Smallest chain of consecutive integers not all coprime Let $t$ be a positive integer. What is the smallest $t$ for which we can find an integer $a$ such that each element of the set $\{a+1,a+2,\dots ,a+t\}$ is not coprime with all other elements of the set?
I think the answer is $t=17$, with $a=2183$, but I'm not sure. This is somewhat related to the discussion here.
 A: Yes, this is true.  I tried all $t \le 17$ and all $0 \le a \le 30030 = 2\times 3\times 5 \times 7 \times 11 \times 13$.
A: I am expanding on Robert Israel's answer.  I do not think there is a nice way to show that $t=17$ is the minimum without computer search.  Note that it suffices to check whether the required property holds for $a$ within $\left\{0,1,2,\ldots,p_n\#-1\right\}$, where $p_k$ is the $k$-th prime number, $n$ is the largest integer such that $p_n<t$, and $p_n\#$ is the primorial $$p_n\#=p_1\cdot p_2\cdot\ldots\cdot p_n\,.$$  That is because, if $\left\{a+1,a+2,\ldots,a+t\right\}$ satisfies the requirement, then the greatest common divisor of $a+i$ and $a+j$, for $i\neq j$ is either $1$ or divisible by some prime $p_j$ with $j\leq n$.  Thus, if $a$ meets the condition, then $a+p_n\#$ also does.
A: The OP has shown that $t\leqslant 17$. It remains to show that, of any 16 consecutive integers, at least one is coprime to the others.
Suppose the theorem is false; consider a counterexample. $S$.
Of $S$'s 16 consecutive integers, eight are odd. Let the lowest be $k$. Then the highest is $k+14$.
If $7\nmid k$, $7$ divides at most one odd member of $S$, so at least seven are coprime to $14$. Of these, $3$ divides at most three, $5$ divides at most two, $11$ divides at most one, and $13$ divides at most one. These would have to be all eight of $S$'s odd elements once each, for $S$ to be a counterexample.
There are two possible arrangements of the multiples of $3$ and $5$: either $3\mid k$ and $5\mid k+14$, or vice versa. I consider the former; the latter is handled similarly. Then $5\mid k+4$, $3\mid k+6$ and $3\mid k+12$. The multiple of $13$ is coprime to the rest of $S$ unless $13\mid k+2$ and $k+15\in S$. The multiple of $11$ is coprime to the rest of $S$ unless $11\mid k+10$ and $k-1\in S$. But now $S$ has 17 integers $k-1,\dots, k+15$, contradiction.
If $7\mid k$, then $7\mid k+14$. Consider the six odd numbers $k+2,\dots,k+12$ in between.
Case 1. If $5\mid k+2$ and $3\mid k+4$ then $3\mid k+10$ and $5\mid k+12$, so this leaves only $k+6$ and $k+8$. Neither is divisible by any prime $p<11$, so for any other number $x$ sharing a prime factor with either, either $x\leqslant k+8-11=k-3$ or $x\geqslant k+6+11=k+17$; in each case $x\notin S$. Thus in this case both $k+6$ and $k+8$ are coprime to the other integers in $S$.
Case 2. Otherwise, of the six odd numbers $k+2,\dots,k+12$, at most three are divisible by $3$ or $5$, so at least three are not divisible by any prime $p<11$. $11$ divides at most one, $13$ divides at most one, leaving at least one not divisible by any prime $p<17$ and thus coprime to the other integers in $S$.
