No odd perfect number of Dris index $3$ when $k+1$ has at most one odd prime factor

Dris index $3$ 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 the index $s = 3$ for every special exponent $k \ge 1$ whose successor $k+1$ has at most one odd prime divisor, i.e. $k + 1 = 2^{a} q^{b}$ for a prime $q$. In particular it settles the index-three case for the special exponents $k = 1, 5, 9, 13$, where $k+1 = 2, 6, 10, 14$.

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

The proof combines two ingredients. Fixing $s = 3$ forces $3 \mid m$ and makes $\sigma(m^2) = 3p^k$ a number with only two prime divisors; a lifting-the-exponent argument in the style of Dandapat–Hunsucker–Pomerance then shows that a prime $q \mid m$ with $q \ne 3$ either divides $k+1$ or satisfies $3 \mid \sigma(q^{2a})$, where $q^{a} \,\|\, m$. Since $v_3(\sigma(m^2)) = 1$, at most one prime of the second kind exists, so $m$ has at most three distinct prime divisors and $N$ at most four — contradicting Sylvester's bound $\omega(N) \ge 5$ for odd perfect numbers.