# What is this triangle inside an octagon symbol and square bracket notation?

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}]


• Welcome to MSE. It is a DifferenceRoot. Sep 23 at 19:07
• 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. Sep 23 at 21:20
• When you see a new symbol that you don't know, try putting \\FullForm at the end of the line and running it again
– Joe
Sep 24 at 7:33
• That's an octagon, not a hexagon. Sep 24 at 10:05

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]


As noted in the comments, this is a DifferenceRoot object. If you click on the oblong, you get a few more details:

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.

• Any ideas as to why it's represented as a circle with a triangle in the middle?
– Joe
Sep 24 at 7:34
• @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. Sep 24 at 10:52
• Oh I see that's interesting thank you. I've not seen them called O$\Delta$E before.
– Joe
Sep 24 at 13:57