Update is at bottom of my post.
I saw on YouTube (https://www.youtube.com/watch?v=_rJdkhlWZVQ) a way to approximate $\pi$ starting with a hexagon inscribed inside a circle of unit radius. It uses formulas and relies on doubling the numbers of sides of the inscribed polygon (from $6$ to $12$ to $24$ to $48$ to $96$). Here is a list of the values for the approximation of $\pi$ based on those:
$5$ tuple is:
(# of polygon sides, estimate of $\pi$, diff. in prev $\pi$ estimate, d=prev. diff./curr. diff., $4$-d)
$$(6, 3.00000000000000, 3.00000000000000, xxxxxxxxxxxxxx, xxxxxxxxxxxxxx)$$
$$(12, 3.10582854123025, .10582854123025, xxxxxxxxxxxxxx, xxxxxxxxxxxxxx)$$
$$(24, 3.13262861328124, .02680007205099, 3.94881554903670, .05118445096330)$$
$$(48, 3.13935020304687, .00672158976563, 3.98716270785101, .01283729214899)$$
$$(96, 3.14103195089051, .00168174784364, 3.99678809818983, .00321190181017)$$
I added most of my extrapolated data here for ease of viewing but I explain later how I got it. x1 = extrapolation # $1$, x2 = extrapolation # $2$…
$$(x1, 3.14145247226851, .00042052137800, 3.99919702454746, .00080297545254)$$
$$(x2, 3.14155760788934, .00010513562083, 3.99979925613686, .00020074386314)$$
$$(x3, 3.14158389212432, .00002628423498, 3.99994981403422, .00005018596578)$$
$$(x4, 3.14159046320368, .00000657107936, 3.99998745350855, .00001254649145)$$
$$(x5, 3.14159210597481, .00000164277113, 3.99999686337714, .00000313662286)$$
$$(x6, 3.14159251666767, .00000041069286, 3.99999921584428, .00000078415572)$$
$$(x7, 3.14159261934089, .00000010267322, 3.99999980396107, .00000019603893)$$
$$(x8, 3.14159264500919, .00000002566831, 3.99999995099027, .00000004900973)$$
$$(x9, 3.14159265142627, .00000000641708, 3.99999998774757, .00000001225243)$$
…
$$(x18, 3.14159265356529,.00000000000002, 3.99999999999995, .00000000000005)$$
I believe Archimedes stopped at a $96$ sided inscribed polygon because the calculations by hand became too tedious at that point so the data above is the end of the data we get from the $96$ sided polygon but you can see that there is a predictable pattern to these tuples. To get the approximation of $\pi$, I simply sum up (accumulate) the $3$rd entries from each tuple. Notice how much quicker this is converging on the real value of $\pi$ and with minimal work.
Just playing around in my Excel spreadsheet with these values, I noticed that these approximations of $\pi$ converge on the real value of $\pi$ in a predictable way so I encoded this in the spreadsheet and I came up with an extrapolated estimate of $\pi$ to be $3.14159265356529$ which I am quite happy with because it is as accurate as using about a $400,000$ sided polygon using Archimedes method but my method is much simpler.
In a nutshell, I looked at the difference between the estimate from $12$ to $6$ and from $24$ to $12$… and I see that it is approximately "quartering" each time. I took that a step farther and analyzed how close to $1/4$ the difference is at each doubling and encoded that as well and just from those two simple calculations, I got a pretty decent $\pi$ accurate to $10$ digits after the decimal point. If I had the data on the circumscribed polygons, I suppose the # of accurate digits would increase even more. It makes sense that there is predictability since we are doubling the # of sides each time and not some random increase like from $6$ to $10$ to $25$ sides.
So in my extrapolation method, my first step would be to take the last known "delta" of $.00321190181017$ and "quarter" it to $0.00080297545254$ so that the next tuple entry would be (extrapolation 1, $3.14145247226851
, 0.00042052137800, 3.99919702454746, 0.00080297545254$)
Also it is worth mentioning that using only data from up to $24$ inscribed polygons, my extrapolation method yields a quite good approximation of $3.14159265308398$ which is accurate to $9$ places after the decimal and that is as good as using about a $100,000$ sided polygon using Archimedes method.
So does anyone know if there is documentation that Archimedes knew about this extrapolation method (or similar) during his lifetime?
UPDATE:
I revisited my spreadsheet and got a more accurate version of $ \pi$ using this method. The tweak is in the determination/extrapolation of the 5th parameter. I am not sure how or why this tweak works (maybe someone can explain the math behind it), but it gives an even better approximation of $ \pi$.
