Since d(m) < 2*sqrt(m), we need only test values of m < (2n2k/d(nk))^2.
Additional comments from Roy Maulbogat, Jan 22 2008: (Start)
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 followfollows:
crowded[nk_] := Module[{},
* If[OddQ[nk], Return [False]];*
div = DivisorSigma[0,nk]/nk;
For [ *m=nk+2, m<=2n2k, m+=2*, If[
On second thought, it might be wise to use Min[2n2k, stop] as the stopping condition of the loop ("stop" being the variable defined in the original algorithm). (End)
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