20180926, 10:33  #1 
May 2016
2×3^{4} Posts 
Is this (prime numbers) formula known ?
Good morning,
Is this formula known? does it work or does it not work? The formula is : Thank you. . 
20180926, 10:47  #2 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}×5×307 Posts 

20180926, 11:00  #3 
May 2016
2×3^{4} Posts 

20180926, 11:08  #4 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}×5×307 Posts 

20180926, 11:16  #5 
May 2016
2·3^{4} Posts 

20180926, 11:36  #6 
May 2016
2×3^{4} Posts 
I found a counterexample that does not work.

20180926, 13:50  #7 
Aug 2006
3^{2}·5·7·19 Posts 
With the primes up to 10^5 this works 10240881 out of 91996872 times (about 11%).

20180926, 13:57  #8 
Romulan Interpreter
Jun 2011
Thailand
24E5_{16} Posts 

20180926, 15:53  #9 
Aug 2006
3^{2}·5·7·19 Posts 
559526371/6161857506 for primes up to a million. (I'm done.)

20180926, 18:25  #10 
Feb 2017
Nowhere
2^{3}×3^{4}×7 Posts 
I'm not sure what the or means [possibilities include the bitwise OR of p1 and p2, which in PariGP would be bitor(p1,p2)].
For the sake of discussion I will assume it means p1*p2 + p1  1 or p1*p2 + p2  1. It occurred to me that one could make both expressions divisible by the same prime q. This would require that p1 == p2 (mod q). Calling the common residue class x, we have x^2 + x  1 == 0 (mod q). This quadratic has two solutions (mod q) when 5 is a quadratic residue (mod q), i.e. when q == 1 or 9 (mod 10). For q = 11, the two values of x are 3 (mod 11) and 7 (mod 11). Thus, if p1 and p2 are both congruent to 3 (mod 11) or both are congruent to 7 (mod 11), both p1*p2 + p1  1 and p1*p2 + p2  1 will be divisible by 11. For example, we could take p1 = 3 and p2 = 47, or p1 = 7 and p2 = 29. 
20180927, 02:23  #11 
Feb 2017
Nowhere
4536_{10} Posts 
I was thinking about the possibility that "p1 or p2" meant bitor(p1, p2), and noticed that, if 2 < p1 < p2, and 2^(r1) < p1 < 2^r, then
bitor(p1,p2) = p2 when p2 == p1 (mod 2^r). This would make the expression equal to p2*(p1 + 1)  1. If q is a prime which does not divide p1 + 1, p1 + 1 has a multiplicative inverse (mod q), and the expression is divisible by q when p2 == (p1 + 1)^{1} (mod q). OK, so it's easy to construct counterexamples in which bitor(p1,p2) = p2 and q divides p1*p2 + p2  1, provided q does not divide p1 + 1. Then, I noticed something curious. The first 2 odd primes are 3 and 5. The congruence classes 3 (mod 4) and 5 (mod 8) cover the congruence classes 3, 5, and 7 (mod 8). That only leaves the class 1 (mod 8) uncovered. So it occurred to me to wonder whether it ever does get covered by classes p1 (mod 2^r). Since the first prime congruent to 1 (mod 8) is 17 and the next is 41, things start off badly. You need to cover 8 residue classes (mod 64) and only 3 of them (17, 49, and 41) are covered. That leaves 5 classes (mod 64). Make that 10 classes (mod 128). You have to cover 1/4 of the odd residue classes (mod 2^r) for some r (those congruent to 1 (mod 8)), and the primes congruent to 1 (mod 8) are about a quarter of the primes, and the primes keep getting thinner on the ground. So I sort of doubt the class 1 (mod 8) ever gets covered. 
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