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Could someone please point me in the direction of a proof for Landau's asymptotic formula for k-almost primes:

$$\pi_k(n) \sim \left( \frac{n}{\log n} \right) \frac{(\log\log n)^{k-1}}{(k - 1)!}$$

I realise that it was derived directly from the PNT - would like to see the steps involved though.

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    $\begingroup$ I'm not sure, but it might be in Davenport's Multiplicative Number Theory. $\endgroup$ – Gerry Myerson Jan 2 '14 at 18:51
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    $\begingroup$ @martin: ALso, have you checked the reference here? en.wikipedia.org/wiki/Almost_prime, that is: ^ Tenenbaum, Gerald (1995). Introduction to Analytic and Probabilistic Number Theory. Cambridge University Press. ISBN 0-521-41261-7. $\endgroup$ – Amzoti Jan 2 '14 at 18:58
  • $\begingroup$ Thanks for the suggestions - can't seem to find it in Davenport though - do you have a chapter ref? $\endgroup$ – martin Jan 2 '14 at 19:17
  • $\begingroup$ See also the discussion at mathoverflow.net/questions/35927/… $\endgroup$ – Gerry Myerson Jan 3 '14 at 1:09
  • $\begingroup$ I note that just a couple of weeks ago, you claimed to have a better formula than Landau's; math.stackexchange.com/questions/607895/… $\endgroup$ – Gerry Myerson Jan 3 '14 at 1:13
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The original source, as far as I know, is Landau's Handbuch der Lehre von der Verteilung der Primzahlen. An approachable modern version (in English!) is

Gerald Tenenbaum, Introduction to Analytic and Probabilistic Number Theory. Cambridge University Press (1995).

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  • $\begingroup$ Thanks - yes, English is preferable! $\endgroup$ – martin Jan 2 '14 at 19:17
  • $\begingroup$ Try as I might, I can't find a proof of it. At the start of chapter II.6, Tenenbaum says that"It is easy to show that ....", but I can't find where he shows it! $\endgroup$ – martin Jan 2 '14 at 19:56
  • $\begingroup$ @martin: I'm pretty sure I read Tenenbaum's proof of the theorem, so either you're missing it or it's in another of his works. I'll see if I can find anything, though my uni library is snowed out at the moment (!). $\endgroup$ – Charles Jan 2 '14 at 20:05
  • $\begingroup$ Great, thanks - much appreciated! :) $\endgroup$ – martin Jan 2 '14 at 20:50

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