Solving Recurrence Relations
A recurrence relation is a function that tells how to find the n th element of a sequence in terms of one
or more of its predecessor. By solving a recurrence relation we mean that we will find a Closed
form solution--a function that can be used to compute a n without knowledge of an-1, an-2, etc.
Homogeneous Linear Recurrences
The easiest type of recurrence to solve is the linear homogeneous recurrence with constant
coefficients. In other words, the recurrence does not have any of the following forms:
an = an-1• an-2, an = an-1 + 5 or an = n• an-1.
Example 1: an = 2 an-1 + 3 an-2 and a0 = a1 = 1
Solution by use of the characteristic equation:
1. Substitute rn for an (rn-1 for an-1, etc.) and simplify the result. For this example the characteristic
equation is rn = 2rn-1 + 3rn-2 which simplifies to: r2 = 2r + 3
2. Find the roots of the characteristic equation:
r2 - 2r - 3 = 0 factors as (r - 3)(r + 1) giving roots r1 = 3 and r2 = -1
When there are two distinct roots, the general form of the solution is: a n = 1•r1n + 2•r2n
where 1 and 2 are constants. In this case, we have: an = 1•3n + 2•(-1)n
3. Using the initial conditions we can find the constants 1 and 2
a0 = 1 = 1 + 2
a1 = 1 = 31 - 2
so 1 = 1/2 and 2 = 1/2 and the final solution is an = (1/2)•3n + (1/2)•(-1)n.
If a characteristic equation has equal roots, (i.e. r1 =r2), then the general solution has the form:
an = 1•rn + 2•n•rn
Example 2: an = 2an-2 – an-4 with a0 = 0, a1 = -2, a2 = 4 and a3 = -4.
In this case, we have a degree four linear homogeneous recurrence whose characteristic equation is
r4 = 2 r2 - 1 or r4 -2 r2 + 1 = 0. This factors as: (r2 - 1)( r2 - 1) = 0 so the roots are r1 = r2 = 1 and
r3 = r4 = -1. (The order of the roots is arbitrary.)
Note there are two pairs of equal roots so there will be two terms each with n as a factor.
The form of the general solution is: an = 1•r1n + 2•n•r2n + 3•r3n + 4•n•r4n
Setting up the equations we have:
a0 = 0 = 1 + 3
a1 = -2 = 1 + 2 - 3 - 4
a2 = 4 = 1 + 22 + 3 + 24
a3 = -4 = 1 +32 - 3 - 34
so 1 = -1/2, 2 = 1/2, 3 = 1/2 and 4 = 3/2 and the final solution is:
an = (-1/2)•1n + (1/2)•n•1n + (1/2)• (-1)n + (3/2)•n•(-1)n = -1/2 + n/2 + (1/2)•(-1)n + (3/2)•n•(-1)n
Other Solution Methods
Example 3: Draw n straight lines on a paper so that every pair of lines intersect, but no three
lines intersect at a common point. Determine how many regions the plane is divided into if n lines
are used. (Draw yourself some pictures.)
If n = 0 then there is 1 region.
If n = 1 there are 2 regions.
If n = 2 there are 4 regions.
If n = 3 there are 7 regions.
In general, the nth line intersects n -1 others and each intersection subdivides a region, so the
number of regions that are subdivided by the n th line is: 1 before the first line, 1 after the last line,
and n - 2 regions between the n -1 lines. This gives us the following recurrence relation:
an = an-1 + n
This can be solved iteratively by backward substitution. Specifically,
= an-2 + (n - 1) + n
= an-3 + (n - 2) + (n - 1) + n
•
•
•
n
= a0 + 1 + 2 + ... + (n - 2) + (n - 1) + n = a0 +  i = 1 + n(n + 1)/2
i=1
Divide and Conquer Recurrence Relations
Example 4: Find the largest and smallest numbers in a list. The following approach is used:
divide the list in half, find the largest and smallest numbers in each half, and compare the large
values to each other and the small values to each other. The measure of complexity is the
number of comparisons of list elements. The base case is when a list has length 2. In this case,
it takes one comparison to determine which is the larger number. In general, for this approach,
the amount of work required is the work needed to solve each of the sublists plus the amount of
work needed to put the solutions from the sublists together. Here, each sublist is half as large as
the original, and it requires two comparisons to determine the largest and smallest values in the
combined list. This leads to the recurrence relation below. Each time, the underlined term is
replaced by the quantity in brackets in the line below.
