Gap between composite numbers that aren't divisible by 2,3 or 5

I am not sure how $42$ became the answer to every question in the Universe, but it is an answer to this question. Given any composite number $N$ that is prime to $2×3×5=30$, the next larger composite number prime to $30$ must be no greater than $N+42$. The bound is proven to be a sharp one, but requires a large value of $N$ to saturate it.

By hypothesis, $N\in\{1,7,11,13,17,19,23,29\}\bmod 30$. Suppose $N\equiv 1\bmod 30$. Then we examine the next several qualifying numbers, by subtracting $1$ from each allowed residue to obtain the difference with $N$. Thus

$N+6\equiv7\bmod30$

$N+10\equiv11\bmod30$

$N+12\equiv13\bmod30$

$N+16\equiv17\bmod30$

$N+18\equiv19\bmod30$

$N+22\equiv23\bmod30$

$N+28\equiv29\bmod30$

Now watch what happens when we take residues $\bmod 7$:

$N+6\equiv N+\color{blue}{6}\bmod7$

$N+10\equiv N+\color{blue}{3}\bmod7$

$N+12\equiv N+\color{blue}{5}\bmod7$

$N+16\equiv N+\color{blue}{2}\bmod7$

$N+18\equiv N+\color{blue}{4}\bmod7$

$N+22\equiv N+\color{blue}{1}\bmod7$

$N+28\equiv N+\color{blue}{0}\bmod7$

Note that the increments shown cover all the different residues $\bmod 7$. Perforce if $N$ is one greater than a multiple of $30$, then some larger number prime to $30$ but less than or equal to $N+28$ must be composite by being a multiple of $7$. Also, since the incremental residues $1,2,3,4,5,6$ all occur earlier than $28$, that gap for $N\equiv1\bmod30$ can be achieved only if $N$ is a multiple of $7$.

What if we start with $N$ having a different residue $\bmod30$, such as $7$? If $N$ has that residue, then the increments to larger numbers prime to $30$ will be different from those above, leading to a different pattern of residues $\bmod 7$. Thus a different bound on the gap to a guaranteed multiple of $7$. Put in numbers:

$N+4\equiv11\bmod30$&$N+4\bmod 7$

$N+6\equiv13\bmod30$&$N+6\bmod 7$

$N+10\equiv17\bmod30$&$N+3\bmod 7$

$N+12\equiv19\bmod30$&$N+5\bmod 7$

$N+16\equiv23\bmod30$&$N+2\bmod 7$

$N+22\equiv29\bmod30$&$N+1\bmod 7$

$N+24\equiv1\bmod30$&$N+3\bmod 7$

$N+30\equiv7\bmod30$&$N+2\bmod 7$

$N+34\equiv11\bmod30$&$N+6\bmod 7$

$N+36\equiv13\bmod30$&$N+1\bmod 7$

$N+40\equiv17\bmod30$&$N+5\bmod 7$

$N+42\equiv19\bmod30$&$N+0\bmod 7$

In this case we cover all incremental residues $\bmod 7$, and thus assure a multiple of $7$, at $N+42$ making $42$ the maximum possible gap in this case. Here again the full gap requires $N$ to be a multiple of $7$. Given all the "extra" iterations in this case we might suppose that a different divisor, such as $11$, might produce a lower limit, but we find that covering all eleven incremental residues $\bmod 11$ also requires allowing a maximum gap of $42$.

We do a similar analysis with $N\equiv11\bmod30,N\equiv13\bmod30$, etc, and in all cases a multiple of $7$ is forced on or before $N+42$. A gap of $42$ requires $N$ to be both $\in\{7,11\}\bmod 30$ and a multiple of $7$. So if every question in the Universe is ultimately governed by gaps between composite numbers having no factors of $2,3,$ or $5$, then $42$ is indeed a universal answer.

Probalistic considerations would seem to suggest that the largest gaps would occur with smaller numbers, but gradually the maximum observed gap increases beyond the value of $28$ in the OP's original list. The smallest case that exceeds a gap of $28$ is $N=1273$, which gives a gap of $36$ up to $N+36=1309$. The numbers $1273$ and $1309$ collectively contain the prime factors $7,11,17,19$ leaving relatively few factors that might have broken up an intermediate string of four-digit primes. In the comments a gap of $38$ is reported between $113141$ and $113179$ by Joffan. Finally, Empy2 in another answer gives a gap of $42$, saturating the bound, with $N=1418575498571$ along with larger values.

These numbers are prime

1418575498609, 1418575498607, 1418575498603, 1418575498601, 1418575498597, 1418575498591, 1418575498589, 1418575498583, 1418575498579, 1418575498577, 1418575498573

which gives a gap of 42 between 1418575498571 and 1418575498613

They are 614101948*2310 - 1260 -[11,13,17,19,23,29,31,37,41,43,47]

Edit Suppose we look at numbers near $N$, which might be $10^{13}$.

The chance a number $x$ is prime is $1/\ln(N)$. Prime factors up to 11 are ruled out by the 2310 formula, so the chance of being prime is increased to
$$P=\frac{\frac21\frac32\frac54\frac76\frac{11}{10}}{\ln N}$$ The chance all eleven numbers are prime is $P^{11}$.
The 2310 formula gives two opportunities every 2310, namely $2310k+1260+[11,...]$ and $2310k-1260-[11,...]$
So an estimate for the number of solutions within $N/2$ of $N$, so between $N/2$ and $3N/2$, is
$$\frac{P^{11}N}{1155}$$ That suggests about 4 near $N=10^{12}$, but I have only found one.