## TWIN PRIME S conjectu re (Dimitr is Valiana tos GREECE ) 23/02/20 12 3:53 mm ‏

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• Hi Thank you for your interest. (Warren D. Smith, David Broadhurst, Mark, Jack Brennen, Maximilian Hasler,....) Today not found my son and I try to translate
Message 1 of 1 , Feb 22, 2012
Hi
(Warren D. Smith, David Broadhurst, Mark, Jack Brennen, Maximilian Hasler,....)

He is 22 years old and talented piano soloist.
Ends the Bachelor and preparing Master at the Juilliard School of Music in New York with scholarship.
His name is Konstantine Valianatos his tel. 6463632612
and email: kon_val@...
He helps me to contact you.

You are right, probably this is a coincidence and tricked me.
I tried to answer and reject the hypothesis of my own powers.
But as you see from the results was impossible.
These results regress (palindromic).
So I turned to you for help.
More minds are better than one and that old!

At the beginning of the calculations I was disappointed (at least I said, ok finished, the conjecture is invalid)
because the product seemed to be stabilizing
below the pi. But I insisted until 10^15 and showed that turns up.
(Very mysterious numbers are the twin primes).
It seems that the number pi knows who is the next pair of twin primes,
in the same way he knows who are the prime numbers.

I will try to analyze in details, giving you the whole story of my results.

The reason that made me believe it,
is that the product Π p1/p2 (where p1 mod 4 = 1) I believe that converges?
at sqrt (pi) / C, perhaps at sqrt (pi) / (3 * pi) = 1 / (3 * sqrt (pi)) = 0.1880631945159187622
and the other
Π p2/p1 (where p1 mod 4 = 3 or 2 for pair (2,3)) converges? at C * sqrt (pi),
perhaps at 3 * pi * sqrt (pi) = 16.7049839904951235354.
where (9 < C <= 3 * pi <= pi^2), maybe C=3*pi
Thus Π p1/p2 * Π p2/p1 = sqrt(pi)/C * sqrt(pi)*C =pi (maybe (1 / (3 * sqrt (pi))) * (3 * sqrt (pi) * pi) = pi).
I have calculated the two products separately from one another.

My results for product Πp1/p2 are: (p1,p2) is twin prime

0.5102040816326530610 to 1e 1
0.3249660412937108680 to 1e 2
0.2662953618212302009 to 1e 3
0.2415133533943608408 to 1e 4
0.2284545798441555610 to 1e 5
0.2200551480500073917 to 1e 6
0.2140714784028322344 to 1e 7
0.2097282124057264264 to 1e 8
0.2064151736456673244 to 1e 9
0.2038017137618014013 to 1e 10
0.2016875621577279403 to 1e 11
0.1999429266128650164 to 1e 12
0.1984784595986039453 to 1e 13
0.1972317491276009178 to 1e 14
0.1961586216569405678 to 1e 15 --> (may be 0.1880631945159187622=sqrt (pi) / (3 * pi) )
I can not doing conjecture for this, but only to intuit.
I need more precision to make sure.

The number (enumerator) of (p1/p2)

1 to 1e 1 (5,7) --> (5/7) ^2 = 0.5102040816326530610
4 to 1e 2
19 to 1e 3
105 to 1e 4
604 to 1e 5
4046 to 1e 6
29482 to 1e 7
220419 to 1e 8
1712731 to 1e 9
13706592 to 1e 10
112196635 to 1e 11
935286453 to 1e 12
7917373154 to 1e 13
67890137706 to 1e 14
588604442555 to 1e 15

My results for product Πp2/p1 are: (p1,p2) is twin prime

6.25 to 1e 1
9.86429430631895492790 to 1e 2
11.7503444317515306555 to 1e 3
12.9866693369556925958 to 1e 4
13.7377580031142458498 to 1e 5
14.2769102586504985344 to 1e 6
14.6763238644224561900 to 1e 7
14.9790899244382275820 to 1e 8
15.2194311635919183042 to 1e 9
15.4146144671518627696 to 1e 10
15.5761346924205558509 to 1e 11
15.7120559533048919150 to 1e 12
15.8279856206147634696 to 1e 13
15.9280363623073028917 to 1e 14
16.0152962728894593514 to 1e 15 --> ( may be 16.7049839904951235354=3 * pi * sqrt (pi) )
I can not doing conjecture for this, but only to intuit.
I need more precision to make sure.

The number of (p2/p1)

2 to 1e 1 (2,3)?( exceptionally) and (3,5) --> ( 3/2)^2*(5/3)^2=6.25
5 to 1e 2
17 to 1e 3
101 to 1e 4
621 to 1e 5
4124 to 1e 6
29499 to 1e 7
219894 to 1e 8
1711776 to 1e 9
13706088 to 1e 10
112179414 to 1e 11
935298768 to 1e 12
7917291719 to 1e 13
67890183960 to 1e 14
588604799750 to 1e 15

I hope that i will slowly gather all my work for the prime numbers in files pdf, now there are
over thousands of papers.

Best wishes,
Dimitris Valianatos

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