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Let $\phi(B,k)$ be the statement that there are more than $k$ many positive $B$-rough integers below $kB$, where both parameters are also positive integers. $\phi$ is false for $B \lt 4$, and is also false for $k=1$. Since there is only one prime between 5 and 10, $\phi$ is also false when $B=5$ and $k=2$. However, there is a simple combinatorial argument which shows that $\phi$ is true for many pairs $(B,k)$.

Warning: the argument extending j past B is flawed. I will see what I can do to repair it.

Let $\phi(B,k)$ be the statement that there are more than $k$ many positive $B$-rough integers below $kB$, where both parameters are also positive integers. $\phi$ is false for $B \lt 4$, and is also false for $k=1$. Since there is only one prime between 5 and 10, $\phi$ is also false when $B=5$ and $k=2$. However, there is a simple combinatorial argument which shows that $\phi$ is true for many pairs $(B,k)$.

With more care we can extend this argument to cover some $k$ below $q/2$. Let $r$ be the largest prime less than $q$ (so note that $r,q,B$ are consecutive primes) and replace $P(q,jB)$ in the equation above by $P(r,jB) - P(r,(jB)/q)$. The point of this is that $B\geq r+6$ for $r\gt 3$, and we can for large enough $r$ start $j$ close to $B/6$, and go up to $j \lt q$. One then works with the quantity $P(r,jB)-2$ for $j\lt q$ to prove $\phi(B,k)$ for even smaller $k$.

With more care we can extend this argument to cover some $k$ below $q/2$. Let $r$ be the largest prime less than $q$ (so note that $r,q,B$ are consecutive primes) and replace $P(q,jB)$ in the equation above by $P(r,jB) - P(r,(jB)/q)$. The point of this is that $B\geq r+6$ for $r\gt 3$, and we can for large enough $r$ start $j$ close to $r/3$, and go up to $j \lt q$. One then works with the quantity $P(r,jB)-2$ for $j\lt q$ to prove $\phi(B,k)$ for even smaller $k$.

A totative counting argument gives us $\phi(5,k)$ for all $k \gt 5/2$. Suppose we have $\phi(q,k)$ for all $k \gt q/2$. Using the relation above and writing $j$ for the smallest integer above $q/2$, we have $$P(B,jB) = P(q,jB) - P(q,j) \gt \lfloor jB/q \rfloor - 1.$$ We use the fact that $j \lt q$. Since $B$ and $q$ are consecutive primes, $B\geq q+2$ and so the right hand side is at least $\lfloor (q+3)/2 \rfloor - 1 = j$. This gives the implication $\phi(q,j)$ implies $\phi(B,j)$ when $j=\lceil q/2 \rceil$.

A totative counting argument gives us $\phi(5,k)$ for all $k \gt 5/2$. Suppose we have $\phi(q,k)$ for all $k \gt q/2$. Using the relation above and writing $j$ for the smallest integer above $q/2$, we have $$P(B,jB) = P(q,jB) - P(q,j) \gt \lfloor jB/q \rfloor - 1.$$ We use the fact that $j \lt B$. Since $B$ and $q$ are consecutive odd primes, $B\geq q+2$ and so the right hand side is at least $\lfloor (q+3)/2 \rfloor - 1 = j$. This gives the implication $\phi(q,j)$ implies $\phi(B,j)$ when $j=\lceil q/2 \rceil$.