MATH 681
Notes
Combinatorics and Graph Theory I
1
Recurrence relations, continued continued
Linear homogeneous recurrences are only one of several possible ways to describe a sequence as a
recurrence. Here are several other situations which may arise.
1.1
Linear nonhomogeneous recurrence relations
A linear nonhomogeneous recurrence relation is one which has a linear homogeneous form as well
as a nonhomogeneous term. here’s an example:
Question 1:
How many strings of
n
numbers from the set
{
0
,
1
,
2
,
3
,
4
}
are there so that there is
at least one “1” and the first “1” occurs before any “0”?
Answer 1:
Let us call the above number
a
n
, and a string satisfying the above description “valid”.
To produce a valid string of length
n
, we can consider every possible first number in the string as
an individual case.
A string beginning with “0” is necessarily invalid. If a string begins with “1”, everything that could
follow that is valid. Thus, of the
5
n

1
strings beginning with “1”, all are valid. If a string begins
with a “2”, “3”, or “4”, then it is valid if and only if the remaining
n

1
terms form a valid string,
which could happen in
a
n

1
ways.
Thus,
a
n
= 3
a
n

1
+ 5
n

1
.
This is quite similar to a firstorder linear homogeneous differential
equation, except it has that inconvenient
5
n

1
term. Unfortunately, we don’t have the tools to solve
that, yet!
In addition, note the initial condition
a
0
= 0
, since the string of length zero contains no “1”s and
is thus invalid. We could, if we choose, work out a few small values from this:
a
1
= 1
,
a
2
= 8
,
a
3
= 49
, and so forth.
We’ll define the class of problems to which this belongs, and discuss solution techniques:
Definition 1.
A sequence
{
a
n
}
is given by a
linear nonhomogeneous recurrence relation of order
k
if
a
n
=
c
1
a
n

1
+
c
2
a
n

2
+
c
3
a
n

3
+
· · ·
+
c
k
a
n

k
+
p
(
n
) for all
n
≥
k
. The recurrence relation
b
n
=
c
1
b
n

1
+
c
2
b
n

2
+
c
3
b
n

3
+
· · ·
+
c
k
b
n

k
is referred to as the
associated linear homogeneous
recurrence relation
One result is as easy to show for LNRRs as for LHRRs; the following can be proven as a very slight
variation of the similar proof for LHRRs.
Proposition 1.
A sequence is uniquely determined by an LNRR of order
k
and the initial values
a
0
, a
1
, a
2
, . . . , a
k

1
.
However, there is one very important difference between LNRRs and LHRRs: linear combinations
of LNRRsatisfying sequences do no, in general, satisfy the LNRR. Howver, we do have the result:
Proposition 2.
If
{
a
n
}
satisfies an LNRR, and
{
b
n
}
satisfies the associated LHRR, then
{
a
n
+
b
n
}
satisfies the LNRR.
Proof.
We know that
a
n
=
c
1
a
n

1
+
c
2
a
n

2
+
c
3
a
n

3
+
· · ·
+
c
k
a
n

k
+
p
(
n
)
Page 1 of 5
October 1, 2009