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The fact that there are arbitrarily long sequences of consecutive numbers without prime number is well known, the proof is easy and goes like this:

let $n\ge 2$, then the number $n!+k$ is greater then $k$ and divisible by $k$ for $k=2,\dots,n$, so it is composite, so we have at least $n-1$ consecutive composite numbers.

I think that there are arbitrarily long sequences of consecutive numbers without prime numbers and without powers of primes, probably also without any perfect powers.

Examples:

$17,18,19,20,21,22,23,24$ -- 8 consecutive numbers without any perfect power,

$37,38,\dots,48$ -- 12 consecutive numbers without any perfect power,

$50,51,\dots,63$ -- 13 consecutive numbers without any perfect power,

$65,66,\dots,80$ -- 16 consecutive numbers without any perfect power,

$901,902,\dots,960$ -- 60 consecutive numbers without any perfect power,

$1601,1602,\dots,1680$ -- 80 consecutive numbers without any perfect power.

But I'm not able to prove my conjecture. Is it really true? Any proposals will be appreciated.

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  • $\begingroup$ For perfect powers of a specific exponent (any exponent) it is easy. Simply consider the gap between $n^k$ and $(n+1)^k$ as $n$ grows larger. $\endgroup$ – barak manos Feb 14 '16 at 13:06
  • $\begingroup$ According to oeis.org/A001597 we have "a(n) is asymptotic to $n^2$" (a(n) being the n-th perfect power). That should be enough to show that there are arbitrarily long sequences of consecutive numbers that are not perfect powers. $\endgroup$ – Michael Stocker Feb 14 '16 at 15:41
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Take $n=2^{2k}$, so that $n$ is a square. Moreover it is easy to see that for all the perfect powers $1<x^t\le n$, the exponent $t$ cannot exceed $2k$.

How many are the powers (2nd, 3rd,... $2k$-th) smaller or equal to $n$ ?

The squares are $2^k$. The other powers, with exponents from 3 to $2k$ are clearly much less, but since we are interested in a bound, let's just assume there are at most $2^k$ of each of them.

All in all, up to $n=2^{2k}$ there are thus less than $2k \cdot 2^k$ perfect powers. So, on the average there is less than a perfect power every $2^{2k}/(2k \cdot 2^k) = 2^k/(2k)$ numbers and thus there must be at least a gap larger than that.

Since we can make grow $2^k/(2k)$ as large as we want, gaps are unbounded.

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  • $\begingroup$ "Moreover it is easy to see that (apart 1) there are no perfect powers with exponents greater than 2k smaller than n" - shouldn't it be $2^k$ instead of $2k$ since $(2^k)^2 = 2^{2k}$? $\endgroup$ – Michael Stocker Feb 15 '16 at 12:00
  • $\begingroup$ What I wanted to say is that, if $n=2^{2k}$ , then for any integer power $1<x^t \le n$, we know that the $t \le 2k$. I've edited the answer to make it more clear (I hope). $\endgroup$ – Giovanni Resta Feb 15 '16 at 13:22
  • $\begingroup$ Ah ok, $2k$ is the upper bound of the exponents per base, should have realized that. Thanks for the clarification. $\endgroup$ – Michael Stocker Feb 15 '16 at 14:43

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