# arbitrarily long sequences without perfect powers

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.

• 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. Feb 14 '16 at 13:06
• 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. Feb 14 '16 at 15:41

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.

• "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 $(2^k)^2 = 2^{2k}$? Feb 15 '16 at 12:00
• 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). Feb 15 '16 at 13:22
• Ah ok, $2k$ is the upper bound of the exponents per base, should have realized that. Thanks for the clarification. Feb 15 '16 at 14:43