Benoit Cloitre wrote:
>...
>It seems interesting to have more properties of those numbers
>S(n)=sum(1/k, 1<=k<=n, (k,n)=1)
>...
FYI.
S(n) = (1/n) sum{k|n} mu(n/k) k H(k),
where H(k) = sum{j=1 to k} 1/j.
In other words,
sum{k|n} k S(k)
= n H(n).
Easy stuff, but noteworthy.
thanks,
Leroy Quet