There are two possible solutions here, one using recurrences and the
other one using PIE.
Recurrence.
The recurrences use two sequences $\{a_n\}$ and $\{b_n\}$ which count
strings not containing the two-character pattern that end in the first
character of the pattern and that do not. This gives
$$a_1 = 1 \quad\text{and}\quad b_1 = 25$$
and for $n\gt 1$
$$a_n = a_{n-1} + b_{n-1}
\quad\text{and}\quad
b_n = 24a_{n-1} + 25b_{n-1}.$$
These recurrences produce for $1\le n\le 7$ the sequence of sums
$\{a_n+b_n\}$ which is
$$26, 675, 17524, 454949, 11811150, 306634951, 7960697576,\ldots$$
so that the answer to the problem is
(count of no ocurrences)
$$7960697576.$$
Inclusion-Exclusion.
Let $M_{\ge q}$ be the set of strings containing at least $q$
ocurrences of the two-character pattern and let $M_{=q}$ be the set
containing exactly $q$ ocurrences of the pattern. Then by
inclusion-exclusion we have
$$|M_{=0}| =
\sum_{k=0}^{\lfloor n/2\rfloor}
(-1)^k |M_{\ge k}|.$$
Note however that
$$|M_{\ge k}| = 26^{n-2k} {n-2k + k\choose k}
= 26^{n-2k} {n-k\choose k}.$$
This is because when we have $k$ copies of the pattern there are $n-2k$ freely choosable letters that remain. Hence we have a total of $n-2k+k=n-k$ items to permute. We then choose the $k$ locations of the patterns among the $n-k$ items.
This gives the formula
$$|M_{=0}| =
\sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k \times
26^{n-2k} \times {n-k\choose k}.$$
a :=
proc(n)
option remember;
if n=1 then return 1 fi;
a(n-1)+b(n-1);
end;
b :=
proc(n)
option remember;
if n=1 then return 25 fi;
24*a(n-1)+25*b(n-1);
end;
ex_pie :=
proc(n)
option remember;
add((-1)^q*26^(n-2*q)*binomial(n-q,q),
q=0..floor(n/2));
end;
Proof that the two answers are the same.
Introduce the generating functions
$$A(z) = \sum_{n\ge 0} a_n z^n
\quad\text{and}\quad
B(z) = \sum_{n\ge 0} b_n z^n.$$
Observe that the correct intial value pair is
$a_0 = 0$ and $b_0 = 1.$
Multiply the two recurrences by $z^n$ and sum over $n\ge 1$
to get
$$A(z) - 0 = z A(z) + z B(z)
\quad\text{and}\quad
B(z) - 1 = 24 z A(z) + 25 z B(z).$$
Solve these two obtain
$$A(z) = \frac{z}{z^2-26z+1}
\quad\text{and}\quad
B(z) = \frac{1-z}{z^2-26z+1}.$$
This yields the following generating function $G(z)$ for
$\{a_n+b_n\}:$
$$G(z) = \frac{1}{z^2-26z+1}.$$
On the other hand we have
$$G(z) = \sum_{n\ge 0} z^n
\left(\sum_{k=0}^{\lfloor n/2\rfloor}
26^{n-2k} (-1)^k {n-k\choose k}\right).$$
This is
$$\sum_{k\ge 0} 26^{-2k} (-1)^k
\sum_{n\ge 2k} 26^n z^n {n-k\choose k}
\\ = \sum_{k\ge 0} 26^{-2k} (-1)^k \sum_{n\ge 0} 26^{n+2k}
z^{n+2k} {n+2k-k\choose k}
\\ = \sum_{k\ge 0} z^{2k} (-1)^k
\sum_{n\ge 0} 26^n z^{n} {n+k\choose k}
= \sum_{k\ge 0} z^{2k} (-1)^k \frac{1}{(1-26z)^{k+1}}
\\ = \frac{1}{1-26z}
\sum_{k\ge 0} z^{2k} (-1)^k \frac{1}{(1-26z)^{k}}
= \frac{1}{1-26z} \frac{1}{1+z^2/(1-26z)}
\\ = \frac{1}{1-26z+z^2}.$$
This establishes the equality of the generating functions which was to
be shown.
