In what follows, we let $\sigma=\sigma_1$ denote the classical sum-of-divisors function.
Suppose that $N$ is an odd perfect number and $q^\alpha$ is a prime power with $q^\alpha \parallel N$.
(Euler proved that $N$ can be written in the Eulerian form $p^k n^2$ where $p$ is the special/Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,n)=1$.)
Define the index $m = \sigma(N/q^\alpha)/q^\alpha$. Chen and Chen (2014) proved that, if $q = p$, then $m$ cannot take any of the $30$ forms
$$q_1, {q_1}^2, {q_1}^3, {q_1}^4, {q_1}^5, {q_1}^6, {q_1}^7, {q_1}^8, q_1 q_2, {q_1}^2 q_2, {q_1}^3 q_2, {q_1}^4 q_2, {q_1}^5 q_2, {q_1}^2 {q_2}^2, {q_1}^3 {q_2}^2, {q_1}^4 {q_2}^2, q_1 q_2 q_3, {q_1}^2 q_2 q_3, {q_1}^3 q_2 q_3,$$
$${q_1}^4 q_2 q_3, {q_1}^2 {q_2}^2 q_3, {q_1}^2 {q_2}^2 {q_3}^2, q_1 q_2 q_3 q_4, {q_1}^2 q_2 q_3 q_4, {q_1}^3 q_2 q_3 q_4, {q_1}^2 {q_2}^2 q_3 q_4, q_1 q_2 q_3 q_4 q_5, {q_1}^2 q_2 q_3 q_4 q_5, q_1 q_2 q_3 q_4 q_5 q_6, q_1 q_2 q_3 q_4 q_5 q_6 q_7$$
where $q_1, q_2, q_3, q_4, q_5, q_6, q_7$ are distinct odd primes.
INITIAL QUESTION
What numerical lower bound on the index $m$ of an odd perfect number is implied by the results in F.-J. Chen and Y.-G. Chen's 2014 paper, if $q=p$?
MOTIVATION
Broughan, Delbourgo, and Zhou (2013) proved the following result:
If $q=p$, then $m=\sigma(N/q^\alpha)/q^\alpha$ cannot take any of the $11$ forms
$$p_1, {p_1}^2, {p_1}^3, {p_1}^4, {p_1}^5, {p_1}^6, p_1 p_2, {p_1}^2 p_2, {p_1}^3 p_2, {p_1}^2 {p_2}^2, p_1 p_2 p_3$$
where $p_1, p_2, p_3$ are any distinct odd primes.
Therefore, the smallest possible value of the index $m$ in this case is $3^2 \times 5 \times 7 = 315$.
MY ATTEMPT
Following Broughan, Delbourgo, and Zhou's lead, I obtain (from the results of Chen and Chen) that the smallest possible value of the index $m=\sigma(N/q^\alpha)/q^\alpha$ when $q=p$ is
$$3^2 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 = 14549535.$$
FOLLOW-UP QUESTION
Is my attempted answer correct? If not, how can it be mended?
Best Answer
The smallest possible value of the index $m=\sigma(N/q^\alpha)/q^\alpha$ when $q=p$ is $$3^3\times 5^3=3375$$
Proof :
Let $M$ be the number of prime divisors counted with multiplicity.
Using $M$, we can classify the $30$ forms as follows :
$$\small\begin{align}M=1&:q_1 \\\\M=2&:{q_1}^2, q_1 q_2 \\\\M=3&:{q_1}^3, {q_1}^2 q_2,q_1 q_2 q_3 \\\\M=4&:{q_1}^4, {q_1}^3 q_2,{q_1}^2 {q_2}^2,{q_1}^2 q_2 q_3,q_1 q_2 q_3 q_4 \\\\M=5&:{q_1}^5, {q_1}^4 q_2, {q_1}^3 {q_2}^2,{q_1}^3 q_2 q_3,{q_1}^2 {q_2}^2 q_3,{q_1}^2 q_2 q_3 q_4,q_1 q_2 q_3 q_4 q_5 \\\\M=6&:{q_1}^6,{q_1}^5 q_2, {q_1}^4 {q_2}^2,{q_1}^4 q_2 q_3,{q_1}^2 {q_2}^2 {q_3}^2,{q_1}^3 q_2 q_3 q_4,{q_1}^2 {q_2}^2 q_3 q_4,{q_1}^2 q_2 q_3 q_4 q_5,q_1 q_2 q_3 q_4 q_5 q_6 \\\\M=7&:{q_1}^7, q_1 q_2 q_3 q_4 q_5 q_6 q_7 \\\\M=8&:{q_1}^8\end{align}$$
For $M\le 5$, the $18$ forms are all the possible forms, so there is no form that $m$ can take.
For $M=6$, we see that ${q_1}^3{q_2}^3,{q_1}^3{q_2}^2 q_3$ are not included, so the smallest possible value of $m$ is $3^3\times 5^3=3375$.
For $M=7$, the smallest possible value of $m$ is $3^6\times 5=3645$.
For $M=8$, the smallest possible value of $m$ is $3^7\times 5=10935$.
For $M\ge 9$, the smallest possible value of $m$ is $3^9=19683$.
Therefore, it follows that the smallest possible value of the index $m=\sigma(N/q^\alpha)/q^\alpha$ when $q=p$ is $$3^3\times 5^3=3375$$