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 $36n6$ and $36n+6$.

3$\begingroup$ how long is the longest sequence you know of already? $\endgroup$ – Yaakov Baruch Mar 30 '15 at 16:23

2$\begingroup$ You might also mention $1,2,3,4,5$ and $241, 242, 243, 244, 245$. Those and the ones you listed are all the $5$'s less than $10^6$, and there are no sequences of length $>5$ in this range. $\endgroup$ – Robert Israel Mar 30 '15 at 17:11

4$\begingroup$ The problem is that most numbers have some prime factor with exponent $1$, while in an interval $\{36n+7,\dots,36n+29\}$ many numbers already have prime factors $2$, $3$ or $5$ with exponent $1$, and therefore by the condition in the question must have all larger factors with higher exponents. This seems like a pretty unlikely constellation. $\endgroup$ – Stefan Kohl♦ Mar 30 '15 at 17:34

4$\begingroup$ The numbers of the form $4k+2$ in such an interval of length 23 must be two times an odd powerful number. So by dividing by two we obtain at least 10 consecutive powerful odd numbers! $\endgroup$ – Mostafa Mar 30 '15 at 17:43

4$\begingroup$ ...where by $10$ you mean $5$. $\endgroup$ – Adam P. Goucher Mar 30 '15 at 17:47
The answer to this question is almost certainly no.
It is well known that the ABC Conjecture implies that there are only finitely many triples $(n,n+1,n+2)$ which are all powerful. A similar argument should work for tuples of the form $(2n+1,2n+3,2n+5)$. But, as pointed out by Mostafa and Adam (in the comments above), a sequence of 23 consecutive integers with the property you want will give rise to too many odd powerful numbers near each other.
If you want to see the actual ABC computation: consider any list of 12 consecutive integers $n,n+1,n+2,...,n+11$. Then three of these numbers are exactly divisible by $2$, say they are $2k4,2k,2k+4$ for some odd $k$. Then your condition implies $k2,k,k+2$ are powerful. In particular, $(k2)(k+2)=k^24$ is powerful.
Apply the ABC conjecture to the triple $(4,k^24,k^2)$. The radical base is at most $2k^{3/2}$, which doesn't beat $c=k^2$. So, under the conjecture, there can be only finitely many sequences of 12 consecutive integers of the type you want.
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.

8$\begingroup$ You write "Since the abc conjecture is true ...": have you checked Mochizuki's proof?  As far as I know, its correctness is not generally accepted so far (if you have more recent information, you could update the Wikipedia article). $\endgroup$ – Stefan Kohl♦ Mar 30 '15 at 20:04

$\begingroup$ The following paper settles the matter: link.springer.com/article/10.1007/s120440160326z $\endgroup$ – Bernardo Recamán Santos Feb 4 '17 at 13:22