20170128, 18:06  #2  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
Last fiddled with by science_man_88 on 20170128 at 18:08 

20170128, 18:17  #3 
"Sam"
Nov 2016
2^{2}×3^{4} Posts 
That source includes exponents > 2, which is fine. What about exponents > 1?
Is O (x^(1/2) log x) + x/(log (x)) a good estimate? Anyone think of a better one? Thanks sm88 for that? (This is for the number of prime powers < x, not the probability that n is a prime power, what was originally asked which should easily be inferred from the counting of prime powers < x.) Last fiddled with by carpetpool on 20170128 at 18:19 
20170128, 18:31  #4  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:


20170128, 22:46  #5 
Aug 2006
3×1,993 Posts 
The 'probability', in an appropriate sense, that a random number near x is a prime power with exponent >= 1 is 1/log x. For exponent >= 2 the 'probability' is 1/sqrt(x log x). For exponent >= k the 'probability' is (x log x)^(1/k)/x.
Last fiddled with by CRGreathouse on 20170128 at 22:47 
20170130, 01:43  #6  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2×31×101 Posts 
Quote:
1/sqrt(x log x) != (x log x)^(1/k)/x 

20170130, 02:54  #7 
Aug 2006
3×1,993 Posts 
Oops. I think I was wrong on both.
I think the probability should be 1/(x^(1  1/k) log x) which is 1/(sqrt(x) log(x)) for k = 2. You get a different constant term if you look at the numbers up to x rather than near x as I did. 
Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Probability that N is prime given each divisor of N has the form 2*k*p+1  carpetpool  Miscellaneous Math  6  20170901 13:59 
Probability that n is a semiprime or prime  carpetpool  Miscellaneous Math  27  20170119 21:00 
probability a number is prime with a weighted k.  Trilo  Homework Help  12  20140606 19:17 
Sieve depth vs. prime probability  Unregistered  Information & Answers  2  20100525 20:51 
Probability of finding a prime number  Deamiter  Software  4  20021011 16:36 