Counting powerful integers. Lower bounds

Question

Remark:   The upper bounds are perhaps still more interesting; I may address them in another post.

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PROBLEM:   Find simple (numerically efficient) lower bounds for the number of powerful integers (natural numbers) not exceeding an arbitrary positive real number $\ x;\ $ (let them be as precise as possible).

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Of several equivalent definitions of powerful numbers, I like the following one, and it seems helpful in this note:

A powerful integer $\ n\in\mathbb N\ $ is a product $\ n=a^2\cdot d\ $ where $\ a\in\mathbb N,\ $ and divisor $\ d|a\ $ is a square free natural number too.

This way $\ a=A(n)\ $ and $\ d=D(n)\ $ is uniquely determined by $\ n\in\text{POW}\ $ -- the set of all powerful numbers.

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We get two subsets of POW, namely squares and cubes, while the intersection of the two consists of the sixth powers. Thus, let

$$ \text{POW}(x)\ :=\ \{\,n\in\text{POW}:\ n\le x\,\} $$

We get our first lower bound, as weak as it is:

$$ \left\lfloor x^\frac12\right\rfloor+\left\lfloor x^\frac13\right\rfloor - \left\lfloor x^\frac16\right\rfloor\ \le\,\ |\text{POW}(x)| $$

-- just for starters.

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PS. Another equivalent definition of powerful integers:

Definition: natural number $\ n\in\mathbb N\ $ is powerful $\ \Leftarrow:\Rightarrow$

$$ r^2(n)\, |\, n $$

where $\ r(n)\ $ is the radix of $\ n,\ $ i.e. the product of all prime divisors of $\ n.$

Answer 1

Golomb (1970) proved that for all $x\geq 1$ we have $$\frac{\zeta(3/2)}{\zeta(3)}x^{1/2}-3x^{1/3}\ \leq\ \left|\mathrm{POW}(x)\right|\ \leq\ \frac{\zeta(3/2)}{\zeta(3)}x^{1/2}.$$ Note that $$\frac{\zeta(3/2)}{\zeta(3)}=2.1732543125195541382370898404\dots$$

Answer 2

It has occurred to me that I can get a much more decent lower bound but for a somewhat different function $\ \text{PWR}(x);\ $ it is still closely related to $\ \text{POW}(x).\ $ The new definition takes an advantage of function $\ A(n)\ $ as in the OP post, where $$ \forall_{n\in\text{POW}}\qquad n\ =\ A^2(n)\cdot D(n)\qquad $$ with $\ D(n)\ $ being a square-free divisor of $\ A(n),\ $ i.e. $\ D(n)|A(n).$ Thus, let's define:

$$ \text{PWR}(x)\ :=\ \{\,n\in \text{PWR}:\ A^2(n)\le x\,\} $$

Obviously

$$ |\text{POW}(x)|\ \le\ |\text{PWR}(x)|\ \le \ \left|\text{POW}\left(x^\frac32\right)\right| $$

Now we obtain:

$$ 4\cdot\left\lfloor x^\frac12\right\rfloor\ -\ 2\cdot \Pi\left(\left\lfloor x^\frac12\right\rfloor\right) \,\ \le\ \,\ |\text{PWR}(x)| $$

where $\ \Pi(t)\ $ is the very well-behaved function that counts prime powers that do not exceed positive $\ t\in\mathbb R.$

The left-hand side counts only powerful integers $\ n\ $ such that $\ D(n)\ $ is a product of at the most $\ 2\ $ primes $ \ p\ne q\ $ that both divide $\ A(n);\ $ however, for the sake of this simple counting, we select just a single pair of such two primes (when they exist) -- that's how you get $\ 4\ $ and $\ 2\ $ in the above inequality; indeed, for each prime power $\ A(n):=p^k\ $ we get only 2 powerful extensions $\ n=p^{2\cdot k}\ $, and $\ n=p^{2\cdot k+1}.\ $ However, when $\ A(n)\ $ is divisible by at least two different primes, say $\ p\ $ and $\ q\ $, then we get at least $\ 2^2\ $ different extensions: $$ A^2(n)\quad\text{and}\quad A^2(n)\cdot p\quad\text{and}\quad A^2(n)\cdot q\quad\text{and}\quad A^2(n)\cdot p\cdot q $$ Of course, when the number of prime divisors of $\ A(n)=r\ $ then we get $\ 2^r\ $ extensions of $\ A(n).\ $ Thus, each time when $\ r>2\ $ we get more extensions than just $\ 2^2=4\ $ but it's hard (for me) to take a full advantage of this fact.