[seqfan] Re: Sum of omega(k) for k = 1..n

[allouche at math.jussieu.fr]

Hi

If I am not mistaken, this is a classical result
in analytic number theory:
Sum of omega(k) for k = 1..n is equal to
n log log n + cn + O(n/logn)
where c = 0.26149... is equal to \gamma -  c'
where \gamma is the Euler constant and c' is
given by c' = Sum for all the primes p of
(log(1/(1-1/p)) - 1/p) = 0.31571...

see, e.g., p. 62 and 33 of "Introduction \`a la
th\'eorie analytique et probabiliste des nombres",
G. Tenenbaum, Belin 2008. This is certainly also
in the English version of the book (I do not have it
on hand right now, see  
http://www.amazon.fr/Introduction-Analytic-Probabilistic-Number-Theory/dp/0521412617/).

jean-paul

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