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 Things To Do icon means there are Things To do investigations at the end of that section.

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Other starting values for a "Fibonacci" series

The definition of the Fibonacci series is:
Fn+1 = Fn-1 + Fn , if n>1
F0 = 0
F1 = 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 /

  1. 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 F0 = 1 and F1 = 2? What numbers follow?
  2. What if we started with 2 and 3 so that F0 = 2 and F1 = 3?
  3. What other starting values give the same series as the previous two questions?
  4. 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 F0 = 2 and F1 = 1? Does this become part of the FIbonacci series too?
  5. Try some other starting values of your own.
  6. 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 = (sqrt5+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?

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The Lucas series

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 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 Ln, for Lucas:
Ln = Ln-1 + Ln-2 for n>1
L0 = 2
L1 = 1
and here are some more values of Ln together with the Fibonacci numbers for comparison:
n: 0 1 2 3 4 5 6 7 8 9 10 ...
Fn: 0 1 1 2 3 5 8 13 21 34 55 ...
Ln: 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
Lucas ratios tend to Phi

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!

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Two formulae relating the Lucas and Fibonacci numbers

Suppose we add up alternate Fibonacci numbers, Fn-1 + Fn+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 ...
Fn: 0 1 1 2 3 5 8 13 21 34 55 ...
Ln: 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 ...
Fn: 0 1 1 2 3 5 8 13 21 34 55 ...
Ln: 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 /

    1. 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 ...
      Fn: 0 1 1 2 3 5 8 13 21 34 55 ...
      Ln: 2 1 3 4 7 11 18 29 47 76 123 ...
      What is the relationship between F(n-2), and F(n+2) that will give L(n)?
    2. There is also a relationship between F(n-3) and F(n+3) that gives L(n).
      n: 0 1 2 3 4 5 6 7 8 9 10 ...
      Fn: 0 1 1 2 3 5 8 13 21 34 55 ...
      Ln: 2 1 3 4 7 11 18 29 47 76 123 ...
      What is it? Write it down as a mathematical formula.
    3. .. and between F(n-4) and F(n+4) to give L(n)?
    4. Look back at the formula you have just found. Do they work if n is negative (n<0)?
    5. Can you write down a general expression that relates F(n-k) and F(n+k) to give L(n)?
  1. How about the other way round now!
    1. 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) to give F(n)?
    2. And now try using L(n-3) and L(n+3) to get F(n).
    3. .. and how can you use L(n-4) and L(n+4) to derive F(n)?
    4. Look back at the formula you have just found. Do they work if n is negative (n<0)?
    5. Can you write down a general expression that relates L(n-k) and L(n+k) to give F(n)?
  2. 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?

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Here is Fibonacci and Lucas Numbers Calculator to help with the investigations on this page. It opens the calculator in a separate window.

Lucas Factors of Fibonacci Numbers

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)
11211
21433
32684
438217
551055 
6812144 
71314377 
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.

A special case...

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=22, 8=23, 16=24, 32=25, 64=26, 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 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
Mathematically:
Fib(2n) = Luc(2) x Luc(4) x ... x Luc(2n-1)

A formula for the Lucas Numbers involving Phi and phi

Binet's formula for the Fibonacci numbers in terms of Phi and phi is:
Fib(n) = Phi
n
( –phi )
n

sqrt5
Some alternative forms for this equation are:
Binet Formula

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:Phin (-phi)n Phin+(-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 Luc-as if they are (hint!)? Yes! They are the Lucas numbers again:
Lucas(n) = Phin + ( –phi )n
A.W.W.J.M. van Loon noticed that, since Phi-phi=1 and Phi+phi=sqrt5 there is a particularly nice way of writing the Lucas numbers formula that shows a closer relationship with the Fibonacci numbers formula:
F(n) =
Phin – (–phi)n
Phi– (–phi)
 
L(n)=
Phin + (–phi)n
Phi + (–phi)

