If I want to interpret the following summation:
$$\sum_{d|n} (n/d) \tau (d),$$
Where $\tau$ is the number of positive divisors of $n$, i.e. $$\tau (n) = \sum_{d|n} 1.$$
My opinion is:
The above summation, $$\sum_{d|n} (n/d) \tau (d),$$ can be interpreted as $$\sum_{d|n} (n/d)\sum_{d|n} 1 ,$$ but then what?
The given picture below contains another interpretation for the summation(if we let $n = p^k$):
But I do not know how my interpretation and the picture can be the same, may be if we consider the 2 summation signs in my interpretation as 1 sign because the index of the summation is the same , this will make my interpretation and the photo's interpretation close to each other, but still in the interpretation in the picture the numerator is 1, 2, 3, ....., k+1, which I do not understand why, could anyone explain this for me please?
Note that $\tau (n) = \sum_{d|n} 1$ but $\tau (d) = \sum_{m|d} 1$ for $m \leq d$, the number of divisors of $\textbf{d}$. Hence
$$\sum_{d|n} (n/d) \tau (d) = \sum_{d|n} (n/d)\sum_{m|d} 1$$
And if you replace $n$ with $p^k$ for some prime $p$, $$ \sum_{d|n} (n/d) \tau (d) = \sum_{p^i|p^k} (p^k/p^i) \tau (p^i) = p^k\sum^k_{i=0} (1/p^i) \tau (p^i) \\ = p^k\sum^k_{i=0} (1/p^i) (i+1) = \sum^k_{i=0} p^{k-i} (i+1) = \sum^k_{i=0} p^{i} (k+1-i)\\ $$ Note that $\tau (p^i) = i+1$, and the last equality follows from a change of index.