32, 64, ...
is a product of all the Lucas numbers with index numbers before it in the same series
Extend this table by a few more rows.
Do the values look like they are integers
always? What integers do they Luc-as 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 Phi-phi=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
Using the above table,round the powers of
Phi.
What do you notice? Which value does not fit the pattern?
Round( Phi^{n} ) = L(n) except for n=2. For all values of n>2, this method is accurate.
This is an easier method than the formula given above if we are careful about the exception.
Take a Fibonacci number, double it and add this to its neighbour on the right. What
do you notice?
Can you prove that your observation is always true?
[Hint: Use the first formula for the Lucas numbers
given in terms of the Fibonacci numbers.]
2 F(n) + F(n+1)
= F(n) + ( F(n) + F(n+1) )
= F(n) + F(n+2) by the Fibonacci Rule
= L(n+1) by the First Formula for Lucas Numbers above
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
...
In the table above, multiply a Lucas number L(n) by the Fibonacci number in the next column F(n+1).
Can we write this in terms of another Fibonacci number?
n
L(n)
F(n+1)
L(n)F(n+1)
nearest Fib
2
3
2
6
5=F(5)
3
4
3
12
13=F(7)
4
7
5
35
34=F(9)
5
11
8
88
89=F(11)
This suggests L(n) F(n+1) = F(2n+1) –1
Try the previous investigation but with F(n) and L(n+1)
If we sum the first k Fibonacci numbers, the answer is almost another
Fibonacci number. First that a good guess at the exact formula by continuing the
calculating the pattern for a few more terms:
The sum of the first n Fibonacci numbers is F(n+2) – 1
Now try the same pattern as in the previous question, but using Lucas numbers to
sum instead of Fibonaccis. Start from L(0)=2. What is the formula this time?
The sum of the first n Lucas numbers L(n+2) – 1
It is perhaps surprising that almost always when when we try to find a formula derived from Fibonacci
numbers we find the Lucas numbers are there too!
There are many more formulae involving Fibonacci and Lucas numbers and Phi and phi on my
Fibonacci and Phi Formulae page.
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:
^{3}
√
4 + 2 √5
2
=
_{4}
√
7 + 3 √5
2
You can ask them to verify these formulas on their calculators and they will always work out!
The ^{4} by the sign means the fourth-root.
So if
2^{4}
=
16
"2 to the fourth is 16"
then
2
=
^{4}16
"2 is the fourth-root 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:
Phi^{2} = Phi + 1, so, multipying by Phi we have
Phi^{3} = Phi^{2} + Phi^{1} and, continuing to multiply by Phi gives:
Phi^{n} = Phi^{n-1} + Phi^{n-2} Let's call this The Phibonacci Rule
We can use this formula in another way as follows:
1
=
1
Phi
=
Phi
now add these two rows using the Phibonacci Rule
Phi^{2}
=
1
+
Phi
and again, adding the last two rows:
Phi^{3}
=
1
+
2 Phi
and again...
Phi^{4}
=
2
+
3 Phi
and again...
Phi^{5}
=
3
+
5 Phi
so we see (and can prove properly) that
Phi^{n} = Fib(n-1) + Fib(n) Phi
But Phi = (1 + √5)/2
Substituting this in the above table we have
1
=
1
=
(2)/2
Phi
=
Phi
=
(1 + √5)/2
Phi^{2}
=
1
+
Phi
=
(3 + √5)/2
Phi^{3}
=
1
+
2 Phi
=
(4 + 2√5)/2
Phi^{4}
=
2
+
3 Phi
=
(7 + 3√5)/2
Phi^{5}
=
3
+
5 Phi
=
(11 + 5√5)/2
or, generally,
Phi^{n} = ( L(n) + F(n)√5 )/2 which we can write as Phi = ^{n}√^{ } L(n) + F(n)√5 )/2
This is the secret behind the "trick". Choose two different values of n and they are both Phi and so
equal to each other!
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 complicated-looking 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:
^{5}
√
11 + 5 √5
2
–
_{10}
√
123 – 55 √5
2
= 1
In the new example above, I chose two different positions: 5 for the first
expression and 10 for the second, which must always be even.
