This problem can be attacked through the very convoluted use of Inclusion-Exclusion theory and Stars and Bars theory. Personally, I regard the approach that I will describe as inferior to attacking the problem with recursion. The reason is that although this approach will allow a closed form expression for the computation, as a function of $~n,~$ it will generally require the use of computer assistance to perform the computation. So, I regard any advantage of this approach over recursion to be very iffy.
See this article for an
introduction to Inclusion-Exclusion.
Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.
For Stars and Bars theory, see
this article and
this article.
As an illustrative example, I will set $~n = 20,~$ and obtain the corresponding closed form expression, for two illustrative terms that are pertinent to the computation, when $~n = 20.~$ At the end of my answer, I will then describe the overall closed form expression for any value of $~n.~$
Following the syntax in the 2nd Inclusion-Exclusion link above, let $~S~$ denote the collection of all $~20~$ character strings, without any regard for whether the string contains at least one occurrence of ccc.
Then, for $~k \in \{1,2,\cdots,18\},~$ let $~S_k~$ denote the subset of $~S~$ that specifically contains the ccc substring, starting in position $~k.~$ So, for example, an element in $~S_1,~$ will contain ccc, starting in position 1, and may or may not also contain ccc elsewhere in the string.
So, the desired computation is
$$| ~S_1 \cup S_2 \cup \cdots \cup S_{18} ~|. \tag1 $$
The standard Inclusion-Exclusion approach to enumerating the expression in (1) above is:
Let $~T_1~$ denote $~\displaystyle \sum_{1 \leq i_1 \leq 18} | ~S_{i_1} ~|.$
That is, $~T_1~$ represents the sum of $~\displaystyle \binom{18}{1}~$ terms.
For $~r \in \{2,3,\cdots,18\},~$
let $~T_r~$ denote $~\displaystyle \sum_{1 \leq i_1 < i_2 < \cdots < i_r \leq 18} | ~S_{i_1} \cap S_{i_2} \cap \cdots \cap S_{i_r} ~|.$
That is, $~T_r~$ represents the sum of $~\displaystyle \binom{18}{r}~$ terms.
Then, in accordance with Inclusion-Exclusion theory, the computation in (1) above is equivalent to
$$\sum_{r=1}^{18} (-1)^{r+1} T_r. \tag2 $$
By considerations of symmetry, you have that
$~3^{17} = |~S_1~| = | ~S_2 ~| = \cdots = | ~S_{18} ~|.$
Therefore, $~T_1 = 18 \times 3^{17}.$
The difficulty is that symmetrical considerations break down when computing $~T_r ~: ~r \geq 2.$
To attack this (general) problem, I will
Illustrate the manual way that $~T_2~$ can be computed, assuming that $~n=20$.
Illustrate the analytical way of computing $~T_2,~$ assuming that $~n=20$.
Create a helper function $~f(r,z,o).$
Create a helper function $~g(n,r,z,o).$
Illustrate the analytical way of computing $~T_{12},~$ assuming that $~n=20$.
Provide a closed form expression of $~\displaystyle \sum_{r=1}^{n-2} (-1)^{r+1}T_r.$
$\underline{\text{Manual Computation of} ~T_2}$
In this section, it will be assumed that each intersection is of form $~S_{i_1} \cap S_{i_2} ~: ~1 \leq i_1 < i_2 \leq 18.$
Each of the $~\displaystyle \binom{18}{2} = 153 ~$ intersections will fall into one of three categories:
Category-1
An intersection like $~S_1 \cap S_4.~$
This intersection has ccc starting in positions 1 and 4. Therefore, there are $~14~$ unspecified positions. Therefore, $~|~S_1 \cap S_4 ~| = 3^{14}.$
In Category-1, as $~i_1~$ goes from $~1~$ through $~15,~$ there are $~18 - (i_1+2) = 16-i_1 ~$ possible values for $~i_2.~$
Therefore, Category-1 has $~\displaystyle \sum_{i_1=1}^{15} (16 - i_1) = 120~$ separate terms, each of which equals $~3^{14}.$
Therefore, when computing $~T_2,~$ the Category-1 subtotal is
$120 \times 3^{14}.$
Category-2
An intersection like $~S_1 \cap S_3.~$
This intersection has ccc starting in positions 1 and 3. Therefore, there are $~15~$ unspecified positions. Therefore, $~|~S_1 \cap S_3 ~| = 3^{15}.$
In Category-2, as $~i_1~$ goes from $~1~$ through $~16,~$ there is exactly 1 possible value for $~i_2.~$
Therefore, Category-2 has $~16~$ separate terms, each of which equals $~3^{15}.$
Therefore, when computing $~T_2,~$ the Category-2 subtotal is
