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Least k such that sigma(k)/k >= sigma(n)/n.
1

%I #12 May 05 2022 07:17:59

%S 1,2,2,4,2,6,2,6,2,6,2,12,2,4,4,6,2,12,2,12,4,4,2,24,2,4,2,6,2,24,2,6,

%T 2,4,2,36,2,4,2,12,2,12,2,6,4,4,2,48,2,6,2,6,2,12,2,12,2,4,2,60,2,4,4,

%U 6,2,12,2,6,2,12,2,60,2,4,4,6,2,12,2,12,2,4,2,60,2,4,2,12,2,60,2,6,2,4

%N Least k such that sigma(k)/k >= sigma(n)/n.

%H Amiram Eldar, <a href="/A073348/b073348.txt">Table of n, a(n) for n = 1..10000</a>

%F It seems that Sum_{k=1..n} a(k) is asymptotic to C*n*log(n)*log(log(n)) with C>1.

%t a[n_] := Module[{k = 1, r = DivisorSigma[-1, n]}, While[DivisorSigma[-1, k] < r, k++]; k]; Array[a, 100] (* _Amiram Eldar_, May 05 2022 *)

%o (PARI) a(n)=if(n<0,0,s=1; while(sigma(s)/s<sigma(n)/n,s++); s)

%Y Cf. A000203, A017665, A017666.

%K easy,nonn

%O 1,2

%A _Benoit Cloitre_, Aug 23 2002