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Consider a set of integers $Q$ such that the set of all positive integers $\mathbb{Z}$ is equivalent to the span of ever possible power tower

$$a_1^{a_2^{\ldots a_N}}$$ involving $a_i \in Q$.

In simpler terms. Take the integers, remove all square numbers, cube numbers, fourth powers, fifth powers, etc... And this remaining set is $Q$.

What is the density of $Q$ compared to positive $\mathbb{Z}$? Does it obey a theorem similar to the prime number theorem for primes? Are there infinity many numbers $x$, in $Q$ such that both $x$ and $2x$ are members of $Q$? Is there a formula for the elements of $Q$?

This is basically analogous to prime numbers except now it deals with exponents as opposed to multiplication.

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  • $\begingroup$ I guess I'm just curious if any research has been done in this area $\endgroup$
    – Sidharth Ghoshal
    Commented May 9, 2013 at 21:42
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    $\begingroup$ The set of squares, cubes, and higher powers is very thin. Asymptotically it's no bigger than the set of just squares which is $\sqrt{n}$ in size. But I'm not exactly sure what your first construction means. $\endgroup$
    – Erick Wong
    Commented May 9, 2013 at 21:48
  • $\begingroup$ Suppose f(x) and g(x) are functions such that lim x --> log f(x) (g(x)) = 1 what is the word to describe their relationship? Basically the logarithm with base f(x) when applied to g(x) approaches 1 as x approaches infinity $\endgroup$
    – Sidharth Ghoshal
    Commented May 9, 2013 at 21:52
  • $\begingroup$ I would call this "$\log f$ is asymptotic to $\log g$", which is weaker in most cases than $f \sim g$. But how is that related at all to your question? $\endgroup$
    – Erick Wong
    Commented May 9, 2013 at 22:10
  • $\begingroup$ Well here is the deal... when we are sieving primes via say-eratosthenes (or even more complex sieves) we find that 1/2, 1/3, 1/5... 1/pn of the remaining unsieved numbers are sieved during each step. So in a way: 1/2 + 1/2*1/3 + 1/2*2/3*1/5 + 1/2*2/3*4/5*1/7... 1/2*2/3*4/5*6/7...(1)/pn = C(z) is the density of non prime numbers to all numbers. Therefore: z(1 - C(z)) ~ prime counting function... $\endgroup$
    – Sidharth Ghoshal
    Commented May 11, 2013 at 15:03

2 Answers 2

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The density of exponential primes (i.e. non-power numbers) is 1. In fact, there are so few perfect powers that the sum of the reciprocals of perfect powers converges: $$ \frac{1}{2^2} + \frac1{2^3} + \frac1{3^2} + \frac1{2^4} + \frac1{5^2} + ... \approx 0.87446... $$ For more details, see Wikipedia's article.

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Most numbers are not perfect powers (e.g, $0,1$ and $8,9$ are the only examples of two consecutive perfect powers).

If $x$ is has two odd prime divisors and their exponents are relatively prime, then neither $x$ nor $2x$ (in fact any multiple by a factor not a multiple of either of the two given primes) is a perfect power.

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