Determine all positive whole numbers w such that d(w) = √w, where d(w) denotes the number of positive divisors of w.

    5   dim FctCt(33)
   10   for N=1 to 1000000
   20    X=N
   25    FacNo=0:PrevFctr=0
   30    repeat
   40      Pfctr=prmdiv(X):X=X//Pfctr
   50      if Pfctr=PrevFctr then
   60       :FctCt(FacNo)=FctCt(FacNo)+1
   70      :else
   80       :FacNo=FacNo+1
   90       :FctCt(FacNo)=1
   95      PrevFctr=Pfctr
  100    until Pfctr=1
  110    NDivsr=1
  120    for I=1 to FacNo
  130     NDivsr=NDivsr*(FctCt(I)+1)
  140    next
  145    NDivsr=NDivsr//2
  150    if NