I was looking at some of the examples at visualpatterns.org and coming up with explicit and recursive formulas for various aspects of the patterns. Consider the pattern below and the number of cubes at each step.
At step $n$ (starting with $n=1$), the number of cubes $C(n)$ can be found explicitly with $$ C(n) = n^3-(n-1)^3 $$ noting that at each step the "missing" cubes are a cube of side length $(n-1)$. I've stumbled trying to come up with a recursive formula to get $C(n)$ or $C_n$.
An interesting observation is that at each step, the number of missing cubes is equal to the sum of all the cubes from the previous steps. In other words, you could complete, or fill in, the cube at any step by packing all of the previous shapes together. However, I feel like it isn't legitimate to write something like $C_{n+1}=(n+1)^3-\sum_{i=1}^{n}C_n$ where $C_1 = 1$.
So I think I am missing something simple,
What is a recursive formula for calculating $C_n$?
It seems we can use three cuts (front, right, top) to reduce a $C_{n+1}$ body to a $C_n$ body, and reassembling we get \begin{align} C_{n+1} &= C_{n} + [\underbrace{(2n + 1)}_{\text{front "L"}} + \underbrace{(2n + 1)}_{\text{right "L"}} - 1] + \underbrace{2(n-1) + 1}_{\text{top "L"}} \\ &= C_{n} + 6n \\ \end{align} This gives the recurrence: \begin{align} C_1 &= 1 \\ C_{n+1} &= C_n + 6n \end{align}