$16 \times 3^{15}.$
Category-3
An intersection like $~S_1 \cap S_2.~$
This intersection has ccc starting in positions 1 and 2. Therefore, there are $~16~$ unspecified positions. Therefore, $~|~S_1 \cap S_2 ~| = 3^{16}.$
In Category-3, as $~i_1~$ goes from $~1~$ through $~17,~$ there is exactly 1 possible value for $~i_2.~$
Therefore, Category-3 has $~17~$ separate terms, each of which equals $~3^{16}.$
Therefore, when computing $~T_2,~$ the Category-3 subtotal is
$17 \times 3^{16}.$
So,
$$T_2 = \left( ~120 \times 3^{14} ~\right) + \left( ~16 \times 3^{15} ~\right) + \left( ~17 \times 3^{16} ~\right).$$
$\underline{\text{Analytical Computation of} ~T_2}$
Consider the following tableau:
- - - - i-1 - - - - - i-2 - - - - - - -
The above tableau, which has $~[ ~20 - ( ~3-1 ~) ~] = 18~$ positions, assigns two of these positions to $~i_1~$ and $~i_2.~$ Here, the position assigned to $~i_1~$ is to the left of the position assigned to $~i_2.$
The positioning of $~i_1~$ and $~i_2~$ create $~(2 + 1) = 3~$ islands. Examining the islands from left to right, let $~x_1, ~x_2, x_3,~$ denote the size of these islands.
In order to categorize the nature of the $~S_{i_1} \cap S_{i_2}~$ intersection, you can ignore the values of $~x_1~$ and $~x_3,~$ and focus exclusively on $~x_2.~$
Either $~x_2 = 0, ~x_2 = 1, ~$ or $~x_2 > 1.$
$\mathbf{x_2 = 0}$
This represents what I describe as a fully compressed intersection. Because it is fully compressed, instead of their being $~[ ~20 - ( ~2 \times 3 ~) ~] = 14~$ unspecified positions, there are $~[ ~20 - ( ~2 \times 3 ~) ~] + 2 = 16~$ unspecified positions.
The number of such intersections is the same as the number of solutions to
By Stars and Bars theory, with the variable $~x_2 = 0~$ ignored, there are $~\displaystyle \binom{16 + [2 - 1]}{[2 - 1]} = 17~$ such solutions.
So, analytically, the first partial sum when computing $~T_2~$ is $~\displaystyle \left( ~17 \times 3^{16} ~\right).$
$\mathbf{x_2 = 1}$
Similarly, when computing the second partial sum of $~T_2,~$ it will be $~3^{15}~$ times the number of solutions to
By Stars and Bars theory, with the variable $~x_2 = 1~$ ignored, and the $~x_1 + x_3~$ sum adjusted to $~15,~$ there are $~\displaystyle \binom{15 + [2 - 1]}{[2 - 1]} = 16~$ such solutions.
So, analytically, the second partial sum when computing $~T_2~$ is $~\displaystyle \left( ~16 \times 3^{15} ~\right).$
$\mathbf{x_2 > 1}$
Finally, when computing the third partial sum of $~T_2,~$ it will be $~3^{14}~$ times the number of solutions to
By Stars and Bars theory, with the change of variable $~y_2 = x-2 \implies y_2 \in \Bbb{Z_{\geq 0}},~$ and the $~x_1 + y_2 + x_3~$ sum equal to $~(16 - 2), ~$ there are $~\displaystyle \binom{14 + [3 - 1]}{[3 - 1]} = \binom{16}{2} = 120~$ such solutions.
So, analytically, the third partial sum when computing $~T_2~$ is $~\displaystyle \left( ~120 \times 3^{14} ~\right).$
$\underline{\text{Helper Function} ~f(r,z,o)}$
This section and the next section will be used to create helper functions that will facilitate expressing the general closed form expression. In this section, assuming that you have $~(r+1)~$ variables, how many ways are there that, of the variables $~\{x_2,x_3,\cdots,x_r\},~$ you can have exactly $~z~$ of these variables equal to $~0,~$ and $~o~$ of these variables equal to $~1~?$
For the function $~f(r,z,o),~$ the allowable range for the variables $~r, ~z, ~$ and $~o,~$ will be as specified, in the next section, for the function $~g(n,r,z,o).$
Given that, I specify that
$$f(r,z,o) = \binom{r-1}{z} \times \binom{r-1-z}{o}.$$
$\underline{\text{Helper Function} ~g(n,r,z,o)}$
In this section, assume that you have
The specific intersection represented by
$~(x_1,x_2,\cdots,x_r,x_{r+1}).~$
$x_1 + x_2 + \cdots + x_r + x_{r+1} = (n-2-r).$
$x_1,x_{r+1} \in \Bbb{Z_{\geq 0}}.$
$x_2,x_3,\cdots,x_{z+1}~$ are all equal to $~0.$
$x_{z+2}, x_{z+3}, \cdots, x_{(z+1+o)}~$ all equal to $~1.~$
$x_{(z+1+o + 1)}, \cdots, x_r, ~$ are all in $~\Bbb{Z_{\geq 2}}.~$
Then, what is the product of the number of solutions possible times $~3^p,~$ where $~p~$ equals the number of unspecified characters in each solution?
