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](http://oeis.org/A130091) in OEIS. 24 such numbers are impossible because of $36n-6$ and $36n+6$. 



Consecutive numbers with mutually distinct exponents in their canonical prime factorization
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 $2k-4,2k,2k+4$ for some odd $k$. Then your condition implies $k-2,k,k+2$ are powerful. In particular, $(k-2)(k+2)=k^2-4$ is powerful.


Apply the ABC conjecture to the triple $(4,k^2-4,k^2)$. 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.