T(n) = 2T(n/2) + 2 with the base case T(2) = 1.
Assuming n = 2 k, we have
T(n) = 2[2T(n/4) + 2] + 2 = 22T(n/22) + 22 + 2
= 22[2T(n/8) + 2] + 22 + 2 = 23T(n/23) + 23 + 22 + 2
•
•
•
= 2k-1T(n/2k-1) + 2k-1 + 2k-2 + … 23 + 22 + 2
= 2k-1T(2) + 2k-1 + 2k-2 + … 23 + 22 + 2
k-1
= 2k-1 +  2i = 2k-1 + 2k - 2
i=1
2k
= + 2k-1 - 2 = 2k + 2k/2 - 2
= (3/2) 2k - 2 = (3/2) n - 2 since n = 2k.
One of the most common recurrences is a geometric progression in which each term is obtained
from its predecessor by multiplying by a constant. For example, 5, 15, 45, ... is described by the
recurrence relation an = 3 an-1. Various methods (including backward substitution) can be used
to obtain the closed form solution which is an = 5•3n.
Example 5: Consider the relation an2 = 5(an-1) 2 where an > 0 and a0 = 2.
Make the following change of variable: let bn = an2. Then, b0 = 4 and bn = 5 bn-1.
Since this is a geometric series, its solution is bn = 4•5n. Now substituting back,
an = √bn = 2√5n for n  0.
Inhomogeneous Recurrence Relations
We now turn to inhomogeneous recurrence relations and look at two distinct methods of solving
them. These recurrence relations in general have the form an = cn-1an-1 + g(n) where g(n) is a
function of n. (g(n) could be a constant)
One of the first examples of these recurrences usually encountered is the tower of Hanoi
recurrence relation: H(n) = 2H(n-1) + 1 which has as its solution H(n) = 2n – 1. One way to solve
this is by use of backward substitution as above. We’ll see another method shortly.
Example 6: Consider the sequence: 0, 2, 6, 12, 20, 30, 42,  .
between successive terms increases:
a1 – a0 = 2
a2 – a1 = 4
a3 – a2 = 6
•
•
•
an – an-1 = 2n
Now, add all the equations above to get:
an – a0 = 2 + 4 + 6 +  + 2n
= 2[1 + 2 + … + n] = 2[n(n+1)/2].
Notice how the difference
Since a0 = 0, we have an = n(n+1).
Generalizing, this is a first-order inhomogeneous recurrence of the form a n = cn-1an-1 + g(n) for n  1.
n
When c = 1, its general solution is an = a0 +  g(i). To see this, notice the right hand side of each
i=1
equation in example 6 above can be replaced by the appropriate function value. That is,
a1 – a0 = g(1)
a2 – a1 = g(2)
a3 – a2 = g(3)
•
•
•
an – an-1 = g(n)
Then, summing these equations gives the formula above. We could also
use backward substitution to obtain the same result.
While there is no one method of solution that is always guaranteed to work, here's a basic
approach that frequently does. We consider inhomogeneous recurrences of degree at most two:
an = c1 an-1 + c2 an-2 + g(n) where g(n) is a function of n. The techniques below apply to
recurrences in which g(n) is a polynomial or a constant. Two steps are needed to obtain a
solution:
1) first find a solution for the homogeneous part (i.e. , a n = c1 an-1 + c2 an-2) and then
2) find one particular solution of the recurrence and combine the two solutions.
