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As we all know, a natural number $n$ is prime if and only if there do not exist natural numbers $x, y$ exclusively between $1$ and $n$ such that $xy = n$.

Is there any generally recognized analogy for primes in powers? For example, a natural number $n$ in a set $S$ that is not the $x$th power of $y$ for two natural numbers $x$ and $y$ exclusively between $1$ and $n$? I recognized that the definition is less elegant than that for multiplication, since powers aren't commutative, but I was curious if this set or any similar set has been explored in mathematics.

For clarity, in my example definition, $1, 2, 3, 5, 6, 7 \in S$, but $4 = 2^2, 8 = 2^3, 9 = 3^2 \not\in S$.

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  • $\begingroup$ Any assistance in better tagging this question would be greatly appreciated. I have very little idea about what branch of mathematics this would involve (number theory?). $\endgroup$ – waiwai933 May 21 '13 at 6:25
  • $\begingroup$ The same as math.stackexchange.com/questions/387024/… $\endgroup$ – Zander May 29 '13 at 10:53
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A natural number satisfies your criterion if and only if it is not a perfect power.

Consider the function $f\colon\mathbb{R}_{>0}\to\mathbb{R}$ defined by $f(x)=\log_2(x)=\ln(x)/\ln(2)$. It satisfies $f(2)=1$, and $$f'(x)=\frac{d(\ln(x)/\ln(2))}{dx}=\frac{1}{\ln(2)}\frac{d(\ln(x))}{dx}=\frac{1}{x\ln(2)}.$$ Because $f'(x)<1$ for all $x>\frac{1}{\ln(2)}\approx 1.442$, it will never "catch up" with the function $g(x)=x$, having already started "behind" at $f(2)=1$. Thus, $f(x)<x$ for all $x\geq 2$.

Let $n>1$ be a natural number. Suppose that there exist natural numbers $a$ and $b$ with $n=a^b$, where $b>1$. Because $n>1$, we must have $a>1$, and hence $a\geq 2$. Thus, $b\leq\log_2(n)< n$.

Thus, if a natural number $n>1$ is not equal to $a^b$ for any natural numbers $a$ and $b$ exclusively between $1$ and $n$, then it is not equal to $a^b$ for any natural numbers $a$ and $b$ with $b>1$, period.

Thus, your numbers are precisely those which are not perfect powers.

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