Recurrence Relation - # of binary strings with given property

Question

Let $a_n$ be the number of binary strings of length $n$ with the property that each entry is adjacent to at least one entry of the same type.

ex: $11000111$ is a valid string but $11011000$ is not valid

$\textbf{(a) Find $a_1,a_2,a_3,a_4,a_5,a_6,a_7$}$

If someone can check that my attempt is correct, I would really appreciate it.

$a_1=0$ since we cannot have just $0$ or just $1$ as there will be no adjacent of the same type

$a_2=2$: either $00$ or $11$

$a_3=2$: either $000$ or $111$

$a_4=4$:

Reasoning:

$\textbf{If we start with a $0$}$: For the second entry we have $1$ choice as we are forced to put a $0$ since we started with a $0$. For the third entry we have $2$ choices, and similarly for the fourth entry we have $1$ choice. So there are $2$ such strings.

$\textbf{If we start with a $1$}$: For the second entry we are forced to put a $1$. For the third entry we have $2$ choices, and for the fourth entry we have $1$ choice. So there are $2$ such strings.

So $a_4=2+2=4$ strings.

Following the same method for the remaining:

$a_5=4$

$a_6=8$

$a_7=8$

$\textbf{(b) Find the recurrence relation for $a_n$}$

$$a_n= \begin{cases} 2a_{n-2}&n \text{ even},\\ a_{n-1}&n \text{ odd} \end{cases}$$

Answer 1

Your argument seems wrong to me. In particular the following part.

If we start with a $0$: For the second entry we have one choice as we are forced to put a $0$ since we started with a $0$. For the third entry we have two choices, and similarly for the fourth entry we have one choice.

That last sentence seems to be true for $n=4$, but not in general for $n>4$. In this case it is only true if you choose $1$ for the third entry, but if you've chosen $0$ then you have two choices.

That analogous happens in the case where you start with $1$.

For the recurrence relation I think the following should work. For any $m$ let $b_m$ and $c_m$ denote respectively the strings of the desired form that start with a $0$ and with a $1$ repectively. Note that $b_m=c_m=a_m/2$. So this is all a bit silly, but let's do it for the sake of keeping the argument clear.

Let's fix $n\geq 3$. I'll calculate $b_n$ in terms on $c_m$ for $m<n$.

How many strings are there that have $0< s < n$ zeroes in a row before having a one? As you've noted if $s=1$ then the answer is zero strings. For $s\geq 2$ then observe that the answer is $c_{n-s}$.

Using this show that $b_m=1+ \Sigma_{2\leq s < n} c_{n-s}$.

Answer 2

Using $z$ for ones and $w$ for zeros we get the generating function

$$F(z, w) = (1+z^2+z^3+\cdots) \times \sum_{q\ge 0} (w^2+w^3+\cdots)^q (z^2+z^3+\cdots)^q \\ \times (1+w^2+w^3+\cdots).$$

This is

$$\left(1+\frac{z^2}{1-z}\right) \times \sum_{q\ge 0} \frac{w^{2q} z^{2q}}{(1-w)^q (1-z)^q} \\ \times \left(1+\frac{w^2}{1-w}\right).$$

Continuing without the distinction between ones and zeros we get

$$\left(1+\frac{z^2}{1-z}\right)^2 \sum_{q\ge 0} \frac{z^{4q}}{(1-z)^{2q}} \\ = \left(1+\frac{z^2}{1-z}\right)^2 \frac{1}{1-z^4/(1-z)^2} \\ = (1-z+z^2)^2 \frac{1}{(1-z)^2-z^4}.$$

The difference of two squares yields $$(1-z+z^2)^2 \frac{1}{(1-z+z^2)(1-z-z^2)}.$$

which simplifies to

$$\bbox[5px,border:2px solid #00A000]{ G(z) = \frac{1-z+z^2}{1-z-z^2}.}$$

From the coefficients of this OGF we get the sequence

$$1, 0, 2, 2, 4, 6, 10, 16, 26, 42, 68, 110, 178, 288, 466, 754, \\ 1220, 1974, 3194, 5168, 8362, \ldots$$

which is OEIS A006355 where these data are confirmed. Now for the coefficients we have

$$[z^0] G(z) (1-z-z^2) = G_0 = [z^0] (1-z+z^2) = 1$$

and hence $G_0 = 1.$ Furthermore

$$[z^1] G(z) (1-z-z^2) = G_1-G_0 = [z^1] (1-z+z^2) = -1$$

so $G_1 = 0.$ Next we find

$$[z^2] G(z) (1-z-z^2) = G_2-G_1-G_0 = [z^2] (1-z+z^2) = 1$$

so $G_2 = 2.$ For $n\ge 3$ we get

$$[z^n] G(z) (1-z-z^2) = G_n - G_{n-1} - G_{n-2} = [z^n] (1-z+z^2) = 0$$

so that for $n\ge 3$

$$\bbox[5px,border:2px solid #00A000]{ G_n = G_{n-1} + G_{n-2}.}$$

The following Maple code documents the problem definition that was used.

ENUM :=
proc(n)
option remember;
local ind, d, res, pos;

    if n=0 then return 1 fi;
    if n=1 then return 0 fi;
    if n=2 then return 2 fi;

    res := 0;

    for ind from 2^n to 2*2^n-1 do
        d := convert(ind, base, 2)[1..n];

        if d[1] = d[2] and d[n] = d[n-1] then
            for pos from 2 to n-1 do
                if d[pos-1] <> d[pos] and
                d[pos] <> d[pos+1] then
                    break;
                fi;
            od;

            if pos = n then
                res := res + 1;
            fi;
        fi;
    end;

    res;
end;


X := n-> coeftayl((1-z+z^2)/(1-z-z^2), z=0, n);

Answer 3

You can take this as a sequence of stretches of at least two equal symbols. In ordinary generating function terms a sequence of two or more would be represented by:

$\begin{align*} z^2 + z^3 + \dotsb &= \frac{z^2}{1 - z} \end{align*}$

A sequence of the above in turn would be:

$\begin{align*} 1 + \frac{z^2}{1 - z} + \left(\frac{z^2}{1 - z}\right)^2 + \dotsb &= \frac{1}{1 - \frac{z^2}{1 - z}} \\ &= \frac{1 - z}{1 - z - z^2} \end{align*}$

Consider the Fibonacci sequence, defined by $F_{n + 2} = F_{n + 1} + F_n$ with $F_0 = 0, F_1 = 1$. It's generating function is:

$\begin{align*} F(z) &= \sum_{n \ge 0} F_n z^n = \frac{z}{1 - z - z^2} \end{align*}$

But also:

$\begin{align*} \sum_{n \ge 0} F_{n + 1} z^n &= \frac{F(z) - F_0}{z} = \frac{1}{1 - z - z^2} \end{align*}$

We see that our generating function is:

$\begin{align*} \frac{1 - z}{1 - z - z^2} &= \sum_{n \ge 0} (F_{n + 1} - F_n) z^n \\ &= \sum_{n \ge 0} F_{n - 1} z^n \end{align*}$

So the number of sequences of length $n$ is $F_{n - 1}$, where we extend to $F_{-1} = 1$ (as it should be to get one sequence of length 0).