<s>[INST] 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  $a_n$?
 [/INST] 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 $a_{i+1} \lt a_i.$
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