Twin primes conjecture and extrapolation method

This is just to present a numerical curiosity. I suspect that it is coincidental, but I feel compelled to share it in case anyone wants to look further.

First some disclaimers: Consider a random walk which starts at a value $x_0$ and, for constants $c_1,c_2$ moves at step $k$ with probability $\frac{c_1}{\log^2 k}$ and, if so, increases or decreases by $\frac{c_2}{k}$ with equal probabilities. Then one would expect a limit $L$ which is not all that far from the positions early on. The same is true for a product which starts at $y_0$ and sometimes gets multiplied by $1 \pm \frac{c_3}{k}.$ If one sees all the steps up to some point $k=N$, one can adjust the starting value $y_0$ to get the values to be reasonably close to any desired value (in the vicinity of $N$, with the same steps.) That description ( up to errors of order $\frac1{k^2}$) seems to apply to the product from the question and another I will give below.

I'll (for my own purposes) restate the astonishing observation mentioned in the question as follows:

Consider the product $$y_0 \frac53 \frac57 \frac{13}{11} \frac{17}{19} \cdots$$ where the terms are $\left( \frac{p}{p+2}\right)^{\pm 1}$ on twin primes and the exponent is chosen according to the value of $p \bmod 4.$ Then for $y_0=\frac32$ the partial products differ from $\sqrt{\pi}$ only in the fifth decimal at $10^j$ for $7 \leq j \leq 15.$

I don't know what happens other than at powers of $10$ but it seems reasonable that it is the same. The starting value $\frac32$ is simple and kind of makes sense in the context of the other values. The values seem to be converging to something ever so slightly below $\sqrt{\pi}$ but one might hope that it crosses to greater much further out. After all, there seems to be a prime race where one of the two options appears to stay in the lead. But perhaps, as with other prime races, the lead changes eventually.

Here is the curious result I have been delaying getting to:

Consider the product $$y_0 \frac{11}5 \frac7{13} \frac{17}{11} \frac{13}{19} \cdots$$ where the terms are $\left( \frac{p}{p+6}\right)^{\pm 1}$ on prime pairs $p,p+6$ and the exponent is chosen according to the value of $p \bmod 6.$ Then for $y_0=\frac32$ the partial products seem kind of close to $\pi.$

Up to the point shown the product is $$\frac32 \frac{11}5 \frac7{13} \frac{17}{11} \frac{13}{19} =\frac{357}{190} \approx 1.8789$$

That corresponds to the first line of this list

$2^4,\, 1.878947368$
$2^{5},\, 2.685490755$
$2^{ 6},\, 3.027471873$
$2^{ 7},\, 2.741208368$
$2^{ 8},\, 2.989201152$
$2^{ 9},\, 3.028013205$
$2^{10}, 3.112518657$
$2^{11}, 3.091760181$
$2^{12}, 3.084789176$
$2^{13}, 3.114315927$
$2^{14}, 3.142868728$
$2^{15}, 3.147232108$
$2^{16}, 3.142821684$
$2^{17}, 3.142827267$
$2^{18}, 3.140499348$
$2^{19}, 3.139682496$
$2^{20}, 3.142046379$
$2^{21}, 3.142116440$
$2^{22}, 3.142199716$
$2^{23}, 3.142444242$
$2^{24}, 3.141721462$
$2^{25}, 3.141603363$

That last line is impressively close to $\pi.$

In the interest of full disclosure though, the next line is not as impressive.

$2^{26}, 3.141825366$

Here is another coincidence:

OP considers a product of the form:

$$\frac{9}{4} \prod_{(p,p+2) twins} \left(\frac{p}{p+2}\right)^{2\chi(p)}$$

Where:

$$\chi(p) = \begin{cases} 1, & \text{if}\ p\equiv 1 \mod 4 \\ -1, & \text{if}\ p\equiv 3 \mod 4 \end{cases}$$

Consider the same product format but with a different $\chi$, mod $8$ instead:

$$\chi(p) = \begin{cases} 1, & \text{if}\ p\equiv 1,7 \mod 8 \\ -1, & \text{if}\ p\equiv 3,5 \mod 8 \end{cases}$$

This new product appears to approach $5 \pi$. Here are some partial products:

$10^3,\,\,15.4657287115890258,\,\,4.9228943459352863$

$10^4 ,\,\,15.6256061945197966 ,\,\,4.9737849293303309$

$10^5 ,\,\,15.7177620190952633 ,\,\,5.0031190393621211$

$10^6 ,\,\,15.7059959568870969 ,\,\,4.9993737854398082$

$10^7,\,\, 15.7070308888808138 ,\,\,4.9997032145249362$

$10^8 ,\,\,15.7078988937411405 ,\,\,4.9999795090532338$

$10^9 ,\,\,15.7074564317562495 ,\,\,4.9998386690291825$

$10^{10},\,\, 15.7074110097452894 ,\,\,4.9998242107540436$

$10^{11} ,\,\,15.7074152334423153 ,\,\,4.9998255551985631$

The first column is $x$, the second column is the product with twin primes up to $x$ and third column is that partial product divided by $\pi$.

I would appreciate if someone with better computational resources could check if this coincidence gets more or less impressive for larger $x$.

See and this example. Let c odd composite number. if c = 1 mod 4 then p = (c-1)/(c+1), p = (c+1)/(c-1) otherwise. Then 4*(Product of p) are equal Pi. 4 * (4/5) * (8/7) * (10/11) * (12/13) * (14/13) * (16/17) * (18/17) * (20/19) * (22/23) * ... = Pi

10^1 3.2000000000000000000000000000000000000, 10^2 3.2794152977678994153059934972279714814, 10^3 3.1559796620065391197135260191310921589, 10^4 3.1501510753296059145709604590165081732, 10^5 3.1437466736474914904946106564058312233, 10^6 3.1416095240706287350146143950829877575, 10^7 3.1417048217877049712559843623819562819, 10^8 3.1416347361057525862432365843056005262, 10^9 3.1416092484172566488533162154330707547, 10^10 3.1415945263787852852755178355994600166

As you see the product converges to Pi.

But convergence is very slow, random, unstable, sensitive and oscillates chaotically.