Jtdylktuy

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 $a_n$}$

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

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 $ngeq 3$.
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 $sgeq 2$ then observe that the answer is $c_{n-s}$.

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

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

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

This is

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

Continuing without the distinction between ones and zeros we get

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

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 $nge 3$ 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 $nge 3$

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

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);