A354768 -id:A354768 - cp's OEIS frontend

A066523 Crowded numbers: for any k in the sequence, d(k)/k is larger than d(m)/m for all m > k.

2, 4, 6, 12, 24, 30, 36, 48, 60, 72, 84, 120, 144, 180, 240, 252, 360, 420, 480, 504, 540, 720, 840, 900, 1008, 1080, 1260, 1440, 1680, 1800, 2520, 2640, 2880, 3360, 3780, 3960, 5040, 5280, 5400, 5460, 5544, 6300, 7560, 7920, 10080, 10920, 12600

Offset: 1

Roy Maulbogat (maulbogat(AT)gmail.com), Jan 05 2002

nonn

Since d(m) < 2*sqrt(m), we need only test values of m < (2k/d(k))^2.
It was briefly conjectured that this sequence was the same as the highly composite numbers (A002182) larger than 1, but this is false: 30 is crowded but not highly composite and 50400 is highly composite but not crowded. Is every super-abundant number (A004394) crowded?
Additional comments from Roy Maulbogat, Jan 22 2008: (Start)
It can easily be shown that all crowded numbers are even and that there is always a crowded number between N and 2N. This allows us to improve the algorithm as follows:
crowded[k_] := Module[{},
* If[OddQ[k], Return [False]];*
div = DivisorSigma[0,k]/k;
For [ *m=k+2, m<=2k, m+=2*, If[
DivisorSigma [0, m] / m<=div, Return [False]]];True];
numlist = Select[Range[1,10^7],crowded]
On second thought, it might be wise to use Min[2k, stop] as the stopping condition of the loop ("stop" being the variable defined in the original algorithm). (End)

Donovan Johnson, Table of n, a(n) for n = 1..300 (first 129 terms from Roy Maulbogat)

Cf. A002182, A000005, A004394, A354768.

Tau := n -> NumberTheory:-tau(n): t := (n, k) -> k*Tau(n) < n*Tau(k):
isCrowded := proc(k) local n; if k::odd then return false fi;
andmap(j -> t(j, k), [seq(k+2..2*k,2)]) end:
aList := n -> select(isCrowded, [seq(1..n)]): aList(1100);  # Peter Luschny, Jan 24 2025

crowded[n_] := Module[{}, stop=(2/(dovern=DivisorSigma[0, n]/n))^2; For[m=n+1, m=dovern, Return[False]]]; True]; Select[Range[1, 13000], crowded]

Edited by Dean Hickerson, Jan 07 2002

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

Original entry on oeis.org

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

Keith J. Bauer, Dec 28 2023

nonn

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.

			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

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

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

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 *)

A354769 Numbers k such that d(k)/k = sup {d(m)/m | m > k}, where d(k) is the number-of-divisors function A000005(k).

Original entry on oeis.org

1, 8, 18, 90, 8400, 1201200

Offset: 1

N. J. A. Sloane, Jun 22 2022

It is not known if this sequence is infinite.

David desJardins, Posting to Math Fun Mailing List, Jun 21 2022.

Cf. A000005, A066523, A354768.

A354770 Numbers k such that d(k)/log(k) sets a new record, where d(k) is the number-of-divisors function A000005(k).

Original entry on oeis.org

2, 60, 120, 180, 240, 360, 720, 840, 1260, 1680, 2520, 5040, 7560, 10080, 15120, 20160, 25200, 27720, 50400, 55440, 83160, 110880, 166320, 221760, 277200, 332640, 498960, 554400, 665280, 720720, 1081080, 1441440, 2162160, 2882880, 3603600, 4324320, 6486480, 7207200, 8648640

Offset: 1

N. J. A. Sloane, Jun 22 2022

nonn

A related sequence, not yet in the OEIS, is "Numbers k such that log(d(k))/log(k) > log(d(m))/log(m) for all m > k". It begins 2, 4, 6, 12, 24, 36, 60, 72, 120, 180, 240, 360, 420, 720, 840, 1260, 1680, 2520, 5040, 7560, ..., and up to this point it agrees with A236021 (except that it doesn't include 1). Does it continue to agree with A236021?

			The values of d(k)/log(k) for k = 2, 3, ... are 2.885390082, 1.820478453, 2.164042562, 1.242669869, 2.232442506, 1.027796685, 1.923593388, 1.365358840, 1.737177928, 0.8340647828, ... and reach record highs at k = 2 (2.885390082...), k = 60 (2.930872040...), and so on.

David desJardins, Posting to Math Fun Mailing List, Jun 22 2022.

Cf. A000005, A066523, A236020, A236021, A354768, A354369.

s = {}; rm = 0; Do[If[(r = DivisorSigma[0, n]/Log[n]) > rm, rm = r; AppendTo[s, n]], {n, 2, 10^5}]; s (* Amiram Eldar, Jun 22 2022 *)

More terms from Amiram Eldar, Jun 22 2022

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A066523 Crowded numbers: for any k in the sequence, d(k)/k is larger than d(m)/m for all m > k.

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

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A354769 Numbers k such that d(k)/k = sup {d(m)/m | m > k}, where d(k) is the number-of-divisors function A000005(k).

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A354770 Numbers k such that d(k)/log(k) sets a new record, where d(k) is the number-of-divisors function A000005(k).

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Mathematica

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