A triangular representation for the divisor summatory function, $D(x)$ 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.
 A: Yes, this is true.  Write $D(x) = \sum_{n \le x} d(n) = \sum_{n \le x} \sum_{d | n} 1 = \sum_{d \le x} \lfloor \frac{x}{d} \rfloor$; this is equivalent to the pattern you observe.  The last step is exchanging the order of summation together with the observation that the number of times a number $d$ appears in the double sum is the number of numbers less than or equal to $x$ it divides.
A: This answer is a bit of a work in progress, but if $n=2^x-1$, then $$\frac{D(n)+u}{2}=\sum_{j\in\mathcal{N}}\sum_{i=1}^{n}{h_{i,j}}  \text{ where } u=\lfloor\sqrt{n}\rfloor$$ 
where $h_{i,j}$ is the value in the corresponding row,column of the matrix described in
https://crypto.stackexchange.com/questions/27003/has-anyone-heard-of-matrix-based-roman-doll-encryption-techniques
Furthermore, letting $r_j=\sum_{i\in\mathcal{N}}{h_{i,j}}$, we write:
$$
D(2^k-1)=u-\xi+2\sum_{l=0}^{k-1}\frac{(k-l+1)(k-l)}{2}\sum_{j=\lfloor 2^{l-1}\rfloor }^{2^l-1}{r_j}
$$
where $\xi$ is computed using the program:
input k 
unsigned  step = 1
unsigned  y = 1
unsigned  $\xi$ = 0
while y < 2^k {
    unsigned bin=(unsigned)log2(y)
    $\xi$ = $\xi$ + (k-bin)(k-bin+1-(k-bin)%2)
    y = y + 8* step
    step = step + 1
} 
output $\xi$
This suggests a slim possibility of computing $D(x)$ in $log_2(x)$ time.
