What is an upper bound for number of prime powers less than $n$?

Question

I mean the numbers in the form $a^b$ in which $b \ge 2$ and $a$ is a prime number.

I have found that $\frac {\log n} {\log 2} + \frac {\log n} {\log 3} + \frac {\log n} {\log 5} + \dots + \frac {\log n} {\log i}$ is an upper bound ($i$ is the greatest prime less than or equal to $n$).

And I guess it is less than $\pi \left( \frac {\log n} {\log 2} \right)$, so $\pi \left( \frac {\log n} {\log 2} \right)$ is an upper bound.

$n$ is an upper bound. (Did you want a better one?) And presumably you want $a$ to be prime. — TonyK, Jan 19 '16 at 16:27
@ttt, $n$ is also an upper bound, although not as sharp as your $\pi(n)$. In my acerbic way, I was implying that if you are not going to put any effort into your question, then I am not going to put any effort into my answer. — TonyK, Jan 19 '16 at 16:38

score 1 · Answer 1 · answered Jan 19 '16 at 16:57

1

We can count the powers of $2,3,5,7,11,13,17$ independently and then take care of the other powers using:

$\pi(x)<\lfloor1.3\frac{x}{\log(x)}\rfloor$, for $n\geq 17$.

Suppose $P$ is the number of prime powers less than or equal to $n$, in other words $\sum\limits_{i=2}^{2^i\leq n}\pi(x^{1/i})$ , Then we have the following inequality for $n>7$:

$P\leq\sum\limits_{i=1}^{i\leq7}\lfloor\log_{p_i}(n)-1\rfloor+\sum\limits_{i=2}^{19^i\leq n}\lfloor\frac{1.3n^{1/i}}{i\log(n)}-7\rfloor$

answered Jan 19 '16 at 16:57

Asinomás

105,651

I need a bound in closed form – ttt Jan 19 '16 at 17:14
Why? We can easily calculate this bound in $\mathcal O(\log n)$ time. – Asinomás Jan 19 '16 at 17:16

score 1 · Accepted Answer · answered Jan 19 '16 at 17:35

The expression you give is a correct computation if you enclose every term on the right in floor brackets. You can distribute out the $\log n$ numerator, so you are really asking for $\log n\sum_{p \text{ prime}}^{p \le n}\frac 1{\log p}$ We can compute the first few terms in the sum explicitly. I arbitrarily did primes up through $19$.
$$\log n\sum_{p \text{ prime}}^{p \le n}\frac 1{\log p}\approx \log n(1.5666 +\sum_{k=10}^{p_k \le n}\frac 1{\log p_k})$$ Then using the prime number theorem, the last term becomes $$\sum_{k=10}^{p_k \le n}\frac 1{\log p_k}\approx \sum_{k=10}^{p_k \le n}\frac 1{\log (k \log k)}=\sum_{k=10}^{p_k \le n}\frac 1{\log k + \log \log k}$$ Now to get an upper bound we can ignore the $\log \log k$ term in the denominator and change the sum to an integral $$\sum_{k=10}^{p_k \le n}\frac 1{\log k + \log \log k}\gt \sum_{k=10}^{p_k \le n}\frac 1{\log k}\approx \operatorname{li}(\pi(n))-\operatorname{li}(10)$$ Putting it all together, your bound is $$\log n\left(1.5666+\operatorname{li}(\pi(n))-\operatorname{li}(10)\right)$$

What is an upper bound for number of prime powers less than $n$?

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