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

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

1,1

Comments

Since d(m) < 2*sqrt(m), we need only test values of m < (2n/d(n))^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 follow:

crowded[n_] := Module[{},

* If[OddQ[n], Return [False]];*

div = DivisorSigma[0,n]/n;

For [ *m=n+2, m<=2n, 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[2n, stop] as the stopping condition of the loop ("stop" being the variable defined in the original algorithm). (End)

Links

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

Mathematica

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

Crossrefs

Cf. A002182, A000005, A004394.

Sequence in context: A085988 A073349 A129301 * A097212 A266228 A362081

Adjacent sequences: A066520 A066521 A066522 * A066524 A066525 A066526

Keyword

nonn

Author

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

Extensions

Edited by Dean Hickerson, Jan 07, 2002.

Status

approved