Search for the nth 3d pentagonal numbers(challenge)

Question

There are square numbers, we can imagine these as a square of dots

Squ(n) $= n^2$

There are triangle numbers with we can imagine as a triangle of dots

tri(n) $=\frac{n(n+1)}{2}$

There are pentagonal numbers we can imgaine these as a pentagon of dots

Pent(n) $= \frac{3n^2 - n}{2}$

Square numbers have a three dimensional counterpart that are cubic numbers, so do triangular numbers.

The question is what is the nth 3D pentagonal number?

What is the three dimensional counterpart of triangular numbers? Tetrahedron? — velut luna, Jun 18 '17 at 20:36

score 4 · Accepted Answer · answered Jun 18 '17 at 20:39

4

In On Regular Polytope Numbers, Kim extends the notion of figurate numbers to 3-D solids. The Dodecahedral numbers seem to be what you describe, and are counted by $$ D_n=\frac{n(3n-1)(3n-2)}{2} $$

answered Jun 18 '17 at 20:39

TomGrubb

12,909

Search for the nth 3d pentagonal numbers(challenge)

1 Answers1