No odd perfect number whose Dris index is an odd prime, when $k+1$ has at most one odd prime factor

A prime Dris index is impossible when $k+1$ has at most one odd prime factor.

In the Dris parametrisation of a hypothetical odd perfect number $N = p^k m^2$ (Euler form: $p$ the special prime, $m$ odd, $p \nmid m$), the index is the number $s$ with

This statement rules out every odd prime index $s$, for every special exponent $k \ge 1$ whose successor $k+1$ has at most one odd prime divisor, i.e. $k+1 = 2^{a}q^{b}$ with $q$ prime. In particular it settles the indices $s = 3, 5, 7, 11, \dots$ for the special exponents $k = 1, 5, 9, 13$, where $k+1 = 2, 6, 10, 14$. (The index of an odd perfect number is always odd, since $\sigma(m^2)$ is odd for odd $m$; the case $s = 1$ is the theorem of Dandapat–Hunsucker–Pomerance.)

The hypothesis on $k+1$ is formalized as $\#\bigl(\mathrm{primeFactors}(k+1)\setminus\{2\}\bigr) \le 1$, and $\sigma$ is written as the sum over the divisor finset.

The proof rests on a counting bound for the prime support of $m$. Because $s$ is prime, $\sigma(m^2) = p^k s$ has only two prime divisors. For a prime $q \mid m$ with $q \ne s$ and $q^{a} \,\|\, m$, the divisor sum $\sigma(q^{2a})$ divides $p^ks$; if it is a pure power of $p$, then a lifting-the-exponent argument in the style of Dandapat–Hunsucker–Pomerance forces $q \mid k+1$, and otherwise $s \mid \sigma(q^{2a})$, which by $v_s(p^ks) = 1$ can happen for at most one prime $q$. Hence the primes of $m$ lie among $s$, the odd prime dividing $k+1$, and one exceptional prime, so $\omega(N) \le 4$ — contradicting Sylvester's bound $\omega(N) \ge 5$ for odd perfect numbers.