Problems & Puzzles: Conjectures

Conjecture 72. Counting twin primes.

Mr. Islem Ghaffor, a young 22 years old from Algeria, sent a "formula" -that he claims to have proved- that counts exactly the quantity of twin primes not greater than a certain given value, 36n2 +60n +21.

Πp;p+2 (36n2 +60n +21) = Gn =The quantity of twin primes not great than 36n2 +60n +21, for n positive integers, 1, 2, ...

More than "a formula" I would say that Islem has produced "an algorithm" to make such count.

In short, the count for the quantity of twin primes not larger than 36n2 +60n +21, for a given integer n value, is equal to 6n2 +10n +4 minus the quantity of distinct terms in 4n arithmetic progressions concisely defined in the Islem's algorithm.

Independently of the practical value of the Islem's algorithm in order to compute Gn, the most striking feature is that the developed algorithm need not to make any primality test, just simple arithmetical operations and discard repeated terms on the 4n AP's involved.

Nowadays Islem only shares with us a paper that shows his formula and three examples effectively computed, but he prefers not to share in this moment the claimed 10 pages proof.

Please see directly the mentioned paper here.

Islem accepts that he has tested his formula only for n=1, 2, 3 & 4.

This is why Carlos Rivera has made some preliminary search in order to verify the Islem's formula for  n values larger than 4, and has found that, as a matter of fact, the Islem's formula gives correct values for n=1 to 18. Due to limitations of his code Rivera can not check the formula for n values larger than 18.

Q1. Would you like to make your own verification, specially beyond the Rivera's limit of search?

Q2. Do you devise any idea about the foundations of this formula? Does this algorithm hides a kind of Eratosthenes mesh?

Q3. Do you devise a way to prove that Gn computed by the Islem's formula goes to infinite as n goes to infinite?

Contributions came from Carlos Rivera, Dmitry Kamenetsky, Fred Schneider, Jeff Heleen and Emmanuel Vantieghem

In short, five different puzzlers have tested positively this conjecture, up to  different extents, and all of us await to see the formal Islem's proof of it. Nobody has an answer to Q2 & Q3 as far as December 12, 2014.

The paper by Francesca Balestrieri, 2011, (submitted by Emmanuel's contribution to this Conjecture) is another example that shows that the twin prime conjecture may be posed in a different terms, providing the expectative of a simpler solution for the old twin primes conjecture

"In this short paper we will show, via elementary arguments, the equivalence of the Twin Prime Conjecture to a problem which might be simpler to prove"

This is the same expectative produced by the Islem´s approach.


Rivera wrote:

I could positively verity this conjecture using a Ubasic code up to n=44


Dmitry wrote:

 I don't know how it works, but it seems to work. I have verified Islem's method for n<=2356. I will be very interested to see the proof.


Fred wrote:

Q1: I proved Mr. Ghaffor's conjecture thru n = 10000 (which translates to  confirming twin primes thru 3600600021).  Here are the last 3 lines:

Thru 3599160045 (n = 9998), we found: 10857111 vs. 10857111 = 599860008 - 589002897

Thru 3599879997 (n = 9999), we found: 10859135 vs. 10859135 = 599980000 - 589120865

Thru 3600600021 (n = 10000), we found: 10861113 vs. 10861113 = 600100004 - 589238891

10861113 is the twin prime count., calculated by checking all the primes through 3600600021.

600100004 is "G" for n=10000. 589238891 is the exclusion count.

My program took about 13 seconds to verify the conjecture after computed the primes. (If you build on the findings of n-1 and only check the new values introduced at the nth stage, this will make the search vastly more efficient.)

I find it very interesting what Mr. Ghaffor did and that it appears to work correctly but I don't think it's much more efficient than the sieve of Eratosthenes. It took be 31 seconds to generate ALL the primes under that threshold and then iterate through the list to get the twin prime count..  

I wonder if his algorithm can be modified to find prime counts of prime pairs with other gaps such as n, n+4.  A general heuristic would be very interesting to see.


Jeff wrote:

For Conjecture 72 Q1. I have no idea how to program this in UBASIC so instead
I made an Excel spreadsheet to do the problem (which I can provide for verification). The spreadsheet shows the Gn, l(vo;r), 4n AP's, calculated and actual counts.

On the spreadsheet I went to n=50. Assuming I made no errors, the algorithm gives correct values up to n=50. 


Emmanuel wrote:

As far as I can see, Islem's procedure is correct for  n = 1  to 100  and als for some bigger  n. In my opinion, the procedure is right for all  n  but I cannot see exactly why.

The method is very similar to an existing one that is based on the following theorem ; If  6n-1  and  6n+1  are simultaneously prime, then  n  is not of the form  6xy+x-y  or  6xy-x-y  or  6xy-x-x  (with x, y natural numbers)

Nevertheless, Islem's numbers (these are the numbers in the given arithmetic progressions) seem to have no direct relation to these forms.

I would like to see his proof !


On December 14, 2014, Islem wrote:

Q3 :Some ideas about how we can calculate lim G_n :

1- may be there is fonction f ( x ) :    G_n+1 = f ( G_n )
    or may be there is fonction f ( x ; y ) :   G_n+1 = f ( G_n ; n )
    if we find this fonction , may be we can calculate lim G_n

2- may be we can prove this conjecture :
    there is always at least one twin prime in ] 36n^2+60n+21 ; 36(n+1)^2+60(n+1)+21 [
    if we prove : always for any n :  G_n+1   >    G_n
                                                          G_n+1 - G_n  >  0

3- we suppose lim G_n = constant

     this mean  G_n+1 - G_n = 0  for any n   ≥  n' 

     and we must find contradiction by prove there is no existence of n'
     so lim G_n = infinite

4- transform the form of G_n :
    - may be by use the inclusion-exclusion principle
    - or may be by use other things

5- may be we can find f ( n ) :    G_n         f  ( n ) 
     with lim f ( n ) = infinite
     so lim G_n = infinite



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