Recurrence relation for quaternary strings length n sum divisible by 3

sunrae

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Oct 29, 2017
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1
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.
 

barrick

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Nov 4, 2017
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27
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.
 
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