I'm trying to find the longest consecutive set of composite numbers Hello and I'm quite new to Math SE.

I am trying to find the largest consecutive sequence of composite numbers. The largest I know is:

$$90, 91, 92, 93, 94, 95, 96$$

I can't make this series any longer because $97$ is prime unfortunately.

I can however, see a certain relation, if suppose we take the numbers like (let $a_1, a_2, a_3,...,a_n$denote digits and not multiplication):

$$a_1a_2a_3...a_n1,\ a_1a_2a_3...a_n2,\ a_1a_2a_3...a_n3,\ a_1a_2a_3...a_n4,\ a_1a_2a_3...a_n5,\ a_1a_2a_3...a_n6,\ a_1a_2a_3...a_n7,\ a_1a_2a_3...a_n8,\ a_1a_2a_3...a_n9,\ a_1a_2a_3...(a_n+1)0$$

The entire list of consecutive natural numbers I showed above can be made composite if:


The number formed by digits $a_1a_2a_3...a_n$ should be a multiple of 3
The numbers $a_1a_2a_3...a_n1$ and $a_1a_2a_3...a_n7$ should be composite numbers


If I didn't clearly convey what I'm trying to say, I mean like, say I want the two numbers (eg: ($121$, $127$) or ($151$, $157$) or ($181$, $187$)) to be both composite.

I'm still quite not equipped with enough knowledge to identify if a random large number is prime or not, so I believe you guys at Math SE can help me out.
You can have a sequence as long as you wish. Consider $n\in\Bbb{N}$ then the set

$$S_n=\{n!+2,n!+3,\cdots,n!+n\}$$

is made of composite consecutive numbers and is of length $n-1$
I wish to point out 2 ideas which arent as good as the primorial idea nor as good as some of the other ideas, but which are enhancements of the idea mentioned that $n!+2, n!+3, n!+4, .... n!+n$ are composite, which is that
$n!-2, n!-3, n!-4, ... n!-n$ will be composite provided $n!-m=(n!/m-1)*m$ has both factors greater than 1 for $m=2,3,...n$.  clearly the second factor $m$ is ok,
for the other factor we need $n!/m-1&gt;=2$ ie $n!/m&gt;=3$ ie $n!&gt;=3m$, this will be true for all the $m$ if $n!&gt;=3n$ ie $(n-1)!&gt;=3$ which is true if $n-1&gt;=3$, ie $n&gt;=4$
so for $n&gt;=4$, the consecutive numbers $n!-2, n!-3, ...., n!-n$ are composite!
The other enhancement is that if $N=lcm(2,3,4,5,6,...,n)$, then $N-2,N-3,...,N-n$ are composite provided $N-m=(N/m-1)*m$ has both factors greater than 1. The second factor is clearly alright.
so we need $N/m-1&gt;=2$ ie $N/m&gt;=3$ ie $N&gt;=3*m$. Now $(m-1)$ and $m$ are coprime, so $N&gt;=(m-1)*m$. Thus for $m-1&gt;=3$ ie $m&gt;=4$ this is true. We thus need to check $m&lt;4$ namely $m=2,3$. for $m=2$ we need $N&gt;=3*2=6$ which will be true for $n&gt;=3$ because 2 and 3 are coprime and thus $N&gt;=2*3=6$. And for $m=3$ we need $N&gt;=3*3=9$ which will be true for $n&gt;=4$ because 3 and 4 are coprime, hence $N&gt;=3*4=12&gt;9$.
so for $n&gt;=4$, $N-2, N-3, ...., N-n$ are composite!
I agree entirely that this is not as good a result as some of the other ones given, but I give this as the idea simply because it is different and somewhere in between the results given.
Here follows an enhancement of the primorial argument:
if we consider $n\#-2, n\#-3,...,n\#-n$, and choose $2&lt;=m&lt;=n$, and let $p$ be a prime factor of $m$, then $p$ is a factor of both $n\#$ and $m$. So $n\#-m=p*((n\#-m)/p)$, and this is composite if the second factor is greater than 1, ie $(n\#-m)/p&gt;1$, ie $n\#&gt;p+m$. If we can show that $n\#&gt;=3n$ for $n&gt;=5$, then $n\#&gt;=3n&gt;p+m$ and thus $n\#-m$ is composite.
In fact via another post, $n\#&gt;=3m$ for $n&gt;=5$, namely:
Prove by elementary means that $n\#\geq 3n$ for $n\geq 5$, where $n\#$ is the primorial function.
so for $n&gt;=5$, $n\#-2, n\#-3,...,n\#-n$ are consecutive composite numbers in descending order. The result is false for $n=4$ because $4\#-3=3$ and $4\#-4=2$ which are primes.