I am looking at the length of the longest sequences of adjacent integers that are not coprime to $N$ for very large $N$.
Let $F_N$ be the set of integers less than $N$ which are not coprime with $N$: $$ F_N = \{x : x \in \mathbb{N}, x \lt N , \gcd(x,N)=1\} $$ Let $g(N)$ be the length of the longest sequence of adjacent integers in $F_N$.
IE something like: $$ g(N) = \max (i\in \mathbb{N}) : \exists a\in F_n : \forall b : a \le b \lt a+i, b\in F_N $$ What is the least upper bound on $g(N)$, i.e. the maximum length of a sequence of adjacent numbers in $F_N$?
Some direct calculation suggests that $g(N)$ steps up when $N$ is primorial, so if $P_i$ is the primorial sequence, then: $$ \forall N \lt P_i, g(N) \lt g(P_i) $$ I have no proof of this, but it suggests $g(P_i)$ may be useful in finding a least upper bound.
There are $N - \phi (N)$ non-coprimes of $N$ less than $N$ (where $\phi$ represents Euler's totient function, labelled 'phi' in the table below). The assumption that these are evenly spread giving $$g(N) \approx \frac N {\phi (N)}$$ is clearly wrong.
Better estimates (for the primorial values of $N$ only) seem to be: $$g(N) \approx 2 \log(N)$$ and $g(N) \approx$ twice the highest prime factor of $N$. However, these are untested outside the data below, and again I have no proof.
Below is a table of $g(N)$ and $\phi(N)$ of the primorials:
N (Primorial) Highest prime phi (N) g(N) 2 2 0 1 6 3 1 3 30 5 7 5 210 7 47 9 2310 11 479 13 30030 13 5759 21 510510 17 92159 25 9699690 19 1658879 33 300690390 21 49766399 39
Edit:
As Gerry Myerson kindly points out in the comments, $g(N)$ is similar to Jacobsthal's function, $j(N)$ series here where $j(N) = g(N)+1$.
Several online sources point to a proof by H. Iwaniec in 'On the problem of Jacobsthal, Demonstratio Math. 11, 225–231, (1978)' (original not available online) that: $$j(N) \ll \log \log (N)$$ Whilst this might be true in asymptotic terms, it's trivially not true for the primorials above, and the sequence appears closer to $\log (N)$ than $\log \log(N)$. So the question remains open.