What is the Dirichlet Convolution of the identity function with itself?

Question

If you have two identity functions, then $f(d) * g(n/d)$ would be just $dn/d = n$. Since we have an $n$ added for each divisor of $n$, would the resulting function just be $n$ times the number of divisors of $n$?

Answer 1

Essentially, yes. Recall

$$(f\ast g)(n) := \sum_{d \mid n} f(d) g \left( \frac{n}{d} \right)$$

Taking $N$ to be the identity function ($N(n) := n$), and $f=g=N$,

$$(N\ast N)(n) = \sum_{d \mid n} N(d) N \left( \frac{n}{d} \right) = \sum_{d \mid n} d \frac n d = \sum_{d \mid n} n = n \sum_{d \mid n} 1$$

hence giving $n$ times the number of its divisors. In terms of the divisor-sum functions, then, we could likewise say

$$(N \ast N)(n) = n \cdot \sigma_0(n) = n \cdot d(n)$$

(whichever notation you prefer).