# Does there exist an inverse function for this summation?

Given the formula $\sum_1^n{i} = \frac{n ( n - 1)}{2}$, does there exist a function $F$ such that $F(n) = i$?

If so, what is it? If not, why not?

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I am looking for a function which runs in $O(1)$, not an algorithm for finding the value of $i$ via a search. –  Kyle Polich Mar 6 '12 at 23:15
What is $F(n)$ and $i$? –  user17762 Mar 6 '12 at 23:18
When you write your sum, do you mean $\sum^{n}_{i=1}i$ (summing as $i$ ranges from 1 to n), $\sum^n_{j=1} i$ (summing $i$ a total of $n$ times), or $\sum^n_{j=1} i(j)$, where $i$ is a function of $j$? –  KReiser Mar 6 '12 at 23:19
Could you give an example, it is a bit vague to me. Are you looking for a Hashing algorithm? –  Emmad Kareem Mar 6 '12 at 23:21
Bit of a necro, but I used this today and found that I had to use the ceiling function to make the answer an integer, and also found out that I can't reply vice adding a new answer. –  Jared Beekman Apr 2 '13 at 16:07

Maybe you are asking, given $X$, how can I find $n$ such that $\sum_1^ni=X$? or maybe not, it's very hard to tell what you are asking. But if that is what you are asking, here's the answer: multiply $X$ by $8$, add $1$, take the square root, subtract $1$, and divide by $2$.
For example, if $X=45$, you go $45\to360\to361\to19\to18\to9$, and indeed $1+2+\cdots+9=45$.
But your $n(n-1)/2$ should be $n(n+1)/2$.