MATH 681
Notes
Combinatorics and Graph Theory I
1
Recurrence relations, continued yet again
One last lousy class of recurrences we should be able to solve:
1.1
Systems of recurrence relations
Sometimes multiple recurrences working in tandem are more effective than a single recurrence.
Let’s see how such a thing might be of assistance.
Question 1:
Let
a
n
represent the number of strings from
{
0
,
1
,
2
}
of length
n
with an even number
of 0s. Find a recurrence for
a
n
.
Answer 1:
Note that recurrences are not actually the best way to solve this particular problem:
it would be a lot easier to do it with the exponential generating function
e
x
+
e

x
2
e
x
e
x
=
e
3
x
+
e
x
2
=
∑
∞
n
=0
3
n
+1
2
x
n
n
!
to get
a
n
=
3
n
+1
2
. But, armed with that knowledge, let’s look at how we might tackle
it as a recurrence
We introduce an auxiliary sequence
b
n
, which counts the bitstrings of length
n
with an odd number
of zeroes. Now, we know:
a
n
= 2
a
n

1
+
b
n

1
b
n
=
a
n

1
+ 2
b
n

1
and
a
0
= 1
and
b
0
= 0
. This uniquely determines
a
n
, but solving it might be tricky. We can have
recourse to either OGFs or EGFs to solve it, depending whether you prefer algebra or differential
equations.
Let’s suppose we use an OGF; letting
f
(
x
) =
∑
∞
n
=0
a
n
x
n
and
g
(
x
) =
∑
∞
n
=0
b
n
x
n
. Then the first
equation in the recurrence can be simplified to
f
(
x
)

a
0
= 2
xf
(
x
) +
xg
(
x
)
; the second will become
g
(
x
)

b
0
=
xf
(
x
) + 2
xg
(
x
)
. Collecting terms, we’ll get the system:
f
(
x
) =
xg
(
x
) + 1
1

2
x
g
(
x
) =
xf
(
x
)
1

2
x
so, solving for
f
(
x
)
in terms of itself,
f
(
x
) =
x
xf
(
x
)
1

2
x
+ 1
1

2
x
=
x
2
f
(
x
)

2
x
+ 1
(1

2
x
)
2
and thus
1

x
2
(1

2
x
)
2
f
(
x
) =
1

2
x
(1

2
x
)
2
so
f
(
x
) =
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 Fall '09
 WILDSTROM
 Combinatorics, Graph Theory, Recurrence relation, Generating function, recurrence

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