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soandos
May25-09, 10:25 PM
Is there a function f(x) that will give the average number of prime factors for x_1 0<x_1<x, in a way similar to the way that Li(x)/x gives the approximate odds that a number from 0 to x is prime?

CRGreathouse
May26-09, 02:04 AM
log log x.

soandos
May26-09, 03:49 AM
I tried that for x = 1000, 10,000, 100,000, and it did not work for any of them.
I got the number of factors for 1000 to be 2.87 on average, 3.19 for 10,000, and 3.43 for 100,000
Did I do something wrong?

CRGreathouse
May26-09, 10:35 AM
There's a constant factor which depends on what you mean by "prime factor". From your numbers I take it you're counting indistinct prime factors, in which case the constant is 1.03465388....

It predicts an average of (2.97, 3.25, 3.48) versus your calculated (2.87, 3.19, 3.43). It will get more accurate as the numbers involved increase. For example, I calculated A071811 (http://www.research.att.com/~njas/sequences/A071811)(9) = 4044220058, which compares favorably to the predicted 4065910904.

It should be possible to work out a second-order term (which would be negative) to correct for the presence of small numbers, if you care about that kind of precision.

soandos
May26-09, 12:06 PM
where would the constant go?

CRGreathouse
May26-09, 01:07 PM
where would the constant go?

Expected number of prime factors per number up to x = 1.03465388... + log log x.

soandos
May26-09, 05:13 PM
how did you arrive at this constant?

CRGreathouse
May26-09, 07:03 PM
how did you arrive at this constant?

I didn't just derive it: the constant is well-known. It's B2, Sloane's A083342 (http://www.research.att.com/~njas/sequences/A083342).