As above, solutions of the homogeneous part will be of the form h(n) = an = 1•r1n + 2•r2n. The
general forms of the particular solutions are described below. Once a form is found, it is
substituted into the recurrence in order to solve for the constants. Call this particular solution
p(n). Then, the solution for the original recurrence is an = h(n) + p(n). All that remains is to use
one or more of the initial conditions to solve for the remaining constants.
Looking again at the first-order inhomogeneous recurrence, an = c•an-1 + g(n), the solution to the
homogeneous part (an = c•an-1) is an = •cn. In finding a particular solution p(n), there are several
cases to consider. These are summarized in the table below. Observe that if g(n) is a polynomial,
then p(n) is a polynomial of the same degree.
g(n)
p(n)
d, a constant

d•n
1n + 0
d•n2
2n2 + 1n + 0
dn
•dn
(See Tucker, Alan, Applied Combinatorics, 2nd ed., p. 286)
Example 7: Consider the Tower of Hanoi recurrence again: H(n) = 2H(n-1) + 1 with H(1) = 1. The
solution to the homogeneous part is •2n. Since g(n) = 1 is a constant, the particular solution has
the form p(n) = . Substituting into the original recurrence to solve for  we get:  = 2 +1 so that
 = -1. Thus, p(n) = -1. This means that H(n) = •2n – 1. Using the fact that H(1) = 1, gives 1 =
•2 –1 so  = 1. Thus, the solution to the original recurrence is H(n) = 2n – 1. (Note that
alternative ways to get this solution are backward substitution and the method shown below.)
Example 8: an = 3an-1 + 5•7n with a0 = 2. (Example from Grimaldi’s Discrete Mathematics book)
The homogeneous part an = 3an-1 has a root of 3, so h(n) has the form •3n. The particular
solution p(n) has the form c•7n. Placing this into the original recurrence relation we get
c•7n = 3c•7n-1 + 5•7n. Dividing through by 7n-1gives 7c = 3c + 35. This gives us c = 35/4 so the
particular solution p(n) = (35/4)•7n or (5/4)•7n+1. Adding the homogeneous and particular
solutions we get an = •3n + (5/4)•7n+1. Using the basis case, a0 = 2 =  + (5/4)•7. Solving for 
we get  = -27/4 and an = (-27/4)•3n + (5/4)•7n+1 or (-1/4)•3n+3 + (5/4)•7n+1.
Example 9: an = 2an-1 + n2 with a1 = 3.
The homogeneous part an = 2an-1 has a root of 2, so h(n) = •2n. The particular solution should
have the form an = 2•n2 + 1•n + 0. Substituting this into the original recurrence gives
2n2 + 1n + 0 = 2[2(n-1)2 + 1(n-1) + 0] + n2. Expanding and combining like terms gives us
0 = n2(2 + 1) + n(-42 + 1) + (22 - 21 + 0). Notice that each parenthesized expression must
evaluate to 0 since the left hand side of the equation is 0. This gives us the following values for
the constants: 2 = -1, 1 = -4 and 0 = -6. Thus, p(n) = -n2 -4n - 6. Then we have
an = •2n -n2 -4n - 6. This gives us  = 7 and the final solution is an = 7•2n -n2 - 4n - 6.
If g(n) consists of two or more terms, then each term is handled separately using the table above.
For example, solving an = 3 an-1 + 6n – 2n, requires finding particular solutions for two
recurrences— an = 3 an-1 + 6n and an = 3 an-1 – 2n.
There is one case in which the particular solutions above will not work. This happens when the
particular solution is a solution of the homogeneous recurrence. Then, the particular solution that
should be tried is a higher degree polynomial. For example. if g(n) = dn and d is a solution of the
homogeneous part then try •n•dn as the particular solution.
Example 10: an = 3an-1 + 2n with a1 = 5.
