Let $d(n)$ represent the divisor function as
$d(n)=\displaystyle\sum\limits_{k|n}1$
and the divisor summatory function as
$D(x)=\displaystyle\sum\limits_{n \leq x}d(n)$
I found the following triangular representation for the values of $D(n)$
$$ \begin{array}{ccccccccc} D(1)=&&&&&&&&& 1 &&&&&&&&&&=1\\ &\\ D(2)=&&&&&&&& 2 &+& 1 &&&&&&&&&=3\\ &\\ D(3)=&&&&&&& 3 &+& 1 &+& 1 &&&&&&&&=5\\ &\\ D(4)=&&&&&& 4 &+& 2 &+& 1 &+& 1 &&&&&&&=8\\ &\\ D(5)=&&&&&5 &+& 2 &+& 1 &+& 1 &+& 1&&&&&&=10\\ &\\ D(6)=&&&&6 &+& 3 &+& 2 &+& 1 &+& 1 &+& 1&&&&&=14\\ &\\ D(7)=&&&7 &+& 3 &+& 2 &+& 1 &+& 1 &+& 1&+& 1 &&&&=16\\ &\\ D(8)=&&8 &+& 4 &+& 2 &+& 2 &+& 1 &+& 1&+& 1&+&1&&&=20\\ &\\ \end{array} $$
The values on the right are the sum of all elements in a row.
EDIT 1:
The above picture is the result of the following observation:
Let $v_{m}(n)$ be the greatest power of $m$ that divides $n$ with $ m,n \in \mathbb{N}$ , so we get that
$D(n)=\displaystyle\sum\limits_{m=2}^{\infty}v_{m}(p^{n}), p \in \mathbb{P}$ where $p$ is a fixed prime number.
I didn't try to prove this. I don't know how to do it, but hopefully some one will have some idea on how to prove or disprove this conjecture.
I'd like to know if this is a known fact. I don't have a proof but I've tested lots of values and woks all the time.
Thanks.