Let $(p\_1, p\_2)$ be a twin prime pair, where we include $(2, 3)$. If $p\_1 \equiv 1$ mod $4$ then we let $t\_{(p\_1, p\_2)} := p\_1 ^ 2 / p\_2 ^ 2$ otherwise, we let $t\_{(p\_1, p\_2)} := p\_2 ^ 2 / p\_1 ^ 2$.


I conjecture that the product 
$$
\prod\_{(p\_1, p\_2): \text{twin primes}}t\_{(p\_1, p\_2)}
=\tfrac{3 ^ 2}{2 ^ 2} \cdot \tfrac{5 ^ 2}{3 ^ 2}\cdot 
\tfrac{5 ^ 2}{7 ^ 2}\cdot\tfrac{13 ^ 2}{11 ^ 2} \cdot\tfrac{17 ^ 2}{19 ^ 2} \cdot\tfrac{29 ^ 2}{31 ^ 2} \cdot\tfrac{41 ^ 2}{43 ^ 2} \cdot \tfrac{61 ^ 2 }{59 ^ 2} \cdot \tfrac{73 ^ 2}{ 71 ^ 2}\cdot \tfrac{101 ^ 2 }{ 103 ^ 2}\cdots
$$


is equal to $\pi$. (If this is true then twin prime numbers are infinity many.)


Some numerical values of partial products:   

\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ $p\_1 \equiv 3$ mod $4$ \_\_\_ $p\_1 \equiv 1$ mod $4$
  
   

3.1887755102040816321 to $10^1$, \_\_\_\_\_\_\_\_\_\_\_ 1 = \_\_\_\_\_\_\_\_\_\_\_\_ 1   

3.2055606708805624550 to $10^2$, \_\_\_\_\_\_\_\_\_\_\_ 4 = \_\_\_\_\_\_\_\_\_\_\_\_ 4   

3.1290622219773513145 to $10^3$, \_\_\_\_\_\_\_\_\_\_ 16 < \_\_\_\_\_\_\_\_\_\_\_ 19   
 
3.1364540609918890779 to $10^4$, \_\_\_\_\_\_\_\_\_ 100 < \_\_\_\_\_\_\_\_\_\_ 105   

3.1384537326021492746 to $10^5$, \_\_\_\_\_\_\_\_\_ 620 > \_\_\_\_\_\_\_\_\_\_ 604   
 
3.1417076006640026373 to $10^6$, \_\_\_\_\_\_\_\_ 4123 > \_\_\_\_\_\_\_\_\_ 4046   

3.1417823471756806475 to $10^7$, \_\_\_\_\_\_\_ 29498 > \_\_\_\_\_\_\_\_ 29482   

3.1415377533170544536 to $10^8$, \_\_\_\_\_\_ 219893 < \_\_\_\_\_\_\_ 220419   

3.1415215264211035597 to $10^9$, \_\_\_\_\_ 1711775 < \_\_\_\_\_\_ 1712731   

3.1415248453830039795 to $10^{10}$, \_\_\_\_13706087 < \_\_\_\_\_ 13706592   

3.1415126339547108140 to $10^{11}$,   

3.1415144504088659201 to $10^{12}$,   

3.1415142045284687040 to $10^{13}$,   

3.1415144719058962626 to $10^{14}$,   

3.1415384423175311229 to $10^{15}$


Can we find a few more decimal places using the extrapolation method?



Twin primes conjecture and extrapolation method
From the product 
$$
\prod\_{(p\_1, p\_2): \text{twin primes}}t\_{(p\_1, p\_2)}
=\tfrac{3 ^ 2}{2 ^ 2} \cdot \tfrac{5 ^ 2}{3 ^ 2}\cdot 
\tfrac{5 ^ 2}{7 ^ 2}\cdot\tfrac{13 ^ 2}{11 ^ 2} \cdot\tfrac{17 ^ 2}{19 ^ 2} \cdot\tfrac{29 ^ 2}{31 ^ 2} \cdot\tfrac{41 ^ 2}{43 ^ 2} \cdot \tfrac{61 ^ 2 }{59 ^ 2} \cdot \tfrac{73 ^ 2}{ 71 ^ 2}\cdot \tfrac{101 ^ 2 }{ 103 ^ 2}\cdots
$$
We keep the terms that are $p\_1 \equiv 5$ mod $8$ or $p\_1 \equiv 7$ mod $8$


I conjecture that 
$$
\tfrac{5 ^ 2}{7 ^ 2}\cdot\tfrac{29 ^ 2}{31 ^ 2} \cdot\tfrac{73 ^ 2}{ 71 ^ 2}\cdot \tfrac{101 ^ 2 }{ 103 ^ 2}\cdot \tfrac{149 ^ 2 }{ 151 ^ 2}\cdot \tfrac{193 ^ 2 }{ 191 ^ 2}\cdot \tfrac{197 ^ 2 }{ 199 ^ 2}\cdot \tfrac{241 ^ 2 }{ 239 ^ 2}\cdots=5^.5/5
$$
or
$$
\tfrac{5 ^ 4}{7 ^ 4}\cdot\tfrac{29 ^ 4}{31 ^ 4} \cdot\tfrac{73 ^ 4}{ 71 ^ 4}\cdot \tfrac{101 ^ 4}{ 103 ^ 4}\cdot \tfrac{149 ^ 4}{ 151 ^ 4}\cdot \tfrac{193 ^ 4 }{ 191 ^ 4}\cdot \tfrac{197 ^ 4 }{ 199 ^ 4}\cdot \tfrac{241 ^ 4 }{ 239 ^ 4}\cdots=0.2
$$


So, Pi (if the product converges to Pi) included in the remaining terms of product.


$$
\tfrac{3 ^ 2}{2 ^ 2} \cdot \tfrac{5 ^ 2}{3 ^ 2}\cdot 
\tfrac{13 ^ 2}{11 ^ 2} \cdot\tfrac{17 ^ 2}{19 ^ 2} \cdot\tfrac{41 ^ 2}{43 ^ 2} \cdot \tfrac{61 ^ 2 }{59 ^ 2}\cdot \tfrac{109 ^ 2 }{107 ^ 2}\cdots=\sqrt {5}\*π
$$
or
$$
\tfrac{3 ^ 4}{2 ^ 4} \cdot \tfrac{5 ^ 4}{3 ^ 4}\cdot 
\tfrac{13 ^ 4}{11 ^ 4} \cdot\tfrac{17 ^ 4}{19 ^ 4} \cdot\tfrac{41 ^ 4}{43 ^ 4} \cdot \tfrac{61 ^ 4 }{59 ^ 4}\cdot \tfrac{109 ^ 4 }{107 ^ 4}\cdots=5\*π^2
$$


