20060124, 16:56  #1 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Random numbers and proper factors
In my current reading of Maths I have come across the following problem which I am at a loss to ascertain.
Take three positive whole numbers at random. What is the chance they have no proper factor in common? Answer around 83% to be precise, 0.83190737258070746868......:surprised Can anyone elucidate? Mally 
20060124, 17:11  #2  
Nov 2003
2^{2}×5×373 Posts 
Quote:


20060124, 17:37  #3 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
I suppose the analysis will be similar to that for two randomly chosen integers. We had that one before in this forum. The result for two random integers was 1/zeta(2), the results for 3 random integers should be 1/zeta(3) which matches your constant.
Alex 
20060124, 17:59  #4  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Random numbers and proper factors.
Quote:
Could you link this number with another well known number in electro dynamics ? Its amazing ! Mally Last fiddled with by mfgoode on 20060124 at 17:59 

20060124, 18:18  #5 
Aug 2002
Buenos Aires, Argentina
3·449 Posts 
The quotient for n even is always rational, but for odd values of n, such as in this case there is no known explicit finite expression using elementary functions and constants.
Last fiddled with by alpertron on 20060124 at 18:19 
20060125, 15:39  #6  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
Quote:
Thanking you, Mally 

20060125, 15:47  #7  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Quote:
We know that Z(2) = (pi^2)/6. Would you call this a rational number? Similarly Z(4) = (pi^4)/90. Mally 

20060125, 17:23  #8  
"Richard B. Woods"
Aug 2002
Wisconsin USA
1111000001100_{2} Posts 
Quote:


20060125, 23:49  #9 
Nov 2005
30_{16} Posts 
What does "two randomly chosen integers" mean? You can't put a uniform distribution on the entire set of integers, so there is some interpretation involved.
Specifically, you seem to be using the following heuristic: Let $n$ be an random integer and $p$ be a prime. Then $Pr(p  n) = 1/p$. This is perfectly sensible and number theorists use this kind of reasoning all the time. I was wondering, though, how you formulate this precisely,. 
20060126, 00:17  #10 
Nov 2005
2^{4}×3 Posts 
Thinking about this some more... you are making some statement over the interval (0, n) and then take a limit as n >\infty. Is there somewhere easily accessible these details are written down? I'm still grappling with whether this really corresponds to "random integers," but I'll do the philosophical musings on my own time.
Simpler question: How do I put this as a postscript to my previous post, rather than a new posting? 
20060126, 16:28  #11  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
Random numbers and proper factors.
Quote:
These rational numbers I take it are of even no.s and no odd has been found to describe the quotient rationally. Mally 

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