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A) Find a recurrence relation for the number of n-digit binary sequences with no pair of consecutive 1s. (A binary sequence only uses the numbers 0 and 1 for those who don't know)

B) Repeat for n-digit ternary sequences. (only uses numbers 0, 1, and 2)

C) Repeat for n-digit ternary sequences with no consecutive 1s or consecutive 2s.

atl
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2 Answers2

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for the binary case the recurrence relation as follows:

$a_{n}=a_{n-1}+a_{n-2}$.

why?

because any n-digit binary sequence has to start with $0$ or $1$ since it is binary.

If it starts with $0$ the rest will be same with the $a_{n-1}$, if it starts with $1$ then next number has to be $0$ meaning that it will start exactly with $10$ so the rest now will be same with $a_{n-2}$.

......................

by using similar idea, for the ternary case we have the following:

$a_{n}=2a_{n-1}+2a_{n-2}$

the sequence can start with $0,1$ or $2$.

if the sequence starts with $0$ or $2$ then the rest will be same with $% a_{n-1}$, if the sequence starts with $1$, next one can be $0$ or $2$, so it can be start with $10$ or $12,$ the rest will be same with $a_{n-2}$.

As a result, since we have two choices for both cases we have recurrence with coefficient $2.$

.......

for part C i refer to Recurrence relation for the number of $n$-digit ternary sequences with no consecutive $1$s or $2$s

the solution also can be obtained by same argument with part A and B.

Community
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pasha
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Let $b_n$ be the number of ways we can produce a ternary sequence of length $n$ with no two consecutive $1$'s. Call such $n$-sequences good.

There are two types of good sequence of length $n$, (i) the ones that don't end in $1$ and (ii) the ones that do end in $1$.

Any Type (i) good sequence of length $n$ can be obtained in $2$ ways from a uniquely determined good sequence of length $n-1$ by appending a $0$ or a $2$. So there are $2b_{n-1}$ of them.

Any Type (ii) good $n$-sequence can be obtained from an arbitrary good sequence of length $n-2$ by appending $01$ or $21$. So there are $2b_{n-2}$ of them.

That yields the recurrence $b_n=2b_{n-1}+2b_{n-2}$.

Though we were not asked, one should add that $b_0=1$ and $b_1=3$.

Problem a) is somewhat easier, and problem c) somewhat harder.

André Nicolas
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