1. Zeroes of a Recurrence Relation
Activity 2-17:
Zeroes of a Recurrence Relation

2. We all know the Fibonacci sequence,
1, 1, 2, 3, 5, 8, 13...
We can write this as
un+1= un + un-1, with u1 = 1, u2 = 1.
So the next term is the sum of the previous two,
and we have some initial conditions.
We have here an example of a
Linear Recurrence Relation (or LRS).

3. un+1 = anun + an-1un-1 + an-2un-2 +......+ an-k+1un-k+1
is the general LRS of order k.
So the next term is a linear combination
of the last k terms (we also need initial conditions).
We will only be looking at the cases
where every ai is an integer:
we will call these ‘integer LRSs’.
Now there is a way to calculate future terms
for an LRS using a matrix.
Let’s demonstrate this
using the Fibonacci sequence.

4. and if we have some computer help in taking the powers of a matrix,
So it is clear that
and if we have some computer help
in taking the powers of a matrix,
we have a good way of calculating future terms of an LRS.

5. Task: use this spreadsheet to find the 21st term,
that’s F21, in the Fibonacci sequence.
Taking Powers of a Matrix spreadsheet
The 21st term
in the
Fibonacci
sequence is

6. Task: given the LRS un+1= un + un-1 + un-2
with u1 = 1, u2 = 1, u3 = 1, find the 22nd term.
Using a 3 x 3 matrix this time:
The 22nd term
in this
order-3
LRS
sequence is

7. Now we can note that it makes perfect sense
to run the Fibonacci sequence backwards.
...5, -3, 2, -1, 1, 0, 1, 1, 2, 3, 5, 8, 13...
So u0 = 0, u-1 = 1, u-2 = -1, and so on: the rule still holds.
Now a further question:
how many 0s are there in the Fibonacci sequence?
It seems clear that there can only be one.
In that case, how many 0s can any order-2 LRS have?
Can it have an infinite number? Or is there a maximum?

8. It is easy to construct an LRS
that does have an infinite number of 0s.
Consider un+2 = 2un, with u1 = 0, u2 = 2.
This gives us the sequence 0, 2, 0, 4, 0, 8, 0, 16...
This clearly has an infinite number of 0s, BUT
we call this type of LRS degenerate.
The LRS un+1 = anun + an-1un-1 has associated with it
the characteristic equation 2 = an + an-1.
This has two roots, 1 and 2, and if their ratio
is a root of unity, we say the LRS is degenerate.

9. So for our example un+2 = 2un, the characteristic equation is
2 = 2, and so 1, 2 are ±√2, and their ratio is -1.
The solution to an LRS is un = A1n+ B2n .
We can find A and B from the initial conditions.
So for our Fibonacci sequence, the characteristic equation is
2  1 = 0,
which gives solutions 1, 2 =
Using u1 = 1, u2 = 1, we find A = , B = , and so
Task: test this out for various n.
Fn =

10. So let’s go back to our question and rephrase it:
how many 0s can a non-degenerate order-2 LRS have?
Theorem: Skolem-Mahler-Lech (proved in 1953).
Any non-degenerate integer LRS
has a finite number of 0s.
It can also be proved that the largest number of 0s
an order-2 non-degenerate integer LRS can have is 1,
(so the Fibonacci sequence has the maximum).
How can we prove this?

11. The order-2 LRS un+1 = anun + an-1un-1 has associated with it
the characteristic equation 2 = an + an-1.
Suppose the roots are 1, 2 .
So we have un = A1n+ B2n .
Suppose un = 0 for n = p and n = q, with p > q.
0 = A1p+ B2p 
Similarly,
Subtracting, we have ,
and thus the ratio of the roots is a root of unity,
and the LRS is degenerate.
So the largest number of 0s
an order-2 non-degenerate integer LRS can have is 1.

12. What is the largest number of 0s
an order-3 non-degenerate integer LRS can have?
Beukers has proved (1991) that the answer is 6.
Is it possible to find an order-3 LRS with six 0s?
The mathematician Berstel
managed to do exactly that.

13. Task: using our matrix spreadsheet,
see if you can find all six zeroes there are to be found.
So we have zeroes
for a0, a1, a4, a6, a13,
and rather surprisingly,
for a52.

14. With thanks to: Graham Everest and Tom Ward.
Carom is written by Jonny Griffiths,