Number of prime factors

soandos · May 25, 2009

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 · May 26, 2009

log log x.

soandos · May 26, 2009

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 · May 26, 2009

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 http://www.research.att.com/~njas/sequences/A071811 [Broken](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 · May 26, 2009

where would the constant go?

CRGreathouse · May 26, 2009

soandos said: ↑

where would the constant go?

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

soandos · May 26, 2009

how did you arrive at this constant?

CRGreathouse · May 26, 2009

soandos said: ↑

how did you arrive at this constant?

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

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