Squarefull Numbers

A positive integer is squarefull if each prime in the factorization occurs at least twice.

We want to count the number of squarefull integers less than $x$ for a large $x$ .

$\displaystyle Q_2(x)=\frac{\zeta(3 / 2)}{\zeta(3)} x^{\frac{1}{2}}+\frac{\zeta(2 / 3)}{\zeta(2)} x^{\frac{1}{3}}+o\left(x^{\frac{1}{6}}\right)$

In general, numbers with each prime occurring at least $h$ times are called powerfull numbers.

The indicator functions of powerfull numbers are multiplicative functions.

Erdos Szekeres first obtained the estimates

$\displaystyle Q_{h}(x)=x^{1 / h} \prod_p\left(1+\sum_{m=h+1}^{2 h-1} p^{-m / h}\right)+O\left(x^{1 /(h+1)}\right)$

A trivial lower bound is ${x^{1/h}}$ coming from the perfect ${h}$ powers.

Every squarefull number can be uniquely written as ${n=a^2b^3}$ , where ${b}$ is squarefree. If a prime divides ${n}$ with odd exponent, it divides at least ${3}$ times, so add that prime to the part ${b}$ .

So we have

$\displaystyle Q_2(x) =\sum_{a^2b^3\le x} \mu^2(b)= \sum_{b=1}^{\infty} \mu^{2}(b)\left[\left(\frac{x}{b^{3}}\right)^{\frac{1}{2}}\right] = \sum_{b=1}^{\infty} \mu^{2}(b)\left(\frac{x}{b^{3}}\right)^{\frac{1}{2}} + O(x^{1/3})$

$\displaystyle \implies Q_2(x) = \frac{\zeta(3 / 2)}{\zeta(3)} x^{\frac{1}{2}} + O(x^{1/3})$

Because

$\displaystyle \sum_{b=1}^{\infty} \frac{\mu^2(b)}{b^s} = \frac{\zeta(s)}{\zeta(2s)}.$

The indicator function ${F_h(n)}$ of these numbers are multiplicative functions and hence the Dirichlet series equals

$\displaystyle \sum_{n=1}^{\infty} \frac{F_h(n)}{n^s} = \prod_p \left(1 +\sum_{k=h}^{\infty} \frac{1}{p^{ks}} \right).$

This helps us to see for instance that the sum of reciprocals of powerfull numbers is

$\displaystyle \sum_{n=1}^{\infty}\frac{F_h(n)}{n} = \prod_p \left(1 +\sum_{k=h}^{\infty} \frac{1}{p^{k}} \right). = \prod_p \left(1 +\frac{1}{p^{h-1}(p-1)} \right) < \infty$

For ${h=2}$ , we have by the factorization

$\displaystyle \sum_{n=1}^{\infty} \frac{F_2(n)}{n^s} =\sum_{a=1}^{\infty}\sum_{b=1}^{\infty} \frac{\mu^2(b)}{(a^2b^3)^s} = \sum_{a=1}^{\infty} \frac{1}{a^{2s}}\sum_{b=1}^{\infty} \frac{\mu^2(b)}{b^{3s}} =\zeta(2s)\frac{\zeta(3s)}{\zeta(6s)}.$

More generally we will have

$\displaystyle \sum_{n=1}^{\infty} \frac{F_2(n)}{n^s} =\zeta(hs)\zeta( (h+1)s) \cdots \zeta((2h-1)s) \frac{1}{\zeta((2h+2)s)} G(s)$

where ${G(s)}$ is Dirichlet absolutely convergent in ${Re(s)> \frac{1}{2h+3}}$

Thus we see that the partial sum of ${F_2(n)}$ (by Perron’s formula/shifting contour till ${Re(s)>1/6+\epsilon}$ has to be

$\displaystyle \sim \frac{\zeta(3 / 2)}{\zeta(3)} x^{\frac{1}{2}} + \frac{\zeta(2 / 3)}{\zeta(2)} x^{\frac{1}{3}} +o(x^{1/6})$

Elementary arguments:

We already saw an argument which gave us the asymptotic with error term ${x^{1/3}}$ , we wrote sum as a convolution of squares ${a^2}$ and cubes ${b^3}$ with ${b}$ square-free. We will use the expansion in terms of the Dirichlet series to find a better way to write the convolution.

We think of ${\zeta(2s)\frac{\zeta(3s)}{\zeta(6s)}}$ as the product of ${\zeta(2s)\zeta(3s)}$ and ${\frac{1}{\zeta(6s)}}$ and hence as Dirichlet series of the convolution of the indicator function of ${a^2b^3}$ and the function ${ \delta(n=m^6) \mu(m)}$

$\displaystyle \zeta(2s)\zeta(3s) =\sum_{a, b} \frac{1}{(a^2b^3)^s}, \quad \frac{1}{\zeta(6s)} =\sum_{m=1}^{\infty} \frac{\mu(n)}{m^{6s}}$

$\displaystyle Q_2(x) = \sum_{m^6 \le x} \mu(m) S(x/m),$ where

$\displaystyle S(x) =\sum_{a^2b^3 \le x} 1 .$

So if we can get an estimate

$\displaystyle S(x) =\zeta(3 / 2) x^{1 / 2}+\zeta(2 / 3) x^{1 / 3}+O\left(x^{1 /6}\right)$ plugging this in the above equation for ${Q_2(x)}$ in terms of ${S(x)}$ , we get an estimate for ${Q_2(x)}$ with error ${O(x^{1/6})}$ .

