# Homework Help: Number of squareful integers less than x

1. Aug 29, 2015

### Boorglar

I'm doing the exercises from Introduction to Analytic Number Theory by A.J. Hildebrand (online pdf lecture notes) from Chapter 2: Arithmetic Functions II - Asymptotic Estimates, and some of them leave me stumped...

1. The problem statement, all variables and given/known data

Problem 2.14:
Obtain an asymptotic estimate with error term $O(x^{\frac{1}{3}})$ for the number of squarefull integers $≤ x$, i.e., for the quantity $S(x) = \left\{n ≤ x : p | n => p^{2} | n\right\}$.

2. Relevant equations

The text describes a method known as the "convolution method" to evaluate sums of arithmetic functions asymptotically. In our case, the arithmetic function would be the characteristic function of the squarefull integers, a(n) = 1 if n is squarefull and 0 otherwise.

If $a = f * g$ (Dirichlet convolution), then $$\sum_{n ≤ x} a(n) = \sum_{d ≤ x} g(d) F\left(\frac{x}{d}\right)$$ where $F(x)$ is the summatory function of $f$.

3. The attempt at a solution
I am not even sure how to start. I tried expressing $a(n) = 1 * (\mu * a)$ thus using $f = 1$ and $g = \mu * a$. But I can't find a way to estimate the sum $$\sum_{d ≤ x} g(d) \left\lfloor{\frac{x}{d}}\right\rfloor$$ Here $\mu$ means the Mobius function.

2. Aug 31, 2015

### Geofleur

Maybe it would be a good idea to investigate S(x) for different values of x. You could plot the number of squarefull integers for x up to some cutoff, for example. It couldn't hurt! Maybe it will reveal some pattern that will give you an idea of how to solve the problem.

3. Sep 16, 2015

### Boorglar

So I finally solved the problem, in a totally different (and way simpler) method -_-

I used an earlier result from another problem, saying that squarefull integers can be written uniquely in the form $a^2b^3$.
So to count the squarefull integers $\leq x$ you can just do $$\sum_{a^2b^3 \leq x} 1 = \sum_{b \leq x^{1/3}} \left\lfloor{\frac{\sqrt x}{b^{3/2}}}\right\rfloor$$ which eventually simplifies to $$\zeta(3/2)\sqrt x + O(x^{1/3})$$

UPDATE:
Ah, turns out I was wrong... The representation $a^2b^3$ is not unique. For example, $p^8 = (p^4)^2 * 1^3 = p^2 * (p^2)^3$. I think we need the additional condition that $b$ is squarefree...
So the sum is actually $$\sum_{a^2b^3 \leq x} µ^2(b) = \sum_{b \leq x^{1/3}}µ^2(b) \sum_{a \leq \frac{x^{1/2}}{b^{3/2}}}1 = \sqrt x \sum_{b \leq x^{1/3}}\frac{µ^2(b)}{b^{3/2}} + O(x^{1/3})$$

which still becomes something of the form $C \sqrt x + O(x^{1/3})$ but $C = \sum_{b = 1}^{\infty} \frac{µ^2(b)}{b^{3/2}}$ which I can't really simplify...

After playing with this expression and using Dirichlet series of $µ(n)$ and $(µ^2 * µ)(n)$, I found that $C = \frac{\zeta(3/2)}{\zeta(3)}$.

Last edited: Sep 16, 2015