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Old 2008-12-30, 00:39   #1
fivemack
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Default 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(fourth-largest prime factor of K > K^d) look like for large d (eg 0.15-0.25)?
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Old 2008-12-30, 03:58   #2
Jens K Andersen
 
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If each prime factor has the same number of digits then they are called brilliant numbers.
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Old 2008-12-30, 15:41   #3
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Quote:
Originally Posted by fivemack View Post
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?
I can't find the thread either (though I haven't seen it, so my ability to find it is limited). But for f(n, k) = 'first (n+1)-digit number with precisely k prime factors' minus 10^n:

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 2008-12-30 at 15:41
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Old 2008-12-30, 16:23   #4
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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 280-digit number having four factors larger than 10^60'
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Old 2008-12-30, 19:22   #5
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Quote:
Originally Posted by fivemack View Post
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 280-digit number having four factors larger than 10^60'
No, the right thing for that would be Dickman's function. (I wrote the Wikipedia article on the topic; check it out!)
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