Palindromes of Length Less Than or Equal to N Suppose we have $n$ slots to fill with words from a regular alphabet of $26$ letters.  We would like to find all possible palindromes of length less than or equal to $n$, where the minimum palindrome is of length $3$ (e.g. aba).  We know that for the case where $n=3$ there are $26^2$ ways to construct such a palindrome.  For $n=4$ there are $(26^2)$ palindromes, all of which are of length $4$ since having a sub-palindrome of length $3$ would result in the total $4$-slot word being non-palindromic.  For the case where $n=5$ we have three degrees of freedom and $26$ additional 'bonus cases' where we have that the outer two characters are equal to the inner two characters, resulting in two sub-palindromes.  As a finally example, skipping $n=6$ we have the case $n=7$ where the maximal palindromes are of the form abcdcba, where if $c=a$ or $d=a$, for example, we get additional sub-palindromes to add to our count.
Example: $n=5$:
abcba means that we have $26^3$ possible palindromes + $3(26^2)$ since we can either fix $c=a$ or $a=b$ or $b=c$
I am looking to generalize this to find all possible palindromes of length less than or equal to $n$ for arbitrary $n$.  I have noticed that The number of degrees of freedom increases every two palindromes, only increasing at odd $n$ (since there is a unique median letter for these cases), and that there are restrictions on the number of shifts for each $k\leq n$ that depend on $n$ but I am not sure how to use these facts or whether they are inherently useful.  Any input is appreciated.
 A: The easiest method is to simply count palindromes of length exactly n. For even n, this is $26^{n/2}$, and for odd n, $26^{(n-1)/2}$.
So, if we include lengths 1 and 2, we want to add $26 +26 +26^2 +26^2 + 26^3 + 26^3 + \ldots$
If n is even, this is just twice the sum of a geometric series. If n is odd, you get one extra term to add at the end. (You can subtract the 52 sequences of length <= 2 at the end if you like).
A: In regards to the accepted answer post of user11977, the formula provided for the number of palindromes is incorrect and for odd $n$ is $\dfrac{26^{n+1}}{2}$, not $n-1$ as posted above on February 2014. The part about $26^{n/2}$ is correct for even $n$.
To rationalize this, in even $n$ scenarios, the first half $(\dfrac{n}{2})$ can be anything, while the rest are set based on the configuration of the first half. In odd $n$ scenarios, the first $\dfrac{n+1}{2}$ characters can be anything, while the rest are set.
A: Use the formula to find even number of palindromes of less than or equal to $n$.

