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I defined a function in Mathematica online and I get this strange output. What does the triangle inside the octagon mean? Also, what is [n]? Also, why is there a y in the output when I haven't defined it?

f[k_,n_]=Sum[k!/(n!(n-k)!),{k,0,n}]

enter image description here

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    $\begingroup$ Welcome to MSE. It is a DifferenceRoot. $\endgroup$ Sep 23 at 19:07
  • $\begingroup$ Did you mean to write $k!/(n!(n-k)!)$? Or did you mean $n!/(k!(n-k)!)$? There's a well-known identity involving the summation of the latter option, but I don't think there's one for the first. $\endgroup$ Sep 23 at 21:20
  • $\begingroup$ When you see a new symbol that you don't know, try putting \\FullForm at the end of the line and running it again $\endgroup$
    – Joe
    Sep 24 at 7:33
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    $\begingroup$ That's an octagon, not a hexagon. $\endgroup$ Sep 24 at 10:05
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Well, I guess it is a shorthand for this elaborate function which I got by evaluating your expression in Mathematica on my desktop.

(1/n!)DifferenceRoot[
  Function[{\[FormalY], \[FormalN]}, {-\[FormalY][\[FormalN]] + (5 + 
         2 \[FormalN]) \[FormalY][
        1 + \[FormalN]] + (-10 - 
         6 \[FormalN] - \[FormalN]^2) \[FormalY][
        2 + \[FormalN]] + (3 + \[FormalN]) \[FormalY][
        3 + \[FormalN]] == 0, \[FormalY][1] == 2, \[FormalY][2] == 7/
     2, \[FormalY][3] == 26/3}]][n]
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As noted in the comments, this is a DifferenceRoot object. If you click on the oblong, you get a few more details:

enter image description here

What this means is that the denominator $y(n)$ satisfies the relationships $$ (n+3) y(n+3) - (n^2 + 6n + 10) y(n+2) + (2n+5) y(n+1) - y(n) = 0 \\ y(1) = 2 \qquad y(2) = \frac{7}{2} \qquad y(3) = \frac{26}{3}. $$ These relations are sufficient to determine all of the coefficients. Usually this also means that Mathematica can't simplify the result any more than this.

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    $\begingroup$ Any ideas as to why it's represented as a circle with a triangle in the middle? $\endgroup$
    – Joe
    Sep 24 at 7:34
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    $\begingroup$ @Joe: My best guess is that the triangle is supposed to be a delta (∆), which is a symbol for differences. People who work on ordinary difference equations or partial difference equations often abbreviate them as "O∆E"s and "P∆E"s to distinguish them from differential equations. $\endgroup$ Sep 24 at 10:52
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    $\begingroup$ Oh I see that's interesting thank you. I've not seen them called O$\Delta$E before. $\endgroup$
    – Joe
    Sep 24 at 13:57

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