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

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, 11, 22, 72, 13, 26, 27, 54, 56, 84, 44, 45, 90, 120, 80, 96, 33, 66, 25, 50, 17, 34, 52, 35, 70, 108, 64, 19, 38, 144, 39, 78, 180, 63, 126, 168, 88, 132, 100, 112

Offset: 1

Author

Keith J. Bauer, Dec 28 2023

Keywords

• nonn

Comments

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

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

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.

Examples

			tau(1)/1 = tau(2)/2 = 1
tau(4)/4 = 3/4
tau(3)/3 = tau(6)/6 = 2/3
tau(8)/8 = tau(12)/12 = 1/2
tau(5)/5 = tau(10)/10 = 2/5
tau(9)/9 = tau(18)/18 = tau(24)/24 = 1/3
		

Crossrefs

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

Programs

Lua

length = 100
result = {}
for n = 1, length do
  local div_count = 0
  local root_n = math.sqrt(n)
  for d = 1, root_n do
    if n % d == 0 then
      div_count = div_count + 2
    end
  end
  if (root_n == math.floor(root_n)) then
    div_count = div_count - 1
  end
  result[n] = {n, div_count / n}
end
function compare(a, b)
  if a[2] ~= b[2] then
    return a[2] > b[2]
  else
    return a[1] < b[1]
  end
end
table.sort(result, compare)
i = 1
bound = 2 / math.sqrt(length)
while result[i][2] >= bound do
  io.write(result[i][1] .. ',')
  i = i + 1
end

Mathematica

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 *)