The error term in ${Q_2(x)}$ can be improved to ${o(x^{1/6}}$ using cancellation in the mobius sums (prime number theorem). Improvement in the error terms of ${S(x)}$ is possible, but it won’t improve the result for ${Q_2(x)}$ . In fact, any estimate better than ${x^{1/6}}$ will mean that ${\zeta(s)}$ doesn’t vanish on ${Re(s)> 1-\delta}$ for some small ${\delta.}$ (Quasi-RH)

But yes, the problem of getting better error terms in ${S(x)}$ is a problem similar to Dirichlet divisor problem and tools like lattice point counting near curves, exponential pairs are used to get better estimates.

${Q_h(x)}$ can be similarly related to the sums

$\displaystyle S_h(x) =\sum_{a_1^ha_2^{h+1}a_3^{h+2} \cdot a_{h}^{2h-1} \le x} 1.$

to get

$\displaystyle Q_h(x) \sim \gamma_{0 h} x^{1 / h}+\gamma_{1 h} x^{1 /(h+1)}+\cdots+\gamma_{r h} x^{1 /(h+r)}$

The constants here will be in term of zeta and the Dirichlet series ${G(s)}$ that occur in the factorization of ${F_h(s)}$ .

Short Intervals:

We can ask for the number of squarefull (powerfull) integers in a short intervals ${[x, x+y]}$

If ${h=2}$ , if we need ${y \gg x^{1/2}}$ because the difference of the main terms is ${\frac{y}{\sqrt x}}$ , so we assume ${y=x^{1/2}y_1.}$

If we subtract the estimates from ${x}$ and ${x+y}$ we get

$\displaystyle Q_2(x+y)-Q_2(x) \sim \frac{1}{2} \cdot \frac{\zeta\left(\frac{3}{2}\right)}{\zeta(3)} y_1$ provided ${y}$ is larger enough, that is we need ${y}$ has to be bigger than the error terms ${x^{1/6}.}$

But we start directly from the initial expression for ${Q(x)}$ as convolution of ${a^2b^3}$ and ${\delta(n=m^6)\mu(m)}$ to get the estimates for ${y_1 =y/x^{1/2}}$ much shorter than ${x^{1/6}}$

$\displaystyle Q(x) =\sum_{a^2b^3m^6 \le x}\mu(m)$

and subtract to get

$\displaystyle Q_2(x+y)-Q_2(x) =\sum_{x<a^2b^3\le x+y} \mu^2(b)$

Now we relate the sum to the divisor sums $S(x) =\sum_{a^2b^3\le x} 1$ like before. But now improved the error terms in ${S(x)}$ will help us!

Using Dirichlet’s Hyperbola trick we get

$\displaystyle S(x)=\zeta\left(\frac{3}{2}\right) x^{\frac{1}{2}}+\zeta\left(\frac{2}{3}\right) x^{\frac{1}{3}}+\Delta(x)$

where

$\displaystyle \Delta(x)=-\sum_{n \leqslant x^{1 / 5}}\left(\left(\frac{x^{\frac{1}{2}}}{n^{\frac{3}{2}}}\right)\right)-\sum_{n \leqslant x^{1 / 5}}\left(\left(\frac{x^{\frac{1}{3}}}{n^{\frac{2}{3}}}\right)\right)+O(1)$

By writing

$\displaystyle ((x))=-\sum_{0<|h| \leqslant H} \frac{e(x h)}{2 \pi i h}+O\left(\min \left(1, \frac{1}{H\|x\|}\right)\right)$ ,

we transform the problem in to estimation of exponential sums that look like

$\displaystyle \sum_{(m, n) \in D_{0}} e\left(h x^{\frac{1}{2}} m^{-3} n^{-\frac{3}{2}}\right), \quad \sum_{(m, n) \in D_{0}} e\left(h x^{\frac{1}{3}} m^{-2} n^{-\frac{2}{3}}\right)$

Finally we get estimates like

$\displaystyle \Delta(x) \ll x^{0.1318161}$

Splitting the ranges of variables and estimating the exponential sums we finally get

$\displaystyle S\left(x+x^{\frac{1}{2}} y_1\right)-S(x)=\zeta(3 / 2) y_1\left(1+O\left(x^{-\delta}\right)\right)$

$\displaystyle Q\left(x+x^{\frac{1}{2}} y_1\right)-Q(x)=\frac{\zeta(3 / 2)}{\zeta(3)} y_1\left(1+O\left(x^{-\eta}\right)\right)$

for ${y_1 > x^{01318162}}$

There were other improvements using techniques to count points near the curve $\displaystyle x^2y^2 \le N .$

References:
1. P. T. Bateman and E. Grosswald. On a theorem of Erdos and Szekeres.
2. Shiu, P., On square-full integers in a short interval, Glasgow Math.J., 25 (1984), 127-134.
3. Liu Hongquan, On Square-full Numbers in Short Intervals
4. Heath-Brown, Square-full numbers in short intervals

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