## prime powers < X

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• 1. number of prime powers Posted by: Werner D. Sand Theo.3.1415@web.de theo2357 Date: Tue Dec 18, 2007 4:00 pm ((PST)) Who knows a good approximate formula
Message 1 of 2 , Dec 19, 2007
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1. number of prime powers
Posted by: "Werner D. Sand" Theo.3.1415@... theo2357
Date: Tue Dec 18, 2007 4:00 pm ((PST))

Who knows a good approximate formula for the number of prime powers up
to x (without simple prime numbers):
N = sum(1)(p^n <= x), p prime, n>1 ?

*******

Would this not be the number of primes < x
+ the number of primes < x^(1/2)
+ the number of primes < x^(1/3)
+ the number of primes < x^(1/4)
+ etc

approximately equal to

[ x + 2 * x^(1/2) + 3 * x^(1/3) + 4 * x^(1/4) + 5 * x^(1/5)
+ . . . ] / ln(x)

Kermit < kermit@... >
• ... up ... (1/5) ... That s allright. Can you sum it up into a closed form? WDS
Message 2 of 2 , Dec 20, 2007
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--- In primenumbers@yahoogroups.com, Kermit Rose <kermit@...> wrote:
>
>
>
>
> 1. number of prime powers
> Posted by: "Werner D. Sand" Theo.3.1415@... theo2357
> Date: Tue Dec 18, 2007 4:00 pm ((PST))
>
> Who knows a good approximate formula for the number of prime powers
up
> to x (without simple prime numbers):
> N = sum(1)(p^n <= x), p prime, n>1 ?
>
>
> *******
>
>
> Would this not be the number of primes < x
> + the number of primes < x^(1/2)
> + the number of primes < x^(1/3)
> + the number of primes < x^(1/4)
> + etc
>
>
> approximately equal to
>
> [ x + 2 * x^(1/2) + 3 * x^(1/3) + 4 * x^(1/4) + 5 * x^
(1/5)
> + . . . ] / ln(x)
>
>
>
>
> Kermit < kermit@... >

That's allright. Can you sum it up into a closed form?
WDS
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