|
%I #11 Sep 08 2022 08:46:15
%S 1,2,4,6,12,24,30,48,60,120,180,210,240,360,420,720,840,1260,1680,
%T 2520,4620,5040,7560,9240,13860,18480,27720,55440,83160,110880,120120,
%U 166320,180180,240240,360360,720720,1081080,1441440,1801800,2042040,2162160,3063060,3603600,4084080
%N Numbers n such that Sum_{d|n} 1/tau(d) > Sum_{d|m} 1/tau(d) for all m < n.
%C Where record values of Sum_{d|n} 1/tau(d) occur.
%C Terms a(n) are the smallest number from sequences numbers with following prime signatures: {}, {1}, {2}, {1, 1}, {2, 1}, {3, 1}, {1, 1, 1}, {4, 1}, {2, 1, 1}, {3, 1, 1}, {2, 2, 1}, {1, 1, 1, 1}, {4, 1, 1}, {3, 2, 1}, ...
%e For n = 4; a(4) = 6 because 6 is the smallest number such that Sum_{d|a(4)} 1/tau(d) = Sum_{d|6} 1/tau(d) = 9/4 > Sum_{d|a(3)} 1/tau(d) = Sum_{d|4} 1/tau(d) = 11/6.
%o (Magma) a:=1; S:=[a]; for n in [2..25] do k:=0; flag:= true; while flag do k+:=1; if &+[1/NumberOfDivisors(d): d in Divisors(a)] lt &+[1/NumberOfDivisors(d): d in Divisors(k)] then Append(~S, k); a:=k; flag:=false; end if; end while; end for; S;
%o (PARI) lista(nn) = {m = 0; for (n=1, nn, if ((mm = sumdiv(n, d, 1/numdiv(d))) > m, print1(n, ", "); m = mm););} \\ _Michel Marcus_, Dec 22 2015
%Y Cf. A000005, A237350, A253139, A265390, A265391, A265392, A265393.
%K nonn
%O 1,2
%A _Jaroslav Krizek_, Dec 14 2015
%E More terms from _Michel Marcus_, Dec 22 2015
|
