I though at first that there does not exists a set of more than $100$ consecutive integers such that each has a prime factor less than $100$, but I made a list, showing I was wrong. Does there exist a set of more than $120$ consecutive integers ($k$ consecutive integers where $k>120$) (other than mine) such that every integer between $n$ and $n+k$ has a prime factor $s$ less than $100$. If so, what is the greatest $k$ value such that for all numbers $x,\;n<x<n+k$, $x$ has a prime factor $<100$. Thanks for solving this. (Already done). My next question is who can find a chain longer than these two:
After $14002333221094855441238405921422197787$, there are $123$ consecutive integers such that each has a prime factor $s$ $<$ $100$.
After $1610596759123800808688936916463498913$, there are $163$ consecutive integers such that each has a prime factor $s$ $<$ $100$.
Thanks for finding one if possible.
Looks like there is no simple formula. OEIS contains the lengths of longest intervals of consecutive integers such that each is divisible by a "small" prime (A058989). There are 25 primes below 100, hence the 25'th term of that sequence is the answer to your question. Indeed, it is much greater than 121.
OEIS also contains the list of left ends of the said intervals (A049300), which, unfortunately, does not extend to the 25'th term, so we'll have to content ourselves with something smaller than that. Look, for example, at the list of 131 numbers starting at $14\,719\,192\,159\,220\,252\,523\,420$: each of them has a prime factor no greater than 61.