0

I have got an interesting task, but I can't solve it:

We use $d(n)$ as the number of divisors for the positive $n$ integer. We have:

$$a(n)=\sum_{i=1}^n d(i)$$

How much is $a(n)$ asymptotic?

$a(1) = 1, a(2) = 3, a(3) = 5, a(4) = 8,...$

I tried something with Euler's Totient Function, since it counts the numbers which are relative prime to $n$, but it is not really exact, since if one number is not relative prime to the other, it is not in all case a divisor.

Any other ideas, will the formula contain logarithm, square root, or anything? Thanks for any help!

Atvin
  • 3,512

1 Answers1

2

The average order of the divisor function satisfies the following inequality:

$$\mbox{for all } x\geq1, \sum_{n\leq x}d(n)=x\log x+(2\gamma-1)x+O(\sqrt{x})$$ In order to prove this you can first prove a weak result by setting: $$\sum_{n\leq x}d(n)=\sum_{n\leq x}\sum_{d|n}1 $$ and try to invert the sums, This result is a direct consequence of Dirichlet hyperbola method.

Elaqqad
  • 13,983