Re: P(prod_i p_i^{k_i}|n)=?
- From: temptony@xxxxxxxxxxx
- Date: Fri, 30 Nov 2007 08:42:49 -0800 (PST)
On Nov 30, 5:40 pm, tempt...@xxxxxxxxxxx wrote:
On Nov 30, 12:35 pm, "I.N. Galidakis" <morph...@xxxxxxxxxxxx> wrote:
Apologies if this is a stupid question.
What is the probability that n has the form prod_i p_i^{k^i}m, p_i prime?
Not stupid, but more complicated than it needs to be. Let x = prod_i
p_i^k_i. Then if I understand you correctly, you are just asking for
the probability that x divides n.
*snip*
P(n=prod_i p_i^{k_i}m)=prod_i p_i^{-k_i}
In other words, P(x divides n) = 1/x. Which is true, if you use a
reasonable definition of the left-hand side: for instance, the limit
as N tends to infinity of P(x divides n), when x is equal to one of
1,2,...,N with equal probability.
I mean 'when n is equal to one of...'.
.
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