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!

Please help!

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".

We already knew about Triangular Numbers but someone asked,

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

A little scribbling and we derived the formula for Tetrahedral Numbes

. . (a term we thoughtinvented): .we

We had more eggnog and wondered about the sum of Tetrahedral numbers.

. . This turned out to be: .

We called them "Pentahedroidal" numbers.

In 4-space, this figure would be bounded by five tetrahedrons,

. . and would be called apentahedroid.

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 gift.

. . The second year, she gives me gifts.

. . The third year, gifts.

. . . . . . . . . . . . .

. . The twelfth year, gifts.

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