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5. Recurrence Relations
We have already encountered a number of recurrence relations when we talked about
recursion. Now we look at more situations in which they arise and techniques for getting
explicit formulas for the sequence they define.
5.1 Modeling with Recurrence Relations
A recurrence relation is a recursive definition of a sequence of numbers. Recall a
sequence of numbers is just a list of numbers
a0, a1, a2, a3, …, an,
Two examples that we have encountered are the following.
(1) The sequence of factorials of the positive integers.
1, 1, 2, 6, 24, 120, …, n!,
Here an = n!.
(2) The Fibonacci sequence
1, 1, 2, 3, 5, 8, …, fn,
Here each element is the sum of the two previous.
We have seen recursive definitions of each of these sequences. For the sequence of
factorials one has
n! =
n (n-1)!

1
if n  1
if n = 0
For the Fibonacci sequence one has
fn =
fn-1

1
+ fn-2
if n  2
if n = 0 or n = 1
In general, a recurrence relation is a recursive definition of a sequence
a0, a1, a2, …, an, …where we are given a formula for the an in terms of some or all of the
aj for j  n – 1. We also need one or more starting values that define the first one or
several aj. For example, for n! we only needed one starting value was 0! = 1, since n!
was defined in terms of (n-1)!. However for the Fibonacci sequence we needed two
5.1 - 1
starting values since fn was defined in terms of fn-1 and fn-2. A recurrence relation is also
called a difference equation.
In this section we look at various situation that can be modeled in terms of recurrence
relations.
Many situations that involved permutations and combinations can also be modeled with
recurrence relations if we take the right point of view.
Example 1. Let Sn be the number of n bit strings that don't contain two consecutive 1's.
Find a recurrence relation for Sn.
Note that S1 = 2 since neither of the one bit strings 0 or 1 contain two consecutive 1's.
However S2 = 3 since only the three two bit sequences 00, 01 and 10 do not contain two
consecutive 1's.
To get a recurrence relation for Sn consider an n bit string b1b2bn. We can divide these
in two parts. The first case is when b1 = 0, i.e. the string is of the form 0b2bn. In this
case b2bn can be any n-1 bit string that does not contain two consecutive 1's. There are
Sn-1 of these. The other case is when b1 = 1, i.e. the string is of the form 1b2bn. In this
case one must have b2 = 0 since if b2 were equal to 1 there would be two consecutive 1's.
So the string is of the form 10b3bn. In this case b3bn can be any n-2 bit string that
does not contain two consecutive 1's. There are Sn-2 of these. Thus
Sn
Sn-1 + Sn-2
= 2
3
if n  3
if n = 1
if n = 2
Note that the basic recurrence relation Sn = Sn-1 + Sn-2 is the same as the Fibonacci
sequence, but the starting values are different.
Another situation where recurrence relations arise is when counting statements in a
recursive algorithm
Example 2. Let Sn be the number of statements executed when the following recursive
algorithm for computing the Fibonacci numbers is called with the value n.
Fibonacci(n)
If n = 0 or n = 1 then
return 1
else
return Fibonacci(n – 1) + Fibonacci(n-2)
Find a recurrence relation for Sn.
5.1 - 2
If n = 0 or n = 1 then the statements "If n = 0 or n = 1 then" and "return 1" are executed,
so S0 = 2 and S1 = 2. If n  2 then the statements "If n = 0 or n = 1 then" and
"return Fibonacci(n – 1) + Fibonacci(n-2)" are executed which is two statements to start
with. However the statement "return Fibonacci(n – 1) + Fibonacci(n-2)" calls the
Fibonacci algorithm twice. Once with the value n-1 and once with the value n-2. The
call with n-1 requires Sn-1 statements to be executed while the call with n-2 requires Sn-2
statements to be executed. So Sn = 2 + Sn-1 + Sn-2. Thus
Sn =
2

2
if n  2
if n = 0 or n = 1
+ Sn-1 + Sn-2
Note that the basic recurrence relation Sn = 2 + Sn-1 + Sn-2 is similar to the Fibonacci
sequence fn = fn-1 + fn-2 , but there is an additional 2 in the formula.
Example 3 – A Home Mortgage. You get a loan to buy a house. Let
a0 = initial amount of the loan
For example, it might be
a0 = $ 100,000
You payoff the loan as follows. Each month the mortgage company charges you a
certain interest rate on the amount you still owe. Let
r = monthly interest rate on the amount you still owe
For example, it might be
r = ½% = 0.005
(This would be an annual rate of 6%.) If the initial amount of the loan was a0 = 100000
and r = ½% then the interest for the first month would be
ra0 = (0.005)(100000) = 500
This amount is added to the loan, so the amount owed becomes
(1)
a0 + ra0 = (1 + r)a0
In the example, it would be
a0 + ra0 = 100000 + 500 = 100500
Each month you make a monthly payment. Let
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P = monthly payment
For example, it might be
P = $ 700
This is subtracted from the amount you owe. So the amount you owe after the first
month is
a1 = (1 + r)a0 - P
In our example this would be
a1 = 100500 - 700 - 99800
The formula (1) applies to every month. If
an = amount owed after month n
Then
an = (1 + r)an-1 - P
This is a recurrence relation. If we couple it with the initial amount of the loan then
if n  1
(1 + r)an-1 - P
an = 
 initial amount of loan
if n = 0
In our example, it is
if n  1
1.05an-1 - 700
an = 
 100000
if n = 0
Example 4 – Cobweb models in Economics. We are modeling how the price of some
commodity changes over time. Let
n = time (measured in some units)
p = pn = price of some commodity at time n
5.1 - 4