<s>[INST] Turning 500, 1100, 1800, 2600, 3500, etc into 1, 2, 3, 4, 5, etc I would love to have a formula of the form $n = F(m)$ for this progression ...
n      m    
---------
1     500
2    1100
3    1800
4    2600
5    3500
...

where the second-differences in $m$ are constant, on up to $n = 100$.
 [/INST] Divide the elements by $100$ to get the sequence $5,11,18,26,35,...$, which is
1  5
2  5+6
3  5+6+7
4  5+6+7+8
...

which is
1  1+2+3+4+5  -  10
2  1+2+3+4+5+6  -  10
3  1+2+3+4+5+6+7  -  10
...

But these sums are triangle numbers, so the terms are given by $\frac{(n+4)(n+5)}{2} -10$. Now multiply by $100$ to recover the original terms:
$$f(n) = 50(n+4)(n+5)-1000 = 50 n^2 + 450 n$$
which can then be inverted using the quadratic formula to obtain
$$n = \frac{-450 + \sqrt{450^2 + 4\cdot 50\cdot f(n)}} {2\cdot 50}.$$
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