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## Main Question or Discussion Point

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?

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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?

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CRGreathouse

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log log x.

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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?

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CRGreathouse

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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.

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.

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where would the constant go?

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CRGreathouse

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Expected number of prime factors per number up to x = 1.03465388... + log log x.where would the constant go?

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how did you arrive at this constant?

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CRGreathouse

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I didn't just derive it: the constant is well-known. It's Bhow did you arrive at this constant?

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