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Minimum value of abs(prime(n)-k*tau(k)) for k>0.
0

%I #6 Mar 30 2012 18:39:05

%S 1,1,1,1,1,1,3,3,1,2,1,1,1,3,1,3,1,1,5,1,1,1,1,1,3,3,1,1,1,5,5,1,1,1,

%T 2,1,1,3,1,5,1,3,1,1,3,3,3,3,1,1,1,1,1,3,3,1,1,1,1,3,5,3,1,1,1,3,3,1,

%U 1,1,5,1,3,1,1,1,3,1,3,1,3,1,1,1,5,1,1,1,1,3,1,1,1,1,1,1,2,5,3,1,5,3,1,1,1

%N Minimum value of abs(prime(n)-k*tau(k)) for k>0.

%F It seems that if n > 1 then sum(k=1, n, a(kn)) > n*Log(Log(n)).

%o (PARI) a(n)=vecmin(vector(prime(n),k,abs(prime(n)-k*numdiv(k))))

%K nonn

%O 1,7

%A _Benoit Cloitre_, Aug 17 2002