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?

Answer 1

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 $$ D_n=\frac{n(3n-1)(3n-2)}{2} $$