Let $f$ be an arithmetic function defined by $$f(n) = |A_n|$$ where $A_n = \{(a, b) : n = ab^2\}$.Estimate $$\sum_{n \leq x} f(n)$$ where $x \in \mathbb{R}^+$, using Dirichlet hyperbola method. The error term should be $O(x^{1/3})$.

Dirichlet hyperbola method requires the function $f$ to be written as Dirichlet convolution of two functions. The problem is I try to compute the values of $f$, and try to guess $g ,h$ such that $f = g * h$. But it is not successful.

* Update *

According to Adam Hughes answer, $$f(n) = d(n) * \lambda(n).$$ Set $$D(n) = \sum_{n \leq x} d(n), \Lambda(n) = \sum_{n \leq x} \lambda(n).$$

Let $a, b >0$ such that $ab = x,$ then by Dirichlet heperbola method $$\sum_{n \leq x} f(n) = \sum_{c \leq a}d(c)\Lambda(x/c) + \sum_{e \leq b}\lambda(e)D(x/e) - D(a)\Lambda(b).$$

The problem is all of the examples I have seen, eg : $\sum_{n \leq x} d(n) = \sum 1*1$, know explicitly that $\sum_{n \leq x} 1 = [x].$ However, here I try to derive, and then search some information on the internet, the summation fomular for $$\Lambda(n), D(d)$$ are complicated.


I try to estimate it with using big O, and follows other examples, but I cannot do it.

Any help, or hint how to do the estimation ? How to deal with that $\Lambda, D$ if the summation formula is not simple, or sometimes no formula ?

Thank you in advanced.

PS link to a new question Estimate # of order paired function using Dirichlet hyperbola method

  • $\begingroup$ Hey Both, it's good you want to take the problem further, but generally switching the underlying question is frowned upon. Usually you want to close this one out and then open up a new one with the new info and the new goal. $\endgroup$ – Adam Hughes Sep 21 '16 at 20:12
  • $\begingroup$ okay, I vote to closed it. I will set up a new question. $\endgroup$ – Both Htob Sep 22 '16 at 3:22

For each $n$ we are computing $f(n) = \displaystyle\sum_{ab^2=n} 1$. Now the indicator function of the square is just $\sum_{d|n}\lambda(d)$ where here $\lambda(d)$ is the Liouville function, $\lambda(d) = (-1)^{\Omega(d)}$. Writing this out we have

$$f(n) = \sum_{d|n}\sum_{e|d}(-1)^{\Omega(e)}$$

Because this will have a $1$ in the outer sum exactly when the divisor $d$ is a perfect square and note that $a$ is totally determined by $b^2$ so we need only count the number of square factors.

As always, we reverse the order of summation using $ef = d, dg = n$ to produce

$$f(n) = \sum_{efg = n}(-1)^{\Omega(e)}\cdot 1(f)\cdot 1(g)$$

which reveals itself as a triple convolution, $1*1*\lambda$. Since we know $(1*1)(n) = d(n)$ we can write this as $d*\lambda(n)$ with $d$ the number of divisors function. Here you have something neatly written as a convolution of two basic arithmetic functions, as desired.


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