Re: Curiosity

[Karim Belabas <Karim.Belabas@math.u-bordeaux.fr>]

* Andrea Ribuoli [2021-01-07 21:05]:
>   I have installed PariGP and tried -in sequence- the following calculations (manually extending the number of iterations):
> 
>             c=0.0; for (i=1,    1000000, c=c+bigomega(i); print(c/i))  
> 3.626619...
>             c=0.0; for (i=1,   10000000, c=c+bigomega(i); print(c/i))  
> 3.786124...
>             c=0.0; for (i=1,  100000000, c=c+bigomega(i); print(c/i))  
> 3.923512...
>             c=0.0; for (i=1, 1000000000, c=c+bigomega(i); print(c/i))  
> 4.044220...
> 
> 
> Is it possible to use PariGP to guess the limit of (c/i) for i => ∞ ?

If I understand correctly, you're computing

   (1 / X) sum_{n <= X} \Omega(n)

for increasing X. This goes to infinity as loglog X.

In fact, 

* sum_{n <= X} \omega(n) = sum_{n <= X} sum_{p | n, p prime} 1
                         = sum_{p <= X} [X / p]
                         = X sum_{p <= X} (1 / p) + O(X)

* sum_{p <= X} (1 / p) ~ loglog X

* it's an easy exercise to show that sum_{n <= X} (\Omega(n) - \omega(n))
  is O(X)

See, e.g., Gerald Tenenbaum's "Introduction to analytic and probabilistic
number theory".

Cheers,

    K.B.
--
Karim Belabas, IMB (UMR 5251)  Tel: (+33) (0)5 40 00 26 17
Universite de Bordeaux         Fax: (+33) (0)5 40 00 21 23
351, cours de la Liberation    http://www.math.u-bordeaux.fr/~kbelabas/
F-33405 Talence (France)       http://pari.math.u-bordeaux.fr/  [PARI/GP]
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