The homogeneous part an = 3an-1 has a root of 3, so f(n) = •3n. The particular solution should
have the form an = •2n. Substituting this into the original recurrence gives •2n = 3 •2n-1 + 2n.
Solving this equation for  gives us the particular solution p(n) = -2n+1. Thus the solution has the
form an = •3n - 2n+1. Using the initial condition that a1 = 5 we get 5 = 3 - 4 which gives  = 3.
Thus, the final solution is an = 3n+1 - 2n+1.
Another method of solving inhomogeneous recurrence relations
It is sometimes possible to convert an inhomogeneous recurrence relation into a homogeneous
one. Although this method does not widely appear in the literature references to it can be found
at the following URLs: http://www.answers.com/topic/recurrence-relation and
http://mathcircle.berkeley.edu/BMC3/Bjorn1/node10.html.
Let’s consider the Tower of Hanoi recurrence again: H(n) = 2H(n-1) + 1 or H(n) – 2H(n-1) = 1.
Substitute n – 1 for n in this equation to get H(n-1) – 2 H(n-2) = 1. Notice what happens if we
subtract the second equation from the first: H(n) – 2H(n-1) – H(n-1) +2H(n-2) = 1 – 1, or H(n) –
3H(n-1) + 2H(n-2) = 0, a homogeneous linear recurrence relation. The characteristic equation x2
– 3x + 2 = 0 factors as (x-2)(x-1) = 0 so the general form of the solution is
H(n) = 1• 2n + 2•1n = 1• 2n + 2. Using the initial conditions to obtain the constants we find
that 1 = 1 and 2 = -1 and thus we have H(n) = 2n – 1 as we found before.
Example 11: an – an-1 = 2n with a0 = 0 (Example 6)
Using the same method as above we find that an-1 - an-2 = 2(n-1). Now, subtracting we get
an – an-1
= 2n
-an-1 + an-2 = -2(n-1)
an -2an-1 + an-2 = 2
This is not yet a homogeneous recurrence but one more application of
the method will give us a linear homogeneous recurrence relation.
an -2an-1 + an-2
=2
-an-1 + 2an-2 – an-3 = -2
an -3an-1 +3an-2 –an-3 = 0 has characteristic equation x3 – 3x2 + 3x - 1 = 0 that factors as (x – 1)3.
The general form of the answer must be an = 1•1n +2•n•1n + 3•n2•1n or 1 + 2•n + 3•n2.
Since a0=0 we immediately get that 1 = 0. a1 = 2 and a2 = 6 giving us
2 = 2 + 3 and 6 = 22 + 4 3. Solving for 2 and 3 we get 2 = 1 and 3 = 1 so the solution to
the recurrence is an = n + n2 or n(n + 1).
The same method can be used to solve recurrence relations in which g(n) is a higher order
polynomial. The substitute n – 1 for n and subtract may have to be used several times, but
eventually you will obtain a linear homogeneous recurrence relation. (However, factoring the
characteristic equation could turn into a challenge.)
A similar method can be used if the function g(n) is an exponential.
Example 12: an = 3an-1 + 2n with a1 = 5. (Example 9)
Rewriting the recurrence relation gives us an - 3an-1 = 2n. First we multiply by 2 to get
2an - 6an-1 = 2n+1. Now substitute n – 1 for n in this new equation to get
2an-1 - 6an-2 = 2n. Now, subtracting this new recurrence from the original one we get:
an - 3an-1
= 2n
-2an-1 + 6an-2 = -2n
an – 5an-1 + 6an-2 = 0. The characteristic equation is x2 – 5x + 6 = 0 which factors as (x - 3)(x – 2)
= 0, so the answer has the form an = 1•3n + 2•2n. Using the facts that a1 = 5 and a2 = 19 we get
5 = 31 + 2 2 and 19 = 91 + 42. Solving for the constants we get 1 = 3 and 2 = -2 giving us
the solution an = 3•3n + (-2)•2n = 3n+1 – 2n+1.