20081230, 00:39  #1 
(loop (#_fork))
Feb 2006
Cambridge, England
13·491 Posts 
Nomenclature question
I'm sure I recall a thread here about the smallest numbers with K digits and precisely two, three or four prime factors, but I can't remember what they were called and so I can't find the thread. Could anyone help me out?
What does the function p(fourthlargest prime factor of K > K^d) look like for large d (eg 0.150.25)? 
20081230, 03:58  #2 
Feb 2006
Denmark
2×5×23 Posts 
If each prime factor has the same number of digits then they are called brilliant numbers.

20081230, 15:41  #3  
Aug 2006
3^{2}·5·7·19 Posts 
Quote:
f(n, 1) = 1,1,9,7,3,3,19,7,7,19,3,39,37,31,37,61,3,3,51,39,117,9,117,7,13,67,103,331,319,57 f(n, 2) = 0,6,3,1,1,1,1,1,6,3,7,7,15,13,3,3,15,7,1,19,3,19,13,29,7,7,3,27,13,3 f(n, 3) = 2,2,1,2,6,2,5,6,3,1,6,1,1,18,2,6,7,1,7,3,5,7,3,1,3,1,6,13,1,7 f(n, 4) = 6,0,12,4,2,5,2,5,10,12,4,6,3,1,4,2,1,6,5,1,2,1,4,3,5,6,2,1,6,13 f(n, 5) = ,8,20,10,20,4,4,2,1,15,1,12,2,2,11,1,5,4,3,11,14,2,1,2,1,2,4,2,11,2 f(n, 6) = ,44,0,8,40,20,8,4,12,16,16,4,12,30,5,8,2,2,2,4,4,14,2,9,22,12,11,5,2,4 f(n, 7) = ,28,8,32,8,125,16,40,20,8,8,8,8,16,1,16,10,38,16,5,17,24,12,4,2,5,1,4,20,1 f(n, 8) = ,156,296,0,32,8,96,8,32,64,32,32,40,26,8,50,24,20,8,8,1,40,8,12,4,35,36,8,24,16 f(n, 9) = ,412,152,80,320,64,32,32,8,32,62,80,25,8,107,10,32,8,71,89,32,8,144,8,16,20,50,10,16,8 f(n, 10) = ,,24,1664,0,320,80,960,80,152,224,125,32,368,92,32,8,32,64,48,8,10,32,80,8,40,89,224,48,64 for n = 1, 2, ..., 30. I had hoped this would dig something up at Sloane's that would lead you to your thread, but no such luck thus far. Last fiddled with by CRGreathouse on 20081230 at 15:41 

20081230, 16:23  #4 
(loop (#_fork))
Feb 2006
Cambridge, England
6383_{10} Posts 
Brilliant numbers were what I was thinking of, though of course they're not quite the right things for answering the question 'what's the probability of a 280digit number having four factors larger than 10^60'

20081230, 19:22  #5  
Aug 2006
3^{2}×5×7×19 Posts 
Quote:

