# Twin primes conjecture and extrapolation method

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?

• It would be completely amazing if this product were equal to $\pi$! Someone should try to confirm the numerical evidence. – André Henriques Nov 5 '17 at 23:24
• A summary glance at your own numerical evidence seems to indicate that your number starts by 3.141514, and is therefore not equal to $\pi$. – André Henriques Nov 5 '17 at 23:38
• Heuristically, we expect about $N / \log^2 N$ twin primes between $N$ and $2N$ (up to constants), about half of which should start with a number that is 1 mod 4. Each such term modifies your product by $1 \pm O(1/N)$. Square root cancellation heuristics then suggest that your product should fluctuate by about $\frac{1}{\sqrt{N} \log N}$ on this block, and hence your product should converge at a rate $O( 1/\sqrt{N} \log N )$. On the other hand, if one multiplies the discrepancy from $\pi$ in your data by $\sqrt{N} \log N$ for $N = 10^1,\dots,10^{15}$, it looks rather divergent. – Terry Tao Nov 6 '17 at 1:41
• As a general rule of thumb, a single agreement to N decimal places is not terribly surprising or impressive if it takes significantly more than N characters to state the two quantities to which one has agreement. But if one had numerical identities of this form for many moduli, not just modulus 4, the situation would be rather different. – Terry Tao Nov 6 '17 at 1:53
• I definitely do not want to 'diverge' a discussion about 'convergence', however just to share an example on how deceitful the apparent convergence of products over large numbers of primes could be (not saying that is the case for the question!). The following product over primes $p$ seems at first sight to nicely converge at 6 digits accuracy towards $\pi^{-e}$: $$\lim_{N \to \infty} \ln^2(p_N)\cdot\frac12\cdot \prod^N_{n=1} \left( \frac{(p_n-1)^2}{p_n^2-1}-\frac{1}{p_n^2}\right)=\pi^{-e}$$ but then when $N>10^9$ the 6-th digit starts to wander of in the wrong direction... – Agno Nov 6 '17 at 16:11

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.

• After $10^{10}$ terms this is approximate to about $10^{-5}$. That's what one should expect. Your data in the question looks different, as Terry Tao pointed out. – Lucia Nov 8 '17 at 18:26
• From 10^6 to 10^9 changed only 3.1416095240-3.141609248=0.000000275.. – Dimitris Valianatos Nov 8 '17 at 22:42
• This identity can be proved, given Wallis's formula, the product formula for $1/\zeta(2)$, and Liebniz's formula: \begin{align*} \frac\pi2 &= \frac21 \frac23 \frac43 \frac45 \frac65 \frac67 \frac87 \frac89 \cdots \\ \frac6{\pi^2} &= \prod_p \bigg( 1-\frac1{p^2} \bigg) \\ \frac\pi4 &= \prod_p \bigg( 1-\frac{\chi_{-4}(p)}p \bigg)^{-1} = \frac34 \frac54 \frac78 \frac{11}{12} \frac{13}{12} \frac{17}{16} \cdots. \end{align*} ... – Greg Martin Nov 8 '17 at 23:22
• The convergence rate here is expected to be of order $O( N^{-1/2} \log^{-1/2} N )$ (coming from the fluctuations of the primes). And indeed, if one multiplies the discrepancy of the $N^{th}$ product from $\pi$ by $N^{1/2} \log^{1/2} N$, the normalised error stays bounded numerically. This is in contrast to the divergence experienced with the product in the OP. – Terry Tao Nov 8 '17 at 23:40
• Sorry for my bad English. I am "he". For more see "Chebyshev's bias and the magic key" – Dimitris Valianatos Nov 12 '17 at 17:19

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

• $\sqrt{20} / 10 = \sqrt{5} /5$ might be clearer, unless you want to emphasize $\sqrt{20}$... – Joseph O'Rourke Nov 14 '17 at 1:00
• I find the result in OEIS (A010476) 4.47213595499957939 and divided it with 10 equals 0.447213595499957939. – Dimitris Valianatos Nov 14 '17 at 1:14