Consecutive numbers with mutually distinct exponents in their canonical prime factorization

Question

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 $36n+6$.

Answer 1

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.

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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.

Answer 2

We prove that there are only finitely many such intervals.

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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:

$$[900m + 439, 900m + 461]$$

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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).

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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.