A368523 - OEIS

%I #21 Jan 04 2024 17:31:46

%S 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 11,22,72,13,26,27,54,56,84,44,45,90,120,80,96,33,66,25,50,17,34,52,

%U 35,70,108,64,19,38,144,39,78,180,63,126,168,88,132,100,112

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

%C Because tau(k)/k is bounded by 2/sqrt(k), this sequence is well-defined.

%C In the case of ties, terms are sorted from least to greatest.

%C 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 tau(1)/1 = tau(2)/2 = 1

%e tau(4)/4 = 3/4

%e tau(3)/3 = tau(6)/6 = 2/3

%e tau(8)/8 = tau(12)/12 = 1/2

%e tau(5)/5 = tau(10)/10 = 2/5

%e tau(9)/9 = tau(18)/18 = tau(24)/24 = 1/3

%t 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 (Lua)

%o length = 100

%o result = {}

%o for n = 1, length do

%o   local div_count = 0

%o   local root_n = math.sqrt(n)

%o   for d = 1, root_n do

%o     if n % d == 0 then

%o       div_count = div_count + 2

%o     end

%o   end

%o   if (root_n == math.floor(root_n)) then

%o     div_count = div_count - 1

%o   end

%o   result[n] = {n, div_count / n}

%o end

%o function compare(a, b)

%o   if a[2] ~= b[2] then

%o     return a[2] > b[2]

%o   else

%o     return a[1] < b[1]

%o   end

%o end

%o table.sort(result, compare)

%o i = 1

%o bound = 2 / math.sqrt(length)

%o while result[i][2] >= bound do

%o   io.write(result[i][1] .. ',')

%o   i = i + 1

%o end

%Y Cf. A000005, A090387, A090395, A323383, A354768.

%K nonn

%O 1,2

%A _Keith J. Bauer_, Dec 28 2023