For the first expression with position=5, I will then use Fib(5)=5 and Lucas(5)=11.
For the second, with position 10, I will use Fib(10)=123 and Lucas(10)=55.
This second position should always use br even number. .
Just substitute your two sets of values: N, Lucas(N) and Fib(N); K (an EVEN number!), Lucas(K) and Fib(K)
in each expression like this, taking care not to mix up your two sets of numbers:
^{n}
√
L(n) + F(n) √5
2
–
_{k}
√
L(k) – F(k) √5
2
= 1
REMEMBER that the first expression always has a plus(+) inside the root sign and
the second always has a minus (-) inside and the second value, k must be even.
Why does it work?
Follow through the suggestions in the following Investigation section and the secret will
be revealed!
Things to do
Repeat the above but for phi instead of Phi.
Start from phi = Phi - 1 and note that
Phi = 1/phi = phi^{-1} so phi = -1 + phi^{-1}
Starting with the Phibonacci RulePhi^{n} = Phi^{n-1} + Phi^{n-2}
Divide it by Phi^{n}
Then multiply it by phi^{n}
to get a rule for adding two powers of phi to get the next: phi^{n} = phi^{???} + phi^{???}.
For positive n, rewrite it to put the largest power of phi on one side of the equation
and check that you have:
phi^{n+2} = phi^{n} – phi^{n+1}
Let's call this The phibonacci Rule -- note phi instead of Phi here!
Use The phibonacci Rule
of the last question
to complete this table of A±B phi forms,
starting with n=0 and n=1.
Then use phi^{2} = phi^{0} – phi^{1} to get
phi^{2} and so on:
phi^{0}
=
1
phi^{1}
=
phi
now subtract this row the one above:
phi^{2}
=
1
–
phi
and again, subtracting this row above from the one above:
phi^{3}
=
– ...
+
... phi
and again...
phi^{4}
=
...
–
...phi
and again...
phi^{5}
=
– ...
+
...phi
Now add in an extra column using
phi = (√5 – 1)/2
1
=
1
=
( 2)/2
phi
=
phi
=
(√5 – 1)/2
phi^{2}
=
1
–
phi
=
(–√5 + 3)/2
phi^{3}
=
–1
+
2 phi
=
(2√5 – 4)/2
phi^{4}
=
2
–
3 phi
=
(–3√5 + 7)/2
phi^{5}
=
–3
+
5 phi
=
(5√5 – 11)/2
The first numbers are negative whenever the power is negative.
This suggests that using powers of –phi might be worth looking at.
Multipying the rows for odd powers of phi by –1
then using –phi^{odd number} = (–phi)^{odd number},
express the whole table as powers of –phi:
(–phi)^{0}
=
1
=
(2 )/2
(–phi)^{1}
=
–
phi
=
(1 – √5 )/2
(–phi)^{2}
=
1
–
phi
=
(3 – √5 )/2
(–phi)^{3}
=
1
–
2 phi
=
(4 – 2√5 )/2
(–phi)^{4}
=
2
–
3 phi
=
(7 – 3√5 )/2
(–phi)^{5}
=
3
–
5 phi
=
(11 – 5√5 )/2
What does The phibonacci Rule become if we use powers of –phi only?
Write out the rule for powers of –phi that summarises the table in the
previous question:
Returning to our "trick", if we always let the power k be even, what single value for even powers k is
( L(k) – F(k)√5 )/2
Finally, using the result from the first "trick", what is the (constant) value of
^{n}√( L(n) + F(n)√5 )/2
–
^{k}√( L(k) – F(k)√5 )/2
if k is even?
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 unlikely-looking collection of identities see: Incredible Identities by D Shanks in Fibonacci Quarterly vol 12 (1974)
pages 271 and 280.
Here is the alternative form of Pascal's triangle that we referred to above,
with the diagonals
re-aligned 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 × 4 / 4 = 1
3 × 4 / 3 = 4
1 × 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:
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 number-theory 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
cash-card 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.