Andrea Ribuoli on Thu, 07 Jan 2021 21:05:23 +0100


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Curiosity


Good evening,

  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 => ∞ ?


Regards,

  Andrea