Introduction
For any non-negative integer $k$, $x^2+y^2=5^k$ has $4(k+1)$ integer solutions (vertices).
For example, here are 3 circles for $k=1,2,3$. The blue dots represent the vertices (integer solutions), then connect the dots to form polygons (dash lines). The red dots represent the points in-between the vertices.
Tabulated the number of in-between points using this formula
$$\sum_{i=0}^{4(k+1)-1} \left( \gcd \left( |x_{i+1} - x_i|, |y_{i+1} - y_i| \right) - 1 \right)$$
k Vertices 4(k+1) In-between Points
1 8 4
2 12 0
3 16 28
4 20 40
5 24 116
6 28 144
7 32 524
8 36 1272
9 40 3028
10 44 5984
11 48 10780
12 52 23440
13 56 63444
14 60 117480 <--- Used in the example below
Approximation Formula for $\pi$
$$\small\pi \approx \frac{1}{5^k} \left[ 4 \sum_{i=0}^{\infty} \left( \left\lfloor \frac{5^k}{4i+1} \right\rfloor - \left\lfloor \frac{5^k}{4i+3} \right\rfloor \right) - 2(k+1) - \frac{1}{2} \sum_{i=0}^{4(k+1)-1} \left( \gcd \left( |x_{i+1} - x_i|, |y_{i+1} - y_i| \right) - 1 \right) \right] $$
Example
Let $k = 14$, then
$$\small\pi \approx \frac{1}{5^{14}} \left[ 4 \sum_{i=0}^{\infty} \left( \left\lfloor \frac{5^{14}}{4i+1} \right\rfloor - \left\lfloor \frac{5^{14}}{4i+3} \right\rfloor \right) - 30 - \frac{1}{2} \sum_{i=0}^{59} \left( \gcd \left( |x_{i+1} - x_i|, |y_{i+1} - y_i| \right) - 1 \right) \right] $$
Applying the precomputed value (see table above) shows
$$\sum_{i=0}^{59} \left( \gcd \left( |x_{i+1} - x_i|, |y_{i+1} - y_i| \right) - 1 \right) = 117480$$
and reduces to
$$\pi \approx \frac{1}{5^{14}} \left[ 4 \sum_{i=0}^{\infty} \left( \left\lfloor \frac{5^{14}}{4i+1} \right\rfloor - \left\lfloor \frac{5^{14}}{4i+3} \right\rfloor \right) - 58770 \right] $$
Plugging into WolframAlpha gives
$$\pi \approx 3.14158$$
Question
Is it possible to compute the number of in-between points on-the-fly instead of using a precomputed table lookup? I tried WolframAlpha but without success.
I've never used Mathematica but it would be nice run a calculation for $\pi$ by prepping it first and then supplying
$$\small\frac{1}{5^k} \left[ 4 \sum_{i=0}^{\infty} \left( \left\lfloor \frac{5^k}{4i+1} \right\rfloor - \left\lfloor \frac{5^k}{4i+3} \right\rfloor \right) - 2(k+1) - \frac{1}{2} \sum_{i=0}^{4(k+1)-1} \left( \gcd \left( |x_{i+1} - x_i|, |y_{i+1} - y_i| \right) - 1 \right) \right] $$