# Multi-dimensional Pyramidal Numbers

I was walking through main suite of the mathematics offices and glanced over at the board where some students, who I believe are in a mathematics education course, were working on calculating the sum of the first few triangular numbers.

For those of you that don’t know, triangular numbers are the numbers that represent the sum of the first [tex]n[/tex] positive integers

[tex]1 = 1[/tex]
[tex]1+2 = 3[/tex]
[tex]1+2+3 = 6[/tex]
[tex]1+2+3+4=10[/tex]
[tex]1+2+3+4+5=15[/tex]
[tex]1+2+3+4+5+6=21[/tex]
[tex]cdots[/tex]

It might be easier to see why they are referred to as triangular if you look at the table below:

 1 O 3 O  O 6 O  O  O 10 O  O  O  O 15 O  O  O  O  O 21 O  O  O  O  O  O

The number in each row represents the number of “balls” on the row and all previous rows. Note that if you want the [tex]n[/tex]th triangular number, you can see that

[tex]T_n = displaystyle frac{n(n+1)}{2}[/tex]

(You can prove this by a standard mathematical induction argument)

So the question was, “What is the sum of the first [tex]n[/tex] triangular numbers?”

[tex]T_1 + T_2 + T_3 + cdots + T_n[/tex]

Another way of looking at this question is to see it as the number of balls in a triangular pyramid. They are called either tetrahedral numbers or pyramidal numbers.

[tex]1 = 1[/tex]
[tex]1+3 = 4[/tex]
[tex]1+3+6 = 10[/tex]
[tex]1+3+6+10=20[/tex]
[tex]1+3+6+10+15=35[/tex]
[tex]1+3+6+10+15+21=56[/tex]
[tex]cdots[/tex]

It can be shown that [tex]displaystyle P_n = frac{n(n+1)(n+2)}{6}[/tex], where [tex]P_n[/tex] denotes the [tex]n[/tex]-th pyramidal number. Again, this is a straight-forward exercise to prove this by mathematical induction.

This inspired me to double check that adding the first [tex]n[/tex] pyramidal numbers would follow this pattern. I’ll spare you the my lengthy version of this mathematical induction proof (that is, unless you ask).

[tex]displaystyle P_1 + P_2 + P_3 + cdots + P_n = frac{n(n+1)(n+2)(n+3)}{4!}[/tex]

Now, I’ve not really seen this in any of my reading or course work. Someone reading this might know if such numbers have a name but for lack of a better one, I’ll call them “hyper-pyramidal numbers” which basically represent the number of balls it would take to form a 4-dimensional triangular pyramid. We can then extend this generalization to any dimension. You’ve seen the 3-D pyramid in the image above. Now imagine extending a triangular pyramid in some new dimension. That is, we really have a sequence of 3-D pyramids of shrinking size. Just like the 3-D pyramid is a sequence of triangles of shrinking size. If we are dealing with a [tex]4[/tex] dimensional hyper-pyramid, then the first 3-D pyramid would be [tex]n[/tex] rows high. This would have [tex]T_n[/tex] balls. Then the next would be [tex]T_{n-1}[/tex] balls and so one down to a single ball. Then the total number would be

[tex]displaystyle P_1 + P_2 + P_3 + cdots + P_n = frac{n(n+1)(n+2)(n+3)}{4!}[/tex]

For simplicity let me modify my notation so that [tex]P_n^{(k)}[/tex] represents the [tex]n[/tex]-th [tex]k[/tex]-D hyper pyramidal number. Thus my notation would say,

[tex]P_n^{(1)} = n[/tex]
[tex]displaystyle P_n^{(2)} = T_n = frac{n(n+1)}{2}[/tex]
[tex]displaystyle P_n^{(3)} = P_n = frac{n(n+1)(n+2)}{3!}[/tex]
[tex]displaystyle P_n^{(4)} = P_1 + P_2 + cdots + P_nfrac{n(n+1)(n+2)(n+3)}{4!}[/tex]
[tex]displaystyle vdots[/tex]
[tex]displaystyle begin{array}{rl}P_n^{(k)} &= P_1^{(k-1)} + P_2^{(k-1)} + cdots + P_n^{(k-1)}\[2ex] &= displaystyle frac{n(n+1)(n+2)cdots(n+k-1)}{k!}end{array}[/tex]

Unfortunately, I did not take the time to verify the general formula for [tex]P_n^{(k)}[/tex] by induction on [tex]k in mathbb{Z}^{+}[/tex]. I do have other things to do you know. Give it a shot if you’re ever bored.

## 3 thoughts on “Multi-dimensional Pyramidal Numbers”

1. Corin Holloway says:

I was working this out too out of boredom and for the four dimensional one, I did it as the 2d one squared.

4d= ((n^2 +n)/2)^2

I also worked out the 5d one, but not as neatly as the ones you’ve worked out.

n^5/5+n^4/2+n^3/3-n/30

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2. Cuc says:

To inspire you a bit more, let P(k,n) denote the k-dimensional n-th pyramidal number.
Then P(k,n) = P(n-1,k+1) for k,n>1.

Example.
In particular P(2,n) = P(n-1,3).
For n=2: P(2,2) = 1 + 2 = 3 = P(1,3)
For n=3: P(2,3) = P(2,3) tautology
For n=4: P(2,4) = P(3,3), because 1 + 2 + 3 + 4 = 10 = P(3,3) = P(2,1) + P(2,2) + P(2,3) = 1 + 3 + 6.
For n=5: P(2,5) = P(4,3) because 1 + 2 + 3 + 4 + 5 = 15 = P(4,3) = P(3,1) + P(3,2) + P(3,3) = 1 + 4 + 10, etc.

Example.
So, because 153 = P(2,17) = P(16,3) = P(15,1) + P(15,2) + P(15,3) = 1 + 16 + 136 it follows that P(15,3) = 136 = P(2,16) = 8.17 as is easily verified.

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