/ Things to do /

  1. Make a table of the first few powers of Phi=(sqrt5+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.
  2. 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.]
  3. Take F2 and multiply it by the Fibonacci number after it:
    F2=1 and F3=2 and 1x2=2.
    Do this with F4,
    with F6,
    with F8 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: Fn+m = Fn-1Fm + FnFm+1
  4. 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:
    F1 = F? - !
    F1 + F2 = F? - !
    F1 + F2 + F3 = 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.
  5. 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:
    L1 = L? - @
    L1 + L2 = L? - @
    L1 + L2 + L3 = L? - @
    ...
    and so fill in this sentence:
    The sum of the first K Lucas numbers is @ less than the ?-th Lucas number.
  6. Compare F2 with F1.
    Compare F4 with F2.
    Compare F6 with F3.
    Compare F8 with F4.
    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.
  7. We have seen that Lucas Number L(n) is also just F(n-1)+F(n+1).
    So we can ask:
    Is there anything special about F(n-2)+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(n-2)+F(n+2) = 3 × ?
    Try the same thing for
    • F(n-3)+F(n+3) = 2 × ?
    • F(n-4)+F(n+4) = 7 × ?
    • F(n-5)+F(n+5)
    • ...
    Can you put all these results into one formula:
    F(n-k) + F(n+k) = ?? x ??
    Hint: consider even values of k then look at the odd values of k.
  8. Surprisingly, there is a similar formula for the Lucas numbers L(n-k)+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.

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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:
4
sqrt  7 + 3 sqrt5

2
4
sqrt  7 – 3 sqrt5

2
= 1
You can ask them to verify these formulas on their calculators and they will always work out! The 4 by the sqrt sign means the fourth-root. So if
24 = 16 "2 to the fourth is 16" then
2 = 4sqrt16 "2 is the fourth-root of 16"

You will often find a button on your calculator which extracts roots (perhaps marked ysqrtx) near the button which computes the power of a number (marked xy). If there is no ysqrtx button on your calculator, you can compute 4sqrt16 for instance by computing 1/4 first and using this as the y power with x as 16. This is because
ysqrtx = x1/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 ...
Fn: 0 1 1 2 3 5 8 13 21 34 55 ...
Ln: 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 4-th-root-sign 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:

n
sqrt  L(n) + F(n) sqrt5

2
±
n
sqrt  L(n) – F(n) sqrt5

2
= 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 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 where I use the fifth and ninth columns:
5
sqrt  11 + 5 sqrt5

2
+
9
sqrt  76 – 34 sqrt5

2
= 1
The sign in the middle (between the two root-expressions) 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:

n
sqrt  L(n) + F(n) sqrt5

2
±
k
sqrt  L(k) – F(k) sqrt5

2
= 1
REMEMBER that the first expression always has a plus(+) inside the root sign and the second always has a minus (-) inside its root-sign but the sign in-between 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 /

  1. (a) See what happens in the first n-th-root expression if we let n=2. The first expression is just:
    sqrt  3 + 1 sqrt5

    2
    Use your calculator and find its value.
    (b) Now try the second expression with n (or k) =2:
    sqrt  3 – 1 sqrt5

    2
    Use your calculator and find this value.
    (c) Adding the numbers in (a) and (b) should give 1. Does it?
  2. Repeat the above for n=3 finding the two values:
    3
    sqrt  4 + 2 sqrt5

    2
    and
    3
    sqrt  4 – 2 sqrt5

    2
    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.
  3. You can try n=4, if you like:
    4
    sqrt  7 + 3 sqrt5

    2
    and
    4
    sqrt  7 – 3 sqrt5

    2
  4. What do you notice about the values of the separate square-, cube- and fourth-roots in all the questions above?
  5. Look at the Table of relationships between Phi, phi and sqrt5 and see if you can spot the two expressions in each of questions. So when we take the square-roots in question (1) and the cube-roots in question (2), and the fourth-roots in question (3), what are the results for each expression?
  6. 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:
Article: Problem 402 in The College Mathematics Journal, vol. 21, No. 4, September 1990, page 339.

For a similar unlikely-looking collection of identities see:
Article: Incredible Identities by D Shanks in Fibonacci Quarterly vol 12 (1974) pages 271 and 280.

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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 re-aligned as columns and the sums of the new columns are the Fibonacci numbers:

 0123456789
01.........
1.11.......
2..121.....
3...1331...
4....146 41.
5.....1510105
6......1 61520
7....... 1721
8........18
9.........1
 11235813213455
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 ...

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References

Book: 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 Each of these systems must send the information in a secure way, free from tampering by fraudsters.

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updated 26 March 2007