It's known that for any sequence of positive integers $a_1$, $a_2$, ..., $a_n$ the following equality holds: $$ \sum_{i=1}^n a_i=\sum_{k\in \mathbb{Z},k\geqslant 1} \left|\{i\colon a_i\geqslant k\}\right|. $$
I think the easiest way to understand this is to draw a set of rectangles $1\times a_1$, $1\times a_2$, ..., $1\times a_n$ and to calculate the area of all of them in two ways:
- "Riemann style": sum their areas.
- "Lebesgue style": calculate the number of squares $1\times 1$ in the first row ($\left|\{i\colon a_i\geqslant 1\}\right|$), add the number of squares $1\times 1$ in the second row ($\left|\{i\colon a_i\geqslant 2\}\right|$) and so on.
Do you know how this transformation is called?
It looks like the special case of Abel transformation (for large enough $N$, $f_k=\left|\{i\colon a_i\geqslant k\}\right|$, $g_k=k-1$): $$ \sum_{k=1}^{N} \bigl(k-(k-1))\bigr)\cdot \left|\{i\colon a_i\geqslant k\}\right|=-\sum_{k=2}^N (k-1)\cdot \bigl(\left|\{i\colon a_i\geqslant k\}\right|-\left|\{i\colon a_i\geqslant k-1\}\right|\bigr)=\sum_{k=1}^{N}k\cdot \left|\{i\colon a_i=k\}\right|. $$ But I want to believe that this fact has its own name (SomeGoodAndBigMathematician transformation)
It's the cumulative frequency distribution.
We can convert the $a_i$ into frequencies $f_j$ from which we can create a histogram.
$$f_j=|\{i:a_i=j\}|$$ When we turn them into a cumulative distribution $g_k$, we get the desired sequence. $$g_k=\sum_{j\le k} f_j = |\{i:k\le a_i\}|$$
I guess we could call it a Pearson transform after Karl Pearson who first introduced the histogram.
Unfortunately this transform is not invertible, which for instance the Abel transform is.
Alternatively we might call it the discrete Lesbesgue transform, which is the same thing.
As for beautiful results... does the entire branch of statistics count?