# finding recurrence relation - question #2

• November 25th 2009, 08:56 PM
sbankica
finding recurrence relation - question #2
Find a recurrence relation for an, n>=0 where an is the number of n-digit sequences (each digit taken from {0, 1, 2, . . . , 9}) that contain “666” somewhere in the sequence.
• November 26th 2009, 05:26 AM
Soroban
Hello, sbankica!

Quote:

Find a recurrence relation for $a_n,\:n \geq 0$,
where $a_n$ is the number of n-digit sequences
(each digit taken from {0, 1, 2, ... 9})
that contain “666” somewhere in the sequence.

Consider $a_7$, the number of 7-digit numbers that contain "666".

One of them is: . $N \:=\:ab666cd$

Write it like this: . $\_\:a\:\_\:b\:\_\:6\,6\,6\:\_\:c\:\_\:d\:\_$

There are 6 spaces ( $n-1$ spaces) in which to insert the eighth digit.
. . and there are 10 choices for the eighth digit.

Hence, there are: $(10)(6)$ ways to change $N$ into an 8-digit number.

That is: . $a_8 \;=\;(10)(6)a_7$

In general: . $a_n \;=\;10(n-1)a_{n-1}$

Edit: I suspect that there is some duplication in my solution,
. . . .but I'm unable to eliminate it.
.
• November 26th 2009, 06:45 AM
Shanks
The above solution is obviously wrong!
This problem is very complicated!
• November 26th 2009, 08:19 AM
Shanks
Let b(n) be the number of n-digit sequence which contains no 666
anywhere in the sequence.
Obviously $a(n)=10^{n}-b(n)$.
Divide b(n) into two classes:
Class A: s(n) be the number of sequence such that the n-th digit is not 6.
Class B: t(n) be the number of sequence such that the n-th digit is 6.
Then b(n)=s(n)+t(n),
To find b(n+1), we need to find s(n+1) and t(n+1):
It is very easy to (see)verify that, we have the following recursive definition:
s(n+1)=9b(n);
t(n+1)=s(n)+s(n-1);
And
s(1)=9,t(1)=1,b(1)=10;
s(2)=90,t(2)=10,b(2)=100.