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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.

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  • $\begingroup$ Let me understand this: If you would want a $k$ for $10$, you could use $k=10$? (I know that you want a $k$ for $120$, but I need to understand what you're asking). $\endgroup$ – MonsieurGalois Dec 8 '16 at 8:24
  • $\begingroup$ Try to come up with a list of more than $121$ consecutive integers like I did, but each must have a prime factor $<$ $100$. $\endgroup$ – J. Linne Dec 8 '16 at 8:26
  • $\begingroup$ @MonsieurGalois at least you see how my list of consecutive integers make sense. Can you come up with a longer one than mine? $\endgroup$ – J. Linne Dec 8 '16 at 8:34
  • $\begingroup$ @Levent I just showed you a set with $120$ consecuative integers all integers in that range have a factor less than $100$. It is possible. $\endgroup$ – J. Linne Dec 8 '16 at 8:40
  • $\begingroup$ How 2^3*5*13*31+1=11^2?? $\endgroup$ – Atul Mishra Dec 8 '16 at 8:48
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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.

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  • $\begingroup$ Is there a list of these $257$ consecutive integers similar to the list of $121$ I provided? $\endgroup$ – J. Linne Dec 8 '16 at 8:58
  • $\begingroup$ No, but it does provide smaller lists, which are still longer than yours. $\endgroup$ – Ivan Neretin Dec 8 '16 at 9:06
  • $\begingroup$ Please look at math.stackexchange.com/questions/2052240/… $\endgroup$ – J. Linne Dec 10 '16 at 8:08
  • $\begingroup$ I found two of the numbers for which the congruence holds. $\endgroup$ – J. Linne Dec 10 '16 at 21:50
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Maybe this is useful for your problem (if I understood it):

Notice that $n!+k$ is composite with $2 \leq k\leq n$, so if you want for example, I don't know, $125$ consecutive numbers, you can see that (if $2\leq k \leq 124$)

$$125!+k=k\left(\frac{125!}{k}+1\right)$$

Is an integer, so $k$ divide it and has a prime factor less to $125$. For the number $125!+125$, you can factor the $5$.

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  • $\begingroup$ That would be useful if the factor limit for the smallest prime $s$ dividing $n+k$ had no limit. However, I am looking at consecutive integers such that each divisible by at least one prime less than $100$. I came up with $121$ as shown. $\endgroup$ – J. Linne Dec 8 '16 at 8:56

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