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How can we find a function for a conversion table with the following properties? The range of both inputs and outputs is $[0..100]$; inputs are integers.

$f(0) = 0$ $f(100) = 100$ $f(1) - f(0) = 2(f(100) - f(99))$

So, $f(1)$ should be $1.\bar3$ and $f(99)$ should be $99.\bar3$. The difference between subsequent outputs is linearly decreasing, so $f(50)-f(49)\approx 1$.

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Let $a_n=f(n)-f(n-1)$. Then $a_1=\frac{4}{3}$ and $a_{100}=\frac{2}{3}$. Since the differences are linearly decreasing, we have $a_n=\frac{4}{3}-\frac{n-1}{99}\frac{2}{3}$.

Hence $f(n)=a_n+...+a_1+f(0)=\frac{4n}{3}-\frac{2}{297}(1+...+(n-1))=\frac{4n}{3}-\frac{n(n-1)}{297}=\frac{n(397-n)}{297}$.

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    Thanks for the question edit, although outputs obviously aren't integers. I fixed that.2012-12-13