Closed form and OEIS entry.
The roots of the denominator of the generating function are
$$\rho_{1,2} = 13 \pm 2\sqrt{42}.$$
Writing
$$\frac{1}{1-26z+z^2}
= \frac{1}{(z-\rho_1)(z-\rho_2)}
= \frac{1}{\rho_1-\rho_2}
\left(\frac{1}{z-\rho_1}-\frac{1}{z-\rho_2}\right)
\\ = \frac{1}{4\sqrt{42}}
\left(\frac{1}{\rho_1}\frac{1}{z/\rho_1-1}
-\frac{1}{\rho_2}\frac{1}{z/\rho_2-1}\right)
\\ = \frac{1}{4\sqrt{42}}
\left(-\frac{1}{\rho_1}\frac{1}{1-z/\rho_1}
+\frac{1}{\rho_2}\frac{1}{1-z/\rho_2}\right).$$
We now extract coefficients to get
$$[z^n] G(z) =
\frac{1}{4\sqrt{42}}
\left(\rho_2^{-n-1}-\rho_1^{-n-1}\right).$$
Since $\rho_1\rho_2 = 1$ this finally becomes
$$[z^n] G(z) =
\frac{1}{4\sqrt{42}}
\left(\rho_1^{n+1}-\rho_2^{n+1}\right)$$
which is the sequence
$$26, 675, 17524, 454949, 11811150, 306634951, 7960697576,
\\ 206671502025, 5365498355074, 139296285729899, \ldots$$
This is OEIS A097309 which has additional
material and where in fact we find a copy of the problem statement
that initiated this thread.
Alternative derivation of the closed form of $G(z).$
This uses the following integral representation.
$${n-k\choose k}
= \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(1+w)^{n-k}}{w^{k+1}} \; dw.$$
This gives for the inner sum
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(1+w)^n}{w}
\left(\sum_{k=0}^{\lfloor n/2\rfloor}
26^{n-2k} (-1)^k \frac{1}{(1+w)^k w^k}\right) \; dw.$$
Note that the defining integral is zero when
$\lfloor n/2\rfloor \lt k \le n,$ so this is in fact
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(1+w)^n}{w}
\left(\sum_{k=0}^n
26^{n-2k} (-1)^k \frac{1}{(1+w)^k w^k}\right) \; dw.$$
Simplifying we obtain
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(1+w)^n}{w} 26^n
\frac{(-1)^{n+1}/26^{2(n+1)}/(1+w)^{n+1}/w^{n+1}-1}
{(-1)/26^2/(1+w)/w-1} \; dw$$
or
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon}
(1+w)^{n+1} 26^n
\frac{(-1)^{n+1}/26^{2(n+1)}/(1+w)^{n+1}/w^{n+1}-1}
{(-1)/26^2-w(w+1)} \; dw$$
The difference from the geometric series contributes two terms, the
second of which has no poles inside the contour, leaving just
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(-1)^{n+1}}{26^{n+2}}
\frac{1}{w^{n+1}}
\frac{1}{(-1)/26^2-w(w+1)} \; dw.$$
It follows that
$$G(z) = \sum_{n\ge 0}
z^n [w^n] \frac{(-1)^{n+1}}{26^{n+2}}
\frac{1}{(-1)/26^2-w(w+1)}.$$
What we have here is an annihilated coefficient extractor which
simplifies as follows.
$$\frac{-1}{26^2} \sum_{n\ge 0}
(-z/26)^n [w^n] \frac{1}{(-1)/26^2-w(w+1)}
\\ = \frac{-1}{26^2}
\frac{1}{(-1)/26^2+z/26(-z/26+1)}
= -\frac{1}{-1+z(-z+26)}
\\ = \frac{1}{1-26z+z^2}.$$
This concludes the argument.
There is another annihilated coefficient extractor at this
MSE link.
