# Thread: Shape/number sequences in different dimensions.

1. ## Shape/number sequences in different dimensions.

I have recently been investigating how number sequences relate to each other when put into different dimensions. The simplest example is squares, where the Nth term is T(n)=n^d (d=number of dimensions). So for a square (in two dimensions), the pattern is T(n)=n^2, so we get 1,4,9,16 etc. Then, when in three dimensions, the pattern is T(n)= n^3, so we get 1,8,27,64 etc. This obviously can go on for an infinite number of dimensions.
I then decided to try equilateral triangles. For triangular numbers, I worked out the pattern as T(n)=(n(n+1))/2. Then for tetrahedral numbers, I found that the pattern was T(n)=(n(n+1)(n+2))/6. Now this looks to me like if you add a dimension, you multiply the previous Nth term rule by (n+d-1)/d. However, I have no idea how to write this whole calculation! The furthest I can get is (n...whatever I'm missing...)/d! . Please Help!

2. Hello, ToTheWareMobile!

I have recently been investigating how number sequences relate to each other when
put into different dimensions. The simplest example is squares, where the Nth term is
T(n)=n^d (d=number of dimensions). So for a square (in two dimensions), the
pattern is T(n)=n^2, so we get 1,4,9,16 etc. Then, when in three dimensions,
the pattern is T(n)= n^3, so we get 1,8,27,64 etc.
This obviously can go on for an infinite number of dimensions.

I then decided to try equilateral triangles.
For triangular numbers, I worked out the pattern as T(n)=(n(n+1))/2.
Then for tetrahedral numbers, I found that the pattern was T(n)=(n(n+1)(n+2))/6.

Now this looks to me like if you add a dimension, you multiply the previous Nth term
rule by (n+d-1)/d. However, I have no idea how to write this whole calculation!
The furthest I can get is (n...whatever I'm missing...)/d!

I don't think I can help with The Big Picture but . . .

Many years ago, I played with multi-deimensional "triangular" numbers.

It was this time of year and a few classmates and I were drinking eggnog
. . and singing "The Twelve Days of Christmas".

. . "How many gifts were given during The Twelve Days?"

A little scribbling and we derived the formula for Tetrahedral Numbes
. . (a term we thought we invented): .$\displaystyle \dfrac{n(n+1)(n+2)}{3!}$

We had more eggnog and wondered about the sum of Tetrahedral numbers.
. . This turned out to be: .$\displaystyle \dfrac{n(n+1)(n+2)(n+3)}{4!}$

We called them "Pentahedroidal" numbers.
In 4-space, this figure would be bounded by five tetrahedrons,
. . and would be called a pentahedroid.

We also wondered what scenario would produce this problem.

Suppose My True Love decided to spread the gift-giving over twelve years.

. . The first year, she gives me $\displaystyle \,1$ gift.

. . The second year, she gives me $\displaystyle 1 + 2$ gifts.

. . The third year, $\displaystyle 1 + 2 + 3$ gifts.

. . . . . . . . . . . . . $\displaystyle \vdots$

. . The twelfth year, $\displaystyle 1 + 2 + 3 + \hdots + 12$ gifts.

At this point we ran out of eggnog and brain cells . . .