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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?

Dimitris Valianatos
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    It would be completely amazing if this product were equal to $\pi$! Someone should try to confirm the numerical evidence. – André Henriques Nov 05 '17 at 23:24
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    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 05 '17 at 23:38
  • @AndréHenriques Note the rapid jump in fifth decimal place between $10^{14}$ and $10^{15}$. Maybe the partial product oscillates chaotically Mertens function-style? The numbers have to be verified though. – Mikhail Tikhomirov Nov 05 '17 at 23:41
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    This depend on the distribution of the twin primes. If prevailed in an area the p2 / p1 > 1 then the product will increase and reach (in fifth decimal place) at 9. – Dimitris Valianatos Nov 05 '17 at 23:50
  • See 3.1415377533170544536 to 10e 8 and 3.1415384423175311229 to 10e 15 – Dimitris Valianatos Nov 05 '17 at 23:51
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    Convergence is very slow and unstable. See Brun's constant. – Dimitris Valianatos Nov 05 '17 at 23:54
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    From 2012: https://groups.yahoo.com/neo/groups/primenumbers/conversations/messages/24288 – jeq Nov 05 '17 at 23:56
  • Yes, but no ones help me to go higher or someone who knows best to apply the extrapolation method – Dimitris Valianatos Nov 06 '17 at 00:02
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    @DimitrisValianatos The numbers here are very random and sensitive, so I don't think any kind of extrapolation would work. It is interesting how to compute the (partial) product faster than in linear time (for instance, the $O(n^{2 / 3})$ prime counting method doesn't seem to generalize here). – Mikhail Tikhomirov Nov 06 '17 at 00:28
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    The logarithms are partial sums, which are basically summations with a sharp cut-off. If you replace the sharp cut-off by a smooth cut-off where after the first N terms, you multiply the $N+k$ th term by a weight $\exp(-a k/N)$ where $a$ is your smooth cut-off parameter, then this will behave in a smooth way as a function of $N$. So, you should be able to extrapolate this for fixed $a$, and in these results you can then extrapolate a to infinity. – Count Iblis Nov 06 '17 at 00:30
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    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 06 '17 at 01:41
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    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 06 '17 at 01:53
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    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 06 '17 at 16:11
  • @Agno so it converges but not quite to what one might naively guess and probably not to any otherwise nameable constant. – Aaron Meyerowitz Nov 06 '17 at 22:47
  • @AaronMeyerowitz. It indeed seems to be the case, however I haven't firmly landed on a stable 6-th digit yet since takes an increasing amount of time to compute the primes for $k > 10^{10}$. If you are interested: I used this simple Pari/GP code in the SageMathCloud: gp("1.0/2(log(precprime(20000000000)))^2prodeuler(p=1,20000000000,(p-1)^2/(p^2-1)-1/(p^2))"). This covers roughly $9 \cdot 10^{8}$ primes and yields 0.0445263... instead of 0.0445252... – Agno Nov 07 '17 at 00:00

4 Answers4

13

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.

$2^{26}, 3.141825366$

Community
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Aaron Meyerowitz
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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 $x$.

Rodrigo
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4

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.

Dimitris Valianatos
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    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 08 '17 at 18:26
  • From 10^6 to 10^9 changed only 3.1416095240-3.141609248=0.000000275.. – Dimitris Valianatos Nov 08 '17 at 22:42
  • The line 10^6, 3.141609524070628 is impressively close to π. The next line is not as impressive 10^7, 3.141704821787704. Therefore we must continue to get closer to the truth. – Dimitris Valianatos Nov 08 '17 at 23:02
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    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 08 '17 at 23:22
  • ... multiplying the first two identities above, dividing by $(1-\frac1{2^2})$, and dividing by the square of the third identity shows that the product over odd composite numbers described in the post does equal $\pi/4$. – Greg Martin Nov 08 '17 at 23:23
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    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 08 '17 at 23:40
  • Yes this is the proof I have written to OEIS before. I have mentioned that the Wallis product not needs the prime numbers to converges to PI / 4 for p-primes > = 5. The above example was to show that someone would be disappointed quickly by watching the experimental data. If he did not know the proof. – Dimitris Valianatos Nov 09 '17 at 08:08
  • I do not think you are producing this result. Do not I understand something? (... multiplying the first two identities above, dividing by 3/4, and dividing by the square of the third identity shows that the product over odd composite numbers described in the post does equal Pi/4) = ? (4/5) * (8/7) * (10/11) * (12/13) * (14/13) * (16/17) * (18/17) * (20/19) * (22/23) * ... ? – Dimitris Valianatos Nov 10 '17 at 15:27
  • I would say "if he or she did not know the proof". – Greg Martin Nov 11 '17 at 18:12
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    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
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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 $$ \tfrac{3 ^ 4}{2 ^ 4} \cdot \tfrac{5 ^ 4}{3 ^ 4}\cdot \tfrac{13 ^ 4}{11 ^ 4} \cdot\tfrac{17 ^ 4}{19 ^ 4} \cdot\tfrac{41 ^ 4}{43 ^ 4} \cdot \tfrac{61 ^ 4 }{59 ^ 4}\cdot \tfrac{109 ^ 4 }{107 ^ 4}\cdots=5*π^2 $$

Dimitris Valianatos
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