cp's OEIS Frontend

A368523 Positive integers in decreasing order of tau(k)/k, where tau(k) = A000005(k).

%I A368523 #21 Jan 04 2024 17:31:46
%S A368523 1,2,4,3,6,8,12,5,10,9,18,24,16,20,7,14,15,30,36,28,48,40,60,21,42,32,
%T A368523 11,22,72,13,26,27,54,56,84,44,45,90,120,80,96,33,66,25,50,17,34,52,
%U A368523 35,70,108,64,19,38,144,39,78,180,63,126,168,88,132,100,112
%N A368523 Positive integers in decreasing order of tau(k)/k, where tau(k) = A000005(k).
%C A368523 Because tau(k)/k is bounded by 2/sqrt(k), this sequence is well-defined.
%C A368523 In the case of ties, terms are sorted from least to greatest.
%C A368523 Let c be the j-th distinct value of tau(a(n))/a(n). Terms of this sequence for which tau(a(n))/a(n) >= c are the proper divisors of A059992(j + 1) for 1 <= j <= 4. True for j = 0 if the 0th value of c is taken to be infinity. Pattern breaks for j > 4.
%e A368523 tau(1)/1 = tau(2)/2 = 1
%e A368523 tau(4)/4 = 3/4
%e A368523 tau(3)/3 = tau(6)/6 = 2/3
%e A368523 tau(8)/8 = tau(12)/12 = 1/2
%e A368523 tau(5)/5 = tau(10)/10 = 2/5
%e A368523 tau(9)/9 = tau(18)/18 = tau(24)/24 = 1/3
%t A368523 nmax = 100; s = Sort[Table[{k, DivisorSigma[0, k]/k}, {k, 1, nmax^2}], #1[[2]] >= #2[[2]] &]; Table[s[[j, 1]], {j, 1, nmax}] (* _Vaclav Kotesovec_, Jan 04 2024 *)
%o A368523 (Lua)
%o A368523 length = 100
%o A368523 result = {}
%o A368523 for n = 1, length do
%o A368523   local div_count = 0
%o A368523   local root_n = math.sqrt(n)
%o A368523   for d = 1, root_n do
%o A368523     if n % d == 0 then
%o A368523       div_count = div_count + 2
%o A368523     end
%o A368523   end
%o A368523   if (root_n == math.floor(root_n)) then
%o A368523     div_count = div_count - 1
%o A368523   end
%o A368523   result[n] = {n, div_count / n}
%o A368523 end
%o A368523 function compare(a, b)
%o A368523   if a[2] ~= b[2] then
%o A368523     return a[2] > b[2]
%o A368523   else
%o A368523     return a[1] < b[1]
%o A368523   end
%o A368523 end
%o A368523 table.sort(result, compare)
%o A368523 i = 1
%o A368523 bound = 2 / math.sqrt(length)
%o A368523 while result[i][2] >= bound do
%o A368523   io.write(result[i][1] .. ',')
%o A368523   i = i + 1
%o A368523 end
%Y A368523 Cf. A000005, A090387, A090395, A323383, A354768.
%K A368523 nonn
%O A368523 1,2
%A A368523 _Keith J. Bauer_, Dec 28 2023