Being $k$-rough is the same as being relatively prime to the product of primes up to $k$. Hence if $P(k)$ denotes that product, and $j$ denotes the Jacobsthal function, then $t=j(P(k))-1$ works for every $n$, while $t=j(P(k))-2$ fails for infinitely many $n$'s. In particular, it follows from a result of Iwaniec (1978) that $t=ck^2$ works for some absolute constant $c>0$.

Being $k$-rough is the same as being relatively prime to the product of primes up to $k$. Hence if $P(k)$ denotes that product, and $j$ denotes the Jacobsthal function, then $t=j(P(k))-1$ works for every $n$, while $t=j(P(k))-2$ fails for infinitely many $n$'s. In particular, it follows from a result of Iwaniec (1971) that $t=ck^2$ works for some absolute constant $c>0$.