Consider a three dimensional discrete coordinate system $(x,y,z)$, where $x,y,z\in$ natural numbers.
The number of digits describing an integer coordinate for each dimension is $l_c=\lfloor log(c) \rfloor+1$, where $c$ is $x$, $y$ or $z$.
The total number of digits describing a point in the space is $l=\lfloor log(x) \rfloor+\lfloor log(y) \rfloor+\lfloor log(z) \rfloor+3$.
I am looking for a formula to describe how many points each total coordinate digit length can describe.
Any suggestions?
Example:
As one digit for each dimension can describe a total of $(10)(10)(10)$ points, the total coordinate length $3$ can describe a total of $10^3$ points in the space.
Two digits for any of the dimensions and one digit for the rest gives a total number of points possible to describe as $(3)(10^2-10)(10)(10)$. In other words, a total of four digits can describe a maximum of $27000$ points in the space.
A total of five coordinate digits can describe a maximum of $$(3)(10^3-10^2)(10)(10)+(3)(10^2-10)^2(10)=513000$$ points in the space.
Six digits can describe a total number of $$(3)(10^4-10^3)(10)(10)+(6)(10^3-10^2)(10^2-10)(10)+(10^2-10)^3=8289000$$ points in the space.
And so on.
Any suggestions on how to produce a formula are greatly appreciated.
Best Answer
OPs approach is fine and in fact the examples already present all possible different variants of the number of digits $k_1,k_2,k_3$ of the three dimensions with $k_1+k_2+k_3=n$ digits.
All three dimensions are equal: $\qquad\qquad\ k_1=k_2=k_3$
Two are equal, the third is different: $\qquad\ k_1=k_2, k_1\ne k_3$
All three are pairwise different: $\qquad\qquad\; k_1\ne k_2, k_1\ne k_3, k_2\ne k_3$
As there are some different cases to distinguish we conveniently use Iverson brackets.
We can simplify the three summands (1), (2) and (3) somewhat.
Since $k_1=k_2=k_3$ we have only one case to consider, namely $3k_1=n$ resp. $k_1=\left\lfloor\frac{n}{3}\right\rfloor$. This implies that $n$ has to be a multiple of $3$ which is asserted by $[[3|n]]$, otherwise the sum is zero.
The upper limit $\left\lfloor\frac{n-1}{3}\right\rfloor$ of the left-hand sum in (5) follows from the index region \begin{align*} &1\leq k_1=k_2<k_3\leq n\\ &2k_1+k_3=n\qquad\qquad\qquad\to\qquad k_3=n-2k_1>k_1\qquad\to\qquad k_1<\frac{n}{3}\\ \end{align*}
The lower limit $\left\lceil\frac{n+1}{3}\right\rceil$ of the right-hand sum in (5) follows from the index region \begin{align*} &1\leq k_1<k_2=k_3\leq n\\ &k_1+2k_3=n\qquad\qquad\qquad\to\qquad k_1=n-2k_3<k_3\qquad\to\qquad k_3>\frac{n}{3}\\ \end{align*}
The upper limit $\left\lfloor\frac{n-1}{2}\right\rfloor$ of the right-hand sum in (5) follows from the index region \begin{align*} &1\leq k_1<k_2=k_3\leq n\\ &k_1+2k_3=n\qquad\qquad\qquad\to\qquad n-2k_3\geq 1\qquad\to\qquad k_3\leq \frac{n-1}{2}\\ \end{align*}
The upper limit $\left\lfloor\frac{n-k_1-1}{2}\right\rfloor$ of the sum in (6) follows from the index region \begin{align*} &1\leq k_1< k_2< k_3\leq n\\ &k_1+k_2+k_3=n\qquad\qquad\to\qquad k_3=n-k_1-k_2>k_2\qquad\to\qquad k_2<\frac{n-k_1}{2}\\ \end{align*}