According to Cayley-Hamilton theorem, the power of a $$$n$$$ times $$$n$$$ matrix can be seen as a linear recurrence have atmost $$$n$$$ terms, so let say we have some $$$dp[n][m]$$$ which every $$$dp[i]$$$ is a $$$n$$$ times $$$1$$$ vector, and we can get $$$dp[i]$$$ from the previous $$$dp[j] (j < i)$$$ by multiplying a transition matrix $$$A_{n \times n}$$$, then we can manipulate Cayley-Hamilton theorem like below to get a linear recurrence of a certain entries of $$$dp[i]$$$, then brute-force the first $$$2n$$$ terms of $$$dp[i] (i \leq 2n)$$$, plug all of them into Berlekamp-Massey, and we will get the actual recurrence relation of a certain entries of $$$dp[i]$$$, finally compute the k-th term of linear recurrence in $$$O(n^2lgk)$$$ or $$$O(nlgnlgk)$$$ using FFT.

($$$A$$$ is a $$$n$$$ times $$$n$$$ matrix, $$$b_i$$$ is a $$$n$$$ times $$$1$$$ vector whose entry are all $$$0$$$ except i-th row is $$$1$$$)

Assume we will construct a palidrome $$$S$$$ of length $$$n + |s|$$$, then we can just build the left half of the string then the rest will be fixed.

To make sure $$$S$$$ contain $$$s$$$ as a subsequence, we can maintain two position $$$L, R$$$ which means $$$s[i](L \le i \le R)$$$ have not been inserted to string $$$S$$$, when we add a letter $$$X$$$ at the end of $$$S$$$, if $$$X = s[L]$$$, then we let $$$L$$$ become $$$L + 1$$$(because we already add $$$s[L]$$$ to $$$S$$$, if $$$X = s[R]$$$, then we let $$$R$$$ become $$$R - 1$$$.

So we can get the following dp solution.

$$$dp[i][L][R]$$$ store the number of string have length $$$i$$$, and $$$s[j](L \le j \le R)$$$ have not been inserted to the string. then we will have the following transition.

\begin{equation} dp[i][L][R] \mathrel{+}= \begin{cases} dp[i-1][L][R] & \text{if } L = R + 1 \text{ or } (s[L] \neq j \text{ and } s[R] \neq j)\\ dp[i-1][L-1][R] & \text{if } s[L-1] = j \text{ and } (s[R] \neq j \text{ or } L = R + 1)\\ dp[i-1][L][R + 1] & \text{if } s[R + 1] = j \text{ and } (s[L] \neq j \text{ or } L = R + 1)\\ dp[i-1][L-1][R + 1] & \text{if } s[L-1] = s[R + 1] = j \end{cases}
\end{equation}

time complexity: $$$O(26(n + |s|)|s|^2)$$$