A066523 - OEIS

%I #22 Jul 19 2024 19:03:57

%S 2,4,6,12,24,30,36,48,60,72,84,120,144,180,240,252,360,420,480,504,

%T 540,720,840,900,1008,1080,1260,1440,1680,1800,2520,2640,2880,3360,

%U 3780,3960,5040,5280,5400,5460,5544,6300,7560,7920,10080,10920,12600

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

%C Since d(m) < 2*sqrt(m), we need only test values of m < (2k/d(k))^2.

%C 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?

%C Additional comments from Roy Maulbogat, Jan 22 2008: (Start)

%C 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:

%C crowded[k_] := Module[{},

%C * If[OddQ[k], Return [False]];*

%C div = DivisorSigma[0,k]/k;

%C For [ *m=k+2, m<=2k, m+=2*, If[

%C DivisorSigma [0, m] / m<=div, Return [False]]];True];

%C numlist = Select[Range[1,10^7],crowded]

%C 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)

%H Donovan Johnson, <a href="/A066523/b066523.txt">Table of n, a(n) for n = 1..300</a> (first 129 terms from Roy Maulbogat)

%t 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]

%Y Cf. A002182, A000005, A004394, A354768.

%K nonn

%O 1,1

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

%E Edited by _Dean Hickerson_, Jan 07 2002