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

Composite Dris index, at an exponent whose successor has 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

The index is necessarily odd. This statement asserts that no odd composite index $s \ge 2$ occurs, for a special exponent $k \ge 1$ whose successor has at most one odd prime divisor, i.e. $k+1 = 2^{a}q^{b}$ with $q$ prime; the hypothesis is formalized as $\#\bigl(\mathrm{primeFactors}(k+1)\setminus\{2\}\bigr) \le 1$.

This is the residual index range for such exponents. The index $s = 1$ is excluded by the theorem of Dandapat–Hunsucker–Pomerance, and every odd prime index is excluded by OddPerfectNumber.no_dris_index_odd_prime_of_one_odd_prime; together with the present statement these cover all indices, so proving it settles the Dris problem for the special exponents $k = 1, 5, 9, 13$ (where $k+1 = 2, 6, 10, 14$).

The obstruction used for a prime index does not extend as it stands. For prime $s$ the divisor sum $\sigma(m^2) = p^ks$ has only two prime divisors, which bounds the prime support of $m$ by three ($s$ itself, the odd prime dividing $k+1$, and one exceptional prime whose divisor sum absorbs the factor $s$), contradicting Sylvester's bound $\omega(N) \ge 5$. For composite $s$ the same counting gives only $\omega(m) \le \omega(s) + \Omega(s) + 1$, which is no longer smaller than $4$.