I am trying to come up with $257$ (or $256$) consecutive integers such that each one has a prime factor $<$ $100$. Is my approach a good method?
$1.$ Choose an integer $n$ such that ($100$#) (primorial function) $|$ $n$.
$2.$ Eliminate a small factor from the set of primes less than $100$ (call this prime $p$). $p$ $|$ $n+1$.
$3.$ Eliminate all primes $q$ congruent to $1$ $\pmod p$ since $p$ $|$ $n+q$. This set is {$q, q_1,.... q_n$}.
$4.$ Choose one of these primes $q_n$ such that $q_n$ $|$ $n+1$. Eliminate all primes $s$ congruent $1$ $\pmod {q_n}$.
$5.$ Repeat this process until no more primes $<$ $100$ can be eliminated.
$6.$ For all divisors $d_n$ of $n+1$, use primes $r_n$ that do not divide $n$ or $n+1$ and make the congruence $r_n$ $|$ $n+d_n$ hold.
Example using this idea:
The $162$ consecutive integers after $1610596759123800808688936916463498913$ have a prime factor $s$ $<$ $100$. (It took me a while and multiple times I got the congruence relation wrong.)
Can someone help create a consecutive integer list $n+-k$ such that a prime less than $100$ divides each integer on the list (based on my example) that beats my record. Thanks.
Take a look at https://oeis.org/A058989 for more info on the max length for these sequences.
UPDATE:
Is anyone willing to try and find a better result than this one:
Each of the first $209$ integers after $980048014805329352638322460936985099$ have a prime factor $s$ $>$ $100$.