OFFSET
1,1
COMMENTS
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)
LINKS
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
KEYWORD
nonn
AUTHOR
Roy Maulbogat (maulbogat(AT)gmail.com), Jan 05 2002
EXTENSIONS
Edited by Dean Hickerson, Jan 07 2002
STATUS
approved