Rather than list a bunch more numbers in an already cluttered post, I will just describe the tweak and leave it at that.
$x1$ (extrapolation 1) is exactly the same as in my previous effort (I took the 5th parameter $.00321190181017$ in the "$96$ sided polygon" entry and divided it by $4$ to get $.00080297545254$.
However, for $x2$, I did NOT do a divide by $4$ to get the 5th parameter. Instead, I used the 4th parameter of the "$48$gon" entry which is $3.98716270785101$. The "new" 5th parm of $x2$ is now $.00020139018931
$. I repeat this pattern for getting the next 5th parameter by "dropping down a line" each time.
The "bottom line" is the approximation of $\pi$ this way improves from my previous value of $3.14159265356529$ to $3.14159265358983$. Notice how the previous estimate was on the low side of the actual value of $\pi$ but the new estimate is slightly on the high side.
Also, here is the follow up YouTube video link for the Archimedes method of circumscribed polygon to "sandwich" the value of $\pi$ between. With those two values, the confidence level of $\pi$ goes way up. https://www.youtube.com/watch?v=9zO0-QOcJQ0
Best Answer
This is not an answer, just an alternative exposition of your findings as far as I was able to reproduce them.
First, the video describes Archimedes' method for approximating $\pi$ via the following recurrence for the side length of a regular $3\cdot2^n$-gon inscribed in a circle, which I'll call $\ell_n$: $$ \ell_1=1 \text{ ,}\qquad \ell_{n+1} = 2\sqrt{2-\sqrt{4-\ell_n^2}} \tag1 $$ Then we approximate $\pi$ as half the perimeter of that $3\cdot2^n$-gon, that is, $$ h_n = 3\cdot 2^{n-1} \ell_n $$
What you've noticed numerically is that $$ \frac{h_n-h_{n-1}}{h_{n-1}-h_{n-2}} \approx \frac14 \tag2 $$ (Nice observation, by the way. If you do the trigonometry, it turns out that $$ \frac{h_n-h_{n-1}}{h_{n-1}-h_{n-2}} = \frac1{2\cos\theta_n(1+\cos\theta_n)} \text{ ,}\quad\text{ where } \theta_n = \frac{\pi}{3\cdot2^n} $$ which confirms your finding because $\cos\theta\to 1$ as $\theta\to 0$. This also shows that the LHS is always $\ge\frac14$.)
Next, you started approximating $h_n$ using (2) instead of the Archimedean recurrence (1). Actually, you say you have something more precise than $\frac14$ on the RHS of (2), but you don't say what it is, so I'll just proceed as if you used $\frac14$.
Let's call the values obtained by this method $\hat h_n$. Since you started using this method after the $96$-gon, the sequence is given by the rules $$ \hat h_4 = h_4 \text{ ,}\quad \hat h_5 = h_5 \text{ ,}\quad \frac{\hat h_n-\hat h_{n-1}}{\hat h_{n-1}-\hat h_{n-2}} = \frac14 $$ By telescoping product, we get $$ \hat h_n - \hat h_{n-1} = (\hat h_5 - \hat h_4) \prod_{k=6}^n \frac{\hat h_k-\hat h_{k-1}}{\hat h_{k-1}-\hat h_{k-2}} = \frac{\hat h_5 - \hat h_4}{4^{n-5}} $$ and then, by telescoping sum, $$ \hat h_n = \hat h_5 + \sum_{k=6}^n (\hat h_k-\hat h_{k-1}) = \hat h_5 + (\hat h_5-\hat h_4)\sum_{k=6}^n \frac1{4^{k-5}} = \hat h_5 + \frac13 (\hat h_5-\hat h_4) \Big(1-\frac1{4^{n-4}}\Big) $$ Now you can see that as $n\to\infty$, we have $$ \hat h_n \to \hat h_5 + \frac13 (\hat h_5-\hat h_4) = \frac{4h_5-h_4}{3} \approx 3.1415925335 $$ So, this sequence is not actually tending to $\pi$, but to something a bit smaller.
Update: I don't know whether Archimedes knew your method, but I will note that the summation of the infinite geometric series used here was absolutely something he knew — he did that as part of his computation of the area of a parabolic sector.
Update: It occurs to me that another way to use your idea is to compute $h_n$ via (1) as usual, but rather than report that as your estimate, report $(4h_n-h_{n-1})/3$. This is as if you had used (2) starting from $n$ instead of starting from $6$, and it gives a sharper lower bound for $\pi$. (It's a lower bound because, as noted above, in (2) we have $\ge$.) You still have to use (1), unfortunately.