Number of real roots of irreducible polynomials that are solvable by radicals Let $n \geq 3$ be a natural number. Define the set $X_n$ as the set of natural numbers that appear as the number of real roots an irreducible polynomial of degree $n$ over $\mathbb{Q}$ which is solvable by radicals can have.
Example: In case $n=p$ is a prime, we have $X_p= \{1,p \}$.
Is $X_n$ or the cardinality $|X_n|$ also known for other values?
It would be interesting to see the beginning of the sequence $a_n=|X_n|$ for small values of $n$, maybe it appears in the oeis.
 A: Jensen On the number of real roots of a solvable polynomial
 includes a proof of: 
Loewy’s theorem. Let $K$ be a real number ﬁeld and $f(X)$ an irreducible polynomial in $K[X]$ of odd degree $n$. If $p$ is the smallest prime divisor of $n$ and the Galois group of $f(X)$ over $K$ is solvable, then $r(f) = 1$ or $n$ or satisﬁes the inequalities $p ≤ r(f)≤ n−p +1.$
A: Klueners and Malle have a database of number fields of degree $\leq 19$ that (tries) to include every Galois group and every possible signature. Examining this database shows that for composite $n$ with $4 \leq n \leq 18$, $X_{n} = \{ k : 0 \leq k \leq n \text{ and } k \equiv n \pmod{2} \}$.
EDIT: If $n = 2k$ is even, then $\mathbb{Z}/2\mathbb{Z} \wr \mathbb{Z}/k\mathbb{Z}$ contains elements of order $2$ with any desired even number of fixed points. It seems possible that one can obtain the same thing for $n$'s that are multiples of $3$. The smallest $n$ that does not fall into this category is $n = 25$, and using the classification of transitive subgroups of $S_{25}$, one finds that $X_{25} \subseteq \{ 1, 5, 9, 13, 17, 21, 25 \}$. 
