Inspired by this question:
Define odd-palindroming $X$ in base $b$, or $OP(X,b)$: Take an integer $X$ and write it in base $b$. Then, reverse its digits and concatenate the reversed digits to the end of the number, without repeating the last digit. For example, $OP(11,7)$: $11=14_7$, $OP(11,7)=141_7=78$. If the last few digits of $X_b$ are the same, odd-palindroming does not truncate them. For example, $OP(122_b,b)=12221_b$ not $121_b$.
I started looking at under what conditions $OP(X,b)$ prime.
The rules vary depending on the relative values of $X$ and $b$.
Trivially, because $\forall b>X$, $OP(X,b)=X$, we get that $OP(X,b)$ is prime $\Leftrightarrow$ $X$ is prime. Equally trivially, when $b=X$, we get a prime if and only if $b^2+1$ is prime. The numbers that would generate this result in the primes of A002496.
We therefore are left with only $b\in\{2,3,4,...,X-1\}$. We can start by applying some basic divisibility rules.
If $d$ is a prime factor of $b$, then $Y\equiv (Y\mod b) \mod d$. Applying this, we get that $OP(X,b)$ is divisible by $d$ iff the leading digit of $X_b$ is divisible by $d$. Trivially, this gives us that there is no $X$ such that $OP(X,p)$ is divisible by $p$ for any prime $p$.
If $b\equiv1\mod d$, then the sum of the digits of $Y_b$, denoted $S_b(Y)$, is equivalent to $Y\mod d$. Applying this, we get that $OP(X,b)$ is divisible by $d$ iff $S_b(X)\equiv-(X \mod b)\mod d$.
If we denote $S_b'(Y)$ as the alternating sum of the digits of $Y_b$, then $Y$ is divisible by $b+1$ iff $S_b'(Y)\equiv0\mod (b+1)$. Applying this, we get that $OP(X,b)$ is divisible by $b+1$ iff $(2S_b'(X)\mod(b+1))\equiv X\mod b$.
If every digit of $X_b$ is divisible by a prime $d$, then trivially $OP(X,b)$ is also divisible by $d$.
Some experimenting showed that for most numbers $X$, for some $b\in\{2,3,...,X-1\}$ you get $OP(X,b)$ is prime and for some $OP(X,b)$ is composite. However, for $X=4$ you always get a prime and for $X=5$ you always get a composite.
Looking at the parity of $OP(X,b)$ gives us that for an odd $b$, there are never consecutive values of $X$ such that $OP(X,b)$ is odd, nor are there ever more than two consecutive values of $X$ such that $OP(X,b)$ is even. In fact, for $b=2k+1$, we get the sequence for the parity of $OP(X,b)$, beginning with $X=b+1$ of: (1 odd followed by 1 even) k times, 1 odd, 2 evens, (1 odd followed by 1 even) k times, 1 odd, 2 evens, etc. As a result, for $X>5$, there all always some odd values of $b$ that return an even number, which means that there always will be at least one value of $b$ such that $OP(X,b)$ is composite, meaning that $X=4$ is the last value of $X$ for which $OP(X,b)$ is always prime for $b\in\{2,3,...,X-1\}$.
While I didn't find any values of $X>5$ for which $\forall b\in\{2,3,...,X-1\}, OP(X,b)$ is composite, I can't prove whether or not they exist. Can anyone provide a proof that they do/don't exist?