Strings of consecutive integers divisible by 1, 2, 3, ..., N For each n, let $a_n$ be the least integer, greater than n, such that the numbers $a_n$, $a_n$+ 1, $a_n$+ 2, ..., $a_n$+ (n – 1) are divisible, in some order, by 1, 2, 3, ..., n. For example $a_{12}$ = 110.
What are the best estimates known for  $a_n$?
 A: I do not know what is the best estimate for $a_n$. However, one can easily found an upper bound noticing that an integer $x$ satisfying your condition can be found by solving a system of congruences
$$x +i \equiv 0  \; \; \mathrm{mod}\, (i+1), \quad i=0, \ldots, n-1.$$
A solution is given by $x=1$, so all solutions are of the form $1+kN(n)$, where $k$ is an integer and $N(n)=lcm(1, \ldots, n)$. It follows that $$a_n \leq 1+N(n).$$
A (very time-consuming) algorithm finding $a_n$ is the following. Given a permutation $\sigma \in S_n$, denote by $x_{\sigma}$ the least solution $> n$ of the system of linear congruences $$x +i \equiv 0  \; \; \mathrm{mod}\, (\sigma(i+1)), \quad i=0, \ldots, n-1.$$
Then $$a_n = \min_{\sigma \in S_n} x_{\sigma}.$$
A: I suspect (and would appreciate verification by someone else) that a more time efficient alternative to Francesco Polizzi's suggestion is the following: generate all such sequences less than $N(n)$, and then use them to generate sequences for n+1. Of course, if we have such an interval starting at $a$, then there is another ending at $N(n)-a$
At the very least, one needs a sequence of numbers of length about $n/2$ (and usually longer) adjacent to a multiple of $n+1$ and this sequence forms a consecutive interval of integers none of which are coprime to $ N(n)$. Since most such sequences are of length log n, there should be relatively few of this length (although more than the number of permutations of $\pi(n)$).  I do not have numerics at hand, but my gut feeling is that the number of such intervals should be way less than $n!$. A recent preprint on Arxiv by Mario Ziller distantly relates to this problem; his (with John Morack) published algorithms should be adaptable to this problem.
Edit 2019.04.16:
It is taking me longer than usual to program. I will describe an algorithm which should shed light on this problem, and which may lead to a proof that an upper bound on the desired number is (for $n \gt 3$) much smaller than $N(n)$, and likely smaller (for $n \gt 10$) than $P(n)/2,$ the product of the odd primes smaller than $n$.
As noted elsewhere, an obstruction to $I=[m+1,m+n]$ having a bijection $d()$ to $[1,n]$ so that for each integer $d(l)$ in the range one has $d(l)$ divides $l$ is that there are two primes (with sizes) bigger than $n$ in the domain interval. More generally, there are at least two obstructions: that $I$ contains two numbers coprime to $N(n)$, and that a special multiple occurs in the middle of $I$, for example $m + 2 \lt Kn(n-1) \lt m+n-1$. The special multiple says that there is only one place in $I$ for multiples of two numbers less than $n$. A more general obstruction is that in $I$ the multiples of $j$ numbers less than $n$ occupy together less than $j$ spots, which looks a lot like the first obstruction as $j$ gets large.
I am still writing code for the second obstruction. For the first obstruction, it is easy to write code to determine the sequence of units in the ring of integers mod $P(n)$, and not much harder to determine mod $P(n)$ the count of possible values which avoid the first obstruction. For example, for $n=6$, one looks at a sequence of differences 6 4 2 4 2 4 6 2 and concludes that for $n=6$ and $P(6)=2*3*5=30$ one must have modulo $30$ that $m+1$ must be in $[0,5] \cup [20,25]$. This example does not show the second obstruction clearly, but if we took $n=5$, then $m=17$ mod $60$ would be ruled out by a large multiple of $20=5*4$ occurring in the middle of $I$.
Initial runs of the code modelling the first obstruction suggest that when $n$ is prime less than $ 1/(n\log n)$ of the numbers in $[0,P(n))$ avoid the obstruction and are candidates for $m$, and when $n$ is not prime the number drops from $1/(n \log n)$ to something like $1/n^2$.  I expect the second obstruction to remove about $1/n$ of these candidates, and hope to post numbers later.  I invite others to write and share code for some modification of the first and second obstructions to come up with the density of $m$ satisfying Bernardo's divisibility condition.
End Edit 2019.04.16.
Gerhard "More General Than Jacobsthal's Function" Paseman, 2019.03.29.
A: If $a_n=a \gt n,$ then there can only be one prime
among $a,a+1,\cdots, a+n-1.$ So the well studied topic of gaps between primes could provide upper bounds and might be the  major factor.
Here are the gaps between the first few primes 
$ 1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6$
$, 2, 6, 4, 2, 6, 4, 6, 8$
$4 , 2, 4, 2, 4, 14, 4,$ 
The record gaps indicated are are $14=127-113$ , and $8=97-89$ 
We are here most interested in the gap between the $k$th and $k+2$nd  primes
$3, 4, 6, 6, 6, 6, 6, 10, 8, 8, 10, 6, 6, 10, 12,$
$8, 8, 10, 6, 8, 10, 10, 14$
$ 6, 6, 6, 6, 18, 18$
The entries at least $13$ indicate the intervals that might perhaps contain $a_{12}$
The $14$ seems like a tight fit. We need to use either $84$ or $96$ (which will be the multiple of $12$) along with $85\cdots 95.$ But this fails as $88$ is the only available multiple of $8$ and also of $11.$ So in fact $a_{11}$ has to be further out. 
The $18$’s show more promise : either part of 
$110,111,112,\mathbf{113},\cdots ,126$ or part of
$114,\cdots ,\mathbf{127},128,129,130.$ The multiple of $11$ must be $110$ or $121$ and, since $110$ works, we are done.
It turns out that $a_{10}=a_{11}=a_{12}=110.$ 
It is possible to have $a_{i+1} \lt a_i.$
