Given an odd abundant number $n$, if one replaces each prime factor of $n$ with the preceding prime while maintaining the same multiplicity (which gives A064989($n$)), does one always get another abundant number?
For example, $945=3^3 \cdot 5 \cdot 7$ is the first odd abundant number, and $120=2^3 \cdot 3 \cdot 5$ is also abundant.
This is what Antti Karttunen's comment at A005231 says, and it needs a proof.
$$\sigma(n)/ n = \prod\limits_{p^k \mid \mid n} \frac{p^{k+1}-1}{p^k(p-1)}$$
The factors in the right are decreasing in $p$ as is seen by noting the derivative with respect to $p$ is $= \dfrac{p^{-k-1} (k (p - 1) - p (p^k - 1))}{(p - 1)^2}$ and we have $p(p^k-1) > k(p-1)$ for $p\geq 2$.