This is the followup challenge from this one, if you're confused please check that one out first.
First, let \$m(s, k)\$ be the number of cache misses a sequence \$s\$ of resource accesses would have assuming our cache has capacity \$k\$ and uses a first-in-first-out (FIFO) ejection scheme when it is full.
Then given a ratio \$r > 1\$, return a non-empty sequence of resources accesses \$s\$ such that there exists \$k > j\$ with \$m(s, k) \geq r \cdot m(s, j)\$.
In plain English, construct a sequence \$s\$ of resource accesses so that there's two cache sizes where the bigger cache has (at least) \$r\$ times more cache misses when used to resolve \$s\$.
An example for \$r = 1.1\$, a valid output is the sequence \$(3, 2, 1, 0, 3, 2, 4, 3, 2, 1, 0, 4)\$, as it causes \$9\$ cache misses for a cache size of \$3\$, but \$10\$ misses for a cache size of \$4\$.
It doesn't matter what sequence you return, as long as it meets the requirements.
Shortest code in bytes wins.