Consecutive composite numbers When I took basic number-theory course there was this exercise to find 2000 consecutive numbers. And of course it's well known that the trick to take numbers of the form 
$$
(n+1)!+m, \quad 2 \leq m \leq n+1
$$
does the trick with $n=2000$. But these numbers are really huge which makes me think of the following question.
Is there a better way to construct consecutive composite numbers? With better I mean that is there a way to find smaller consecutive composites for each $n$? Or even the smallest?
Example: Smallest 3 consecutive composites are clearly $8,9,10$ but the formula gives us the numbers $26, 27, 28$ and even $14,15,16$ would be smaller. So I'm asking a way to find these smaller/smallest consecutive composites.
 A: Based on the ideas in Crostul's answer and my comment, I would suggest the following (not necessarily optimal) algorithm:  Let $2000$ consecutive integers be denoted by $n, n+1,..., n+1999$.  We will put conditions on $n$ so as to make all these integers composite.
(1) Require $n$ to be a multiple of $2$ (and greater than $2$).  Then $n$, $n+2$, $n+4$,... are all composite.  This condition is equivalent to $n\equiv 0\pmod{2}$.
(2) The first non multiple of $2$ is now $n+1$.  So we'll require $3\mid n+1$.  Then $n+1$, $n+4$,... are all divisible by $3$.  This condition is equivalent to $n\equiv 2\pmod{3}$.
(3) At each subsequent step we determine the smallest integer in our sequence not yet known to be composite, and we force it to be a multiple of the next available prime.  This will give us a congruence of the form $n\equiv t\pmod{p_k}$, where $p_k$ is the next available prime, and $t$ is some appropriate integer with $0\le t< p_k$.
Eventually all of $n, n+1, ..., n+1999$ will have to be composite.  We'll then have a biggish Chinese Remainder Theorem (CRT) problem to solve for $n$, but this CRT problem will not have anywhere near $2000$ congruences because of identifying so many composites at each step.
A: Well, the tables do not yet go high enough, but look at Prime pair points slope approaches 1
and the websites mentioned. Experience and the Granville-Cramer conjectures suggest that, to get a prime gap of 2000, we should expect the prime at the beginning of the gap to be at least
$$  e^{\sqrt {2000}} \approx e^{44.721} \approx 2.64 \cdot 10^{19},  $$
maybe a little bigger. 
So, wait a few years, if they get the maximal prime gap tables up to something like $10^{21}$ you would expect to get a gap over 2000. Right now the tables are up to $4 \cdot 10^{18}$
EEEDDDIIITTT: for gaps numbered 52 to 75, the ratio $g / \log^2 p$ is no smaller than $3/4.$ So, taking this as a pessimistic bound, we expect gap 2000 by 
$$  e^{\sqrt {8000/3}} \approx e^{51.639} \approx 2.67 \cdot 10^{22},  $$
so maybe best up to $10^{23}$
ESSAY QUESTION: it would appear that finding a gap of fixed size (2000) is nowhere near as difficult as confirming maximal gaps. I would think that there is some low-tech method that could do this, I just do not know whether the entire range from $10^{18}$ to, say, $10^{24},$ can be done in under a year. We need to spend the smallest possible time deciding whether a number is prime or composite; a hierarchy of methods, (I) trial division by primes up to $10,000;$ (II) trial with Fermat's method of difference of squares (III) "probable prime" function from any CAS (IV) prime proof. Actual program to juggle a number of time/storage issues. Need to think about it, but I might do a tiny run with C++/GMP. All can be done because of the  $10^{24}$ bound, but i have no idea how slow it would be. 
A: This is a partial answer.
Let's denote by $p_1, \dots, p_{2000}$ the first 2000 prime numbers ($p_1 = 2, p_2 = 3, p_3=5, p_4=7$ and so on)
Then the Chinese Reminder Theorem tells you that the system
$$\begin{matrix}
n &\equiv0 &\mod p_1 \\
n+1 &\equiv0 &\mod p_2 \\
n+2 &\equiv0 &\mod p_3 \\
&\vdots &\\
n+k-1 &\equiv0& \mod p_k \\
&\vdots&\\
n+1999& \equiv0& \mod p_{2000}
\end{matrix}$$
has a unique solution $n \in \{ 0, \dots, p_1 \cdots p_{2000}\}$.
Now if you permute the $p_k$'s you get another unique solution, one of these may be the minimal solution for your problem.
The trouble with this approach is that all permutations of the $p_k$'s are $2000!$, and I have no idea on how finding the best one.
