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I am in need of a way to represent the sum

$1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33$

in a geometrical way. What I mean by this is that for example, the sum

$1^2 + 2^2 + 3^2 + 4^2 = 1 + 4 + 9 + 16 = 30$

can be represented geometrically as a pyramid with layers consisting of 1, 4, 9 and 16 pieces respectively a regular manner. Image from Wikipedia to illustrate the geometrical construction of the square numbers.

I have tried to find such a regular pattern to construct a geometrical shape from the factorial numbers, but to no avail. How could this be done?

Also, a follow up question: Is there a way to represent the factorial numbers up to an arbitrary number $n!$, instead of ending at $4!$ as stated in this question above? (less important, but interesting nonetheless)

Thanks in advance!

EDIT: The probably most important part is that the 1, 2, 6 and 24 are discrete and somewhat separated from each other, kind of like the different layers in the comparison between te sum of squares (see linked image above).

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  • $\begingroup$ Why not use a triangle? Place a column of $1$s at the right. Starting from the second row, place a column of $2$s to the left of the column of $1$s. Repeat the column construction until you get to $n$ which should be a column of length one. $\endgroup$ – John Douma Nov 30 '18 at 19:45
  • $\begingroup$ @JohnDouma I am not quite sure how this would produce something connected to factorial sums? Thank you for your comment. Would I have first 4 1:s, then 3 2:s, 2 3:s, 1 4:s? That makes a sum of 20. $\endgroup$ – Sigfrid Stjärnholm Nov 30 '18 at 20:13
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The only idea I could come up with was counting the number of vertices in a tree graph that had branching ratios of $2, 3, 4, \ldots$, so that the number of vertices on each level were $1!, 2!, 3!, 4!, \ldots$.

n=4 tree layout

If you need more layers ($n=6$), you might want a different layout (thanks to @HenrikSchumacher):

n=6 layout

Radial embedding is particularly elegant and helpful too ($n=6$):

enter image description here

Perhaps such a three-dimensional representation would be appropriate:

enter image description here

The answer to your question "Also, a follow up question: Is there a way to represent the factorial numbers up to an arbitrary number n!, instead of ending at 4! as stated in this question above?" is:

$$\sum\limits_{n=1}^k n! = (-1)^{k+1} \Gamma (k+2) \text{Subfactorial}[-k-2]-\text{Subfactorial}[-1]-1$$

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  • $\begingroup$ OP wants to know geometrical interpretation. Is there any geometrical interpretation using this? Please explain :) $\endgroup$ – tarit goswami Nov 30 '18 at 19:35
  • $\begingroup$ In no way does this answer OP's question. $\endgroup$ – MPW Nov 30 '18 at 19:39
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    $\begingroup$ @MPW: "Also, a follow up question: Is there a way to represent the factorial numbers up to an arbitrary number n!, instead of ending at 4! as stated in this question above? " $\endgroup$ – David G. Stork Nov 30 '18 at 19:40
  • $\begingroup$ Thank you for your response! This gives me the sum of the factorial numbers up to a certain point k, and in my case I want k = 4. I see your argument that it would be hard to find such a geometric interpretation of factorial sums, but if we limit the value of k to 4, only considering 4!, is it not possible that for such small values of factorial there would exist a beautiful and simple interpretation as it did for the squares? $\endgroup$ – Sigfrid Stjärnholm Nov 30 '18 at 20:02
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    $\begingroup$ @SigfridStjärnholm: Well, not precisely a fractal because the branching ratio changes at each level... hence the graph is not self-similar. $\endgroup$ – David G. Stork Nov 30 '18 at 22:46
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Polynomial functions of degree $d$ can be represented in a space of $d$ dimensions, using segments, squares, cubes then hypercubes. For example, the square pyramidal numbers can be sketeched in 3D as a stack of squares.

This does not generalize to factorials as they are of "unbounded degree" and would require an unbounded number of dimensions.


Even if you find a trick to limit the number of dimensions, the value of the numbers quickly becomes unmanageable ($10!=3628800$).

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  • $\begingroup$ Ah, i see! Would there however be some sort of way to represent just up to 4!, if ignoring the general case? $\endgroup$ – Sigfrid Stjärnholm Nov 30 '18 at 19:59

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