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.


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 $s$ $>$ $100$.

  • 2
    $\begingroup$ You might want to reference the OEIS sequence and your earlier question, otherwise it is not immediately clear to everyone where did 257 come from. $\endgroup$ Commented Dec 10, 2016 at 8:15
  • $\begingroup$ Good point. Still, would you know how to arrange the rest of the factors so my method works. $\endgroup$
    – J. Linne
    Commented Dec 10, 2016 at 8:17
  • $\begingroup$ I'll have to think of it for a while. $\endgroup$ Commented Dec 10, 2016 at 8:18
  • $\begingroup$ For the numbers from $n$ to $n+100$, $n+11$ appears to be the only one without a listed factor from $22$ primes. $\endgroup$
    – J. Linne
    Commented Dec 10, 2016 at 8:24
  • 1
    $\begingroup$ @Peter I searched and found a result with 209 consecutive integers each with a prime factor less than 100, which is close to the 257 maximum limit. Number posted in question. $\endgroup$
    – J. Linne
    Commented Dec 20, 2016 at 4:07

2 Answers 2


It took me much effort, but finally I found an approach leading to high differences! Here the result :

? q=1;while(q>0,n=Mod(0,1);z=vector(252,s,s);p=1;while(p<97,p=nextprime(p+1);mer
ent(n,2);anz=0;for(j=-20,260,if(gcd(n+j,100!)==1,anz=anz+1;print1(j," ")));print
-20 -16 -14 -10 -4 -2 256 260

The displayed number is $N$. The numbers $N-2$ and $N+256$ have no prime factor below $100$ , but all numbers between have one. So the difference is $258$. If I remember an OEIS-entry correctly, this is the maximum possible difference.


Based on A049300, each of the $232$ integers after $990578949024486399796411481983580$ have a prime factor $s<100$.

A table up to $n=24$ is available at http://oeis.org/A049300/b049300.txt


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