Define $\triangle_n$ to be the $n$th triangular number.
Define $$M_n=(2\triangle_n-1)2\triangle_n(2\triangle_n+1)=2\triangle_n(2\triangle_n^2-1).$$
Define $(\ell,k)$-smough numbers to be numbers that have all prime divisors $p$ of form either $p|\ell!$$p<\ell$ or $k<p$.
GivenIs there an $\ell$ is$\ell_0\geq6$ such that for every $\ell>\ell_0$ and for every $k$ are there a maximuminfinite number of $k$ abovefor which there is noalways an $n$ such that $q(n(n+1))$$M_n$ is $(\ell,k)$-smough where $q(x)=x(x^2-1)$?
Is there a fast algorithm to findGiven an $n$ for$\ell$ and a given $\ell,k$$k$ how small can $n$ be such that $q(n(n+1))$$M_n$ is $(\ell,k)$-smough when such $n$ exists or return $n=\infty$?
Note $q(n(n+1))$ is always of form $2\triangle_n(2^2\square_n-1)$ where $\triangle_n$ is $n$-th triangular number (sum of first $n$ natural numbers) and $\square_n=\triangle_n^2$ is sum of cubes of first $n$ natural numbers.
How large can the ratio $\frac k\ell$ be?
Interesting cases at $n<100$ are at $17,31,59,89,97$:
https://www.wolframalpha.com/input/?i=Factor%28%28p%29%28p%2B1%29%28%28%28p%29%28p%2B1%29%29%5E2-1%29%29+at+p%3D17
https://www.wolframalpha.com/input/?i=Factor%28%28p%29%28p%2B1%29%28%28%28p%29%28p%2B1%29%29%5E2-1%29%29+at+p%3D31
https://www.wolframalpha.com/input/?i=Factor%28%28p%29%28p%2B1%29%28%28%28p%29%28p%2B1%29%29%5E2-1%29%29+at+p%3D59
https://www.wolframalpha.com/input/?i=Factor%28%28p%29%28p%2B1%29%28%28%28p%29%28p%2B1%29%29%5E2-1%29%29+at+p%3D89
https://www.wolframalpha.com/input/?i=Factor%28%28p%29%28p%2B1%29%28%28%28p%29%28p%2B1%29%29%5E2-1%29%29+at+p%3D97
I doubt $k$ grows much if $\ell$ is fixed.