Recurrence relation for The number of quaternary strings of length n for which the sum of all the entries is divisible by 3

I am not sure how to get the recurrence relation but I this is how I started.

All mod 3 Sum of entries is 0: a_n= 2a_{n-1}+c_{n-1}+b_{n-1}

Sum of entries is 1: b_n= 2B_{n-1}+c_{n-1}+a_{n-1}

Sum of entries is 2: c_n= 2c_{n-1}+a_{n-1}+b_{n-1}

Also know that 4^n = a_n + b_n + c_n

I believe the initial conditions are: a_0 = 1 and a_1 = 2

Please help! I also tried solving a system of equations but was unsuccessful.

Hi sunrae,

You are almost there !

You have (correctly) written:

\(\displaystyle \displaystyle a_n = 2a_{n-1} + b_{n-1} + c_{n-1}\)

You also know that:

\(\displaystyle \displaystyle b_{n-1} + c_{n-1} = 4^{n-1} - a_{n-1}\)

You can now extract \(\displaystyle \textstyle b_{n-1} + c_{n-1}\) from the second equation and substitute into the first.

There is also another technique, that looks more complicated. However, it is worth knowing it, for the following reasons:

- It can be used in similar problems, when the trick we used is not available.
- It gives you a recurrence with constant coefficients, and there is a lot of theory about these equations.

We start with the equations you wrote:

\(\displaystyle \displaystyle\begin{align}

a_n &= 2a_{n-1} + b_{n-1} + c_{n-1}\\

b_n &= 2b_{n-1} + a_{n-1} + c_{n-1}\\

c_n &= 2c_{n-1} + a_{n-1} + b_{n-1}

\end{align}

\)

If you add the last two equations and write \(\displaystyle d_n = b_n + c_n\) (because of the symmetry), you get:

\(\displaystyle \displaystyle\begin{align}

a_n &= 2a_{n-1} + d_{n-1}\\

d_n &= 2a_{n-1} + 3d_{n-1}

\end{align}

\)

Now, write the equations, replacing \(\displaystyle n\) with \(\displaystyle (n-1)\):

\(\displaystyle \displaystyle

\begin{align}

a_{n-1} &= 2a_{n-2} + d_{n-2}\\

d_{n-1} &= 2a_{n-2} + 3d_{n-2}

\end{align}\)

The idea is to eliminate \(\displaystyle d_{n-1}\) and \(\displaystyle d_{n-2}\) from these four equations.

The last two equations give you \(\displaystyle d_{n-1}= 3a_{n-1} -4a_{n-2}\); substituting into the first, you get:

\(\displaystyle \displaystyle a_n = 5a_{n-1} - 4a_{n-2}\)

which is another recurrence relation.