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In the following equation, $n^s=(n)_s+f(s)$ What is general form for $f(s)$? Understand that, $(n)_s=n(n-1)(n-2)\cdots(n-[s-1])=\text{ The Falling Factorial }$

I have experimented with this equation for $s=\{1,2,3,4\}$. Unless my calculations are horribly mangled, the following table arises: $ \begin{array}{c|c} s & f(s)\\ \hline 1 & 0\\ 2 & n\\ 3 & 3n^2-2n\\ 4 & 6n^3-11n^2+6n \end{array} $

I cannot see any reasonable pattern to these values. There isn't an obvious (to me) reccurence relationship, so that method of solving this seems useless. I would appreciate any help on this; it is a personal curiosity of mine. I'm a freshman in highschool. So, I would appreciate elaboration on any complex or complicated methods.

To be more specific, I would like $f(s)$ defined in the form of a polynomial. This is the particular form I was considering: $f(s)=c_{s-1}n^{s-1}+c_{s-2}n^{s-2}+\dots+c_{1}n$

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    @ThomasAndrews I think you have your indexing backwards - it's $c_1$ that's $(s-1)!$ and $c_{s-1}$ that's $s-1\choose 2$.2012-04-11

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What you want are known as the Stirling numbers of the first kind; the unsigned version is generally denoted as $\displaystyle{s\brack i}$, and the cofefficients of your polynomial are given by $\displaystyle{f_s(n) = \sum_{i=0}^{s-1} (-1)^{s-i+1}{s\brack i}}n^i$. Note that I'm writing this as $f_s(n)$ rather than $f(s)$, since the latter suggests that it's a function of $s$ where it's more properly thought of as a sequence of functions indexed by $s$. There are recurrence formulas for the coefficients, but there's no explicit representation of them. For more details, I'd suggest checking out the Wikipedia page as a starting point; both of the book references in that article (Knuth's The Art Of Computer Programming and Knuth, Graham and Patashnik's Concrete Mathematics) are excellent sources for learning more about them.

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    Ayep, I missed that - thanks for the edit!2012-04-11