Number of prime factors

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SUMMARY

The discussion centers on the function f(x) that estimates the average number of prime factors for integers in the range 0 < x_1 < x, analogous to the prime-counting function Li(x)/x. The average number of prime factors calculated for x values of 1000, 10,000, and 100,000 were 2.87, 3.19, and 3.43, respectively, which were compared to a predicted average of 2.97, 3.25, and 3.48. The constant factor involved in this calculation is approximately 1.03465388, which accounts for indistinct prime factors. The accuracy of the predictions improves with larger numbers, and a second-order term may be necessary for precision corrections.

PREREQUISITES
  • Understanding of prime factorization and its significance in number theory
  • Familiarity with logarithmic functions, specifically log log x
  • Knowledge of mathematical constants and their applications in number theory
  • Experience with mathematical sequences, particularly those cataloged by Sloane's OEIS
NEXT STEPS
  • Research the properties of the prime-counting function Li(x) and its applications
  • Study the derivation and implications of the constant 1.03465388 in prime factorization
  • Explore the concept of second-order terms in mathematical approximations
  • Investigate Sloane's OEIS sequences, particularly A071811 and A083342, for further insights
USEFUL FOR

Mathematicians, number theorists, and students interested in prime factorization and its statistical properties will benefit from this discussion.

soandos
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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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log log x.
 
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?
 
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 (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?
 
soandos said:
where would the constant go?

Expected number of prime factors per number up to x = 1.03465388... + log log x.
 
how did you arrive at this constant?
 
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 .
 
Last edited by a moderator:

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