Determine whether the $2$-densely divisible numbers are the same as the $k$-practical numbers as $k$ goes to infinity.
First of all take $p_1^{a_1}\ldots p_{\omega(n)}^{a_{\omega(n)}}$ to be the canonical prime factorization of an integer $n\geq2$. Then for convenience, let $n_0=1$ and for $i\in[1,\omega(n)]$,
$$n_i=\prod_{j=1}^i p_j^{a_j}$$
Call $n$ a $2$-densely divisible number if
$$\forall i\in[1,\omega(n)],\ p_i\leq 2n_{i-1}$$
Call $n$ a $k$-practical number if
$$\forall i\in[1,\omega(n)],\ p_i^k-1\leq (2^k-1)\sigma_k(n_{i-1})$$
where
$$\sigma_k(n)=\prod_{i=1}^{\omega(n)}\dfrac{p_i^{k(a_i+1)-1}}{p_i^k-1}$$
is the divisor function.
$n$ is $2$-densely divisible iff each divisor is less than or equal to twice the previous divisor.
The $k$-practical numbers are a bit more complicated...
$n$ is $k$-practical iff every positive integer up to $\sigma_k(n)$ has a representation of the form
$$c_1d_1^k+c_2d_2^k+\cdots+c_{\sigma_0(n)}d_{\sigma_0(n)}^k$$
where $d_1,\ldots,d_{\sigma_0(n)}$ are the divisors of $n$ and $c_i\in[0,2^k-1]$
In words, $n$ is $k$-practical if and only if every positive integer less than or equal to the sum of the $k$th powers of the divisors of $n$ is the sum of at most $2^k-1$ of each of the divisors of $n$ raised to the $k$th power.
My belief (based on computational evidence) is that if $n$ is $k$-practical, but not 2-densely divisible, then there must exist some $x>k$ such that $n$ is not $x$-practical. This belief stems from both intuition and computational evidence, but I never made much progress proving it.