TITLE: Consecutive numbers with mutually distinct exponents in their canonical prime factorization
QUESTION [10 upvotes]: Is it possible to find 23 consecutive positive integers each of which has mutually distinct exponents in its canonical prime factorization? Such numbers are sequence A130091 in OEIS. 24 such numbers are impossible because of $36n-6$ and $36n+6$.

REPLY [9 votes]: We prove that there are only finitely many such intervals.

Suppose $[36n + 7, 36n + 29]$ is one such interval.
Now, $36n + 10$ and $36n + 15$ cannot both be divisible by $5$ (since one must occur with multiplicity $1$), hence $n \neq 0 \mod{5}$.

We can say the same about $36n + 21$ and $36n + 26$ (so $n \neq 4 \mod{5}$).
And about $36n + 14$ and $36n + 24$ (so $n \neq 1 \mod{5}$).
And indeed about $36n + 12$ and $36n + 22$ (so $n \neq 3 \mod{5}$).

This implies that $n \equiv 2 \mod{5}$, so $36n + 18$ is divisible by $5$ (and therefore by $25$). So our set must actually be of the form:
$$[900m + 439, 900m + 461]$$

Irrelevant aside: We can do a similar analysis modulo $7$, enabling us to deduce that precisely one of $900m + 445$, $900m + 450$ and $900m + 455$ is divisible by $49$. This doesn't seem to help as much, though.
Indeed, no generalisation of this argument will work, because only the primes $p \leq 23$ are pertinent (we can just shift the interval with the help of the Chinese Remainder Theorem to avoid large primes completely), and it's possible to find a $23$-element interval in which the multiplicities of each of the primes $2, 3, 5, 7, 11, 13, 17, 19, 23$ are distinct for each element (left as an exercise to the reader).

Let us return to the task of proving that there are only finitely many such intervals. For each interval, either:

$900m + 440$ and $900m + 456$ are both congruent to $4 \mod{8}$.
$900m + 444$ and $900m + 460$ are both congruent to $4 \mod{8}$.

In which case we take those two numbers and divide them by $4$ to obtain odd coprime integers $b, b+4$. One of them is divisible by $3$ (but not $3^2$) and the other is divisible by $5$ (but not $5^2$). Any other primes dividing them must do so with multiplicity $\geq 3$, because $2$ divided each of the original numbers with multiplicity $2$.
Whence $(4, b, b+4)$ has a radical less than $k b^{2/3}$ for some universal constant $k$ which I can't be bothered to calculate.
Since the $abc$ conjecture is true, it follows there are only finitely many intervals with your property.