Clarification
The helper function in this section is intended to dovetail with the helper function in the previous section. $~f(r,z,o)~$ is concerned with how many ways that there are of choosing $~z~$ variables to equal $~0,~$ and then choosing $~o~$ variables to equal $~1,~$ from the set $~\{x_2,\cdots,x_r\}.$
$g(n,r,z,o)~$ then assumes that starting from $~x_2,~$ and considering the variables in ascending order by variable index, the first $~z~$ variables are equal to $~0,~$ and then the next $~o~$ variables are equal to $~1.$
Range of Variables
Before specifying the formula for $~g(n,r,z,o),~$ it is important to establish upper and lower bounds (where appropriate) for $~n, ~r, ~z,~$ and $~o.$
For simplicity, I will require that $~n,~$ which represents the length of the string, is an element in $~\Bbb{Z_{\geq 6}}.$
Considering the significance of the variable $~T_r,~$ which requires that $~r \leq (n-2),~$ I will require that $~r \in \{2,3,\cdots,n-2\}.$
In considering the upper and lower bounds for $~z~$ and $~o,~$ note that both $~x_1~$ and $~x_{r+1}~$ are permitted to be any non-negative integers. Therefore, you never have to be concerned that the sum represented by $~x_2 + x_3 + \cdots + x_r~$ is too low. However, you must not have the sum $~x_2 + x_3 + \cdots + x_r > (n-2-r).~$
With this in mind, you can always have $~z~$ be as large as $~(r-1).$
Suppose however, that $~z < r-1.~$ Then, the minimum sum of $~x_2 + x_3 + \cdots + x_r~$ will be achieved when the $~(r-1-z)~$ remaining variables from $~\{x_2,\cdots, x_r\},~$ are each equal to $~1.~$ So, the minimum sum, which will be $~(r - 1 - z),~$ must not exceed $~(n-2-r).~$
Therefore, you must have that
$$(r - 1 - z) \leq (n - 2 - r) \implies z \geq (2r + 1 - n).$$
So, the allowable values for $~z~$ are
$$z \in \Bbb{Z}, ~z \geq \max\left[ ~0, ~(2r + 1 - n) ~\right], z \leq (r-1).$$
Note that $~r \leq (n-2) \implies (2r + 1 - n) \leq (r - 1).$
To determine the allowable values for $~o,~$ assume that $~z~$ is some element in the range shown immediately above. Then, at first glance, you must have $~0 \leq o \leq (r - 1 - z).~$ Further, for a given value of $~z~$ and $~o,~$ the minimum value for $~x_2 + \cdots + x_r~$ will be
$$~o + 2(r - 1 - z - o) = 2r - 2 - 2z - o,$$
which must not exceed $~(n - r - 2).$
Therefore,
$$(2r - 2 - 2z - o) \leq (n - r - 2) \implies o \geq 3r - 2z - n.$$
So, for the variable $~o,~$ you have the lower bounds of $~0,~$ and $~(3r - 2z - n),~$ and the upper bound of $~(r - 1 - z).$
Also, since $~z \geq 2r + 1 - n,~$ you can never have $~(3r - 2z - n) > (r - 1 - z).$
Therefore, the allowable values for $~o~$ are
$$o \in \Bbb{Z}, ~o \geq \max[ ~0, ~(3r - 2z - n) ~], ~o \leq (r - 1 - z).$$
Computation of $~\mathbf{g(n,r,z,o)}$
Assuming that the variables $~n, ~r, ~z, ~$ and $~o~$ are all in range, Stars and Bars theory provides the computation of the number of solutions.
The original sum is $~(n-2-r),~$ and the original number of variables is $~(r + 1).~$ Assuming that the $~z~$ variables are ignored, the sum is still $~(n-2-r),~$ and the number of variables is now $~(r + 1 - z).$
Next, you employ the change of variables $~a_i = x_i - 1,~$ for each of the $~o~$ variables exactly equal to $~1.~$ So, this reduces the sum from $~(n-2-r)~$ to $~(n-2-r-o),~$ and reduces the number of variables from $~(r + 1 - z),~$ to $~(r + 1 - z - o).$
The last thing to deal with is that while the two variables $~x_1,~$ and $~x_{r+1}~$ are permitted to be in $~\Bbb{Z_{\geq 0}},~$ the $~(r - 1 - z - o)~$ other variables must each be in $~\Bbb{Z_{\geq 2}}.$
Therefore, you must employ the change of variables $~y_i = x_i - 2,~$ on these $~(r - 1 - z - o)~$ variables, which reduces the sum to
$$(n - 2 - r - o) - [ ~2 \times (r - 1 - z - o) ~] = n - 3r + 2z + o.$$
So, the number of solutions must correspond to the number of solutions to
- $x_1 + x_2 + \cdots + x_k = M.$
- $x_1, \cdots, x_k \in \Bbb{Z_{\geq 0}}.$
- $k = (r + 1 - z - o), ~M = n - 3r + 2z + o.$
By Stars and Bars theory, the number of solutions is
$$\binom{M + [k-1]}{k-1} = \binom{[n - 3r + 2z + o] + [r - z - o]}{r - z - o}$$
$$= \binom{n - 2r + z}{r - z - o}.$$
To compute the number of unspecified characters, first assume that $~z = 0 = o.$ Then, you will have $~n - 3r~$ unspecified characters. For each variable in $~\{x_2,\cdots,x_r\},~$ that is equal to $~0,~$ you are free-ing up two character positions, because you are having an intersection of two subsets use $~4~$ character positions, rather than $~6.~$ Similarly, for each variable in $~\{x_2,\cdots,x_r\},~$ that is equal to $~1,~$ you are free-ing up one character position.
Therefore, the number of unspecified character positions is
$$n - 3r + 2z + o.$$
Note that since $~o \geq ( ~3r - 2z - n),~$
you have that $~(2z + o) \geq (~3r - n).$
Therefore,
$$g(n,r,z,o) = \binom{n - 2r + z}{r - z - o} \times 3^{( ~n - 3r + 2z + o ~)}.$$
$\underline{\text{Analytical Computation of} ~T_{12}}$
To utilize the helper functions, you simply have to establish the viable values of $~(z,o),~$ for $~r = 12, n = 20,~$ and then plug in the formulas.
Since $~(2r + 1 - n) = 5, ~$ you have that
$~z \in \{5,6,\cdots,11\}.$
Since $~(3r - n) = 16,~$ you have that the lower bound of $~o~$ is
$16 - 2z ~: ~z \leq 8~$ and $ ~0 ~: ~z \geq 9$,
while the upper bound for $~o~$ is $~(11-z).$
Therefore, for $~n = 20,~$ you have that
$$T_{12} = \sum_{z=5}^{8} \left[ ~\sum_{o = 16 - 2z}^{11-z} f(12,z,o) \times g(20,12,z,o) ~\right] $$
$$+ \sum_{z=9}^{11} \left[ ~\sum_{o = 0}^{11-z} f(12,z,o) \times g(20,12,z,o) ~\right]. $$
$\underline{\text{Closed form expression of} ~\displaystyle \sum_{r=1}^{n-2} (-1)^{r+1}T_r}$
For simplicity, it is assumed that $~n \in \Bbb{Z_{\geq 6}}.$
$$T_1 = (n-2) \times 3^{n-3}.$$
The remainder of this section assumes that $~r \in \Bbb{Z}, ~2 \leq r \leq (n-2).$
The allowable range for the variable $~z~$ is
$$z \in \Bbb{Z}, ~z \geq \max\left[ ~0, ~(2r + 1 - n) ~\right], z \leq (r-1).$$
The allowable range for the variable $~o~$ is
$$o \in \Bbb{Z}, ~o \geq \max[ ~0, ~(3r - 2z - n) ~], ~o \leq (r - 1 - z).$$
With $~z~$ and $~o~$ in range, the helper functions are
$$f(r,z,o) = \binom{r-1}{z} \times \binom{r-1-z}{o}$$
and
$$g(n,r,z,o) = \binom{n - 2r + z}{r - z - o} \times 3^{( ~n - 3r + 2z + o ~)}.$$
Then,
$$T_r = \sum_{z ~\text{in range}} \left[ ~\sum_{o ~\text{in range}} f(r,z,o) \times g(n,r,z,o) ~\right].$$
