20080116, 18:47  #1 
May 2004
New York City
2×2,099 Posts 
Connected Primes
Call a trio {a,b,c} of primes singlyconnected
if all three of {a',b',c'} are prime, where a' = 2a+bc, b' = 2b+ac, c' = 2c+ab. For example, {3,5,7} are singlyconnected, because {41,31,29} are prime. Call the trio doublyconnected if {a',b',c'} are singlyconnected, and triplyconnected if {a',b',c'} are doubly connected. The puzzle is: find a triplyconnected trio {a,b,c} of primes. (I present this as an open question because I haven't yet.) 
20080116, 19:31  #2 
Jan 2005
Transdniestr
503 Posts 
Do the primes have to be consecutive?

20080116, 20:02  #3 
May 2004
New York City
2×2,099 Posts 
No, not necessarily consecutive.

20080116, 20:19  #4  
Nov 2003
2^{6}×113 Posts 
Quote:
there should be ininfinitely many such sets. However, they will be very sparse. You want a,b,c, 2a+bc, 2b+ac, 2c+ab, 2(2a + bc) + (2b+ac)(2c+ab), 2(2b+ac) + (2a+bc)(2c+ab), 2(2c+ab) + (2a + bc)(2b+ac), etc. to all be prime. This a lot of conditions on simultaneous primality. If a,b,c are near N, then a', b', c' are near N^2 and a'', b'', c" are near N^4, so the probability of all being prime is P ~ 1/log^3 N * 1/log^3 (N^2) * 1/log^3(N^4) and this is quite small. You want the number of such sets less than N to be at least 1, so we require N*P > 1, on average to find such a set. N will have to be quite big. 

20080116, 20:50  #5 
Jan 2005
Transdniestr
111110111_{2} Posts 
Davar55,
You should raise this puzzle with Carlos Rivera over at www.primepuzzles.net. He loves puzzles like this. 
20080117, 01:38  #6 
Oct 2007
London, UK
2^{2}×3×109 Posts 
Well I've written a 100 line piece of code in JavaScript (it's the language I'm most fluent in) and determined that when all three arguements are less than 1000, there are 736 singly connected triplets, but no doubly or triply connected triplets.
Of those 736, this if the largest one: 3, 991, 997 988033, 4973, 4967 I may translate my code into C which will allow me to check bigger numbers much more quickly. An interesting thing I noticed is that the overwhelming majority of the results (695 out of 736) have 3 as the first arguement, with the remainder (41 out of 736) having 5 as the first. Edit: On a whim I just checked arguements up to 10000 and there are still no doubly connected triplets. Last fiddled with by lavalamp on 20080117 at 01:56 
20080117, 05:02  #7 
Jan 2005
Transdniestr
503 Posts 
I think there's a problem with your code. 7, 11, 13 is singlyconnected. It gives you : 157, 113, 103, all prime
I found a bunch of doublyconnected trios: Examples: 5, 7, 709 7, 11, 3967 Looking for triplyconnected trios now ... P.S. You should try PARI. It's great for number crunching like this. Last fiddled with by grandpascorpion on 20080117 at 05:10 
20080117, 07:47  #8 
Oct 2007
London, UK
2^{2}·3·109 Posts 
Ah, I made a slight error, breaking a loop where I shouldn't have been breaking a loop, all fixed now. The abundance of 3's with a few 5's for the first arguement was a symptom of the problem.
I now find 9815 singly connected and 68 doubly connected triplets when the size limit for the arguements is 1000. I also find 39529 singly and 206 doubly triplets up to 2000. I've not checked for triply triplets because the numbers are simply too big. Here are the last two singles to be outputted: Code:
257, 1993, 1999 3984521, 517729, 516199 613, 1997, 1999 3993229, 1229381, 1228159 Code:
521, 1553, 1997 3102383, 1043543, 813107 848518322867, 2522571421067, 3237471689183 19, 1259, 1999 2516779, 40499, 27919 1135725139, 70266033899, 101927088559 
20080117, 13:41  #9 
Jan 2005
Transdniestr
503 Posts 
I checked through 5000. No triplyconnected trios found.

20080118, 01:40  #10 
"Jason Goatcher"
Mar 2005
5×701 Posts 
Too bad I'm not a programmer, sounds like a Linux and Windows version could be relatively easily made, given sufficient skills.

20080118, 10:52  #11  
"Lucan"
Dec 2006
England
14463_{8} Posts 
Quote:
I note that a,b,c must all be different and odd. I think I follow RDS's post for a change: There are ~N^3/6 triples with a limit of N on each member. The probability of them all being prime is ~(1/ln(N))^3. a',b',c' are ~N^2 but given a,b,c are odd primes, a',b',c' must be odd. So the probability all of them being prime is ~(2/ln(N^2))^3 =(1/ln(N))^3. So the expected number of singly connected triples is N^3/(6(ln(N))^6) Does this tally with the data tested so far? David Last fiddled with by davieddy on 20080118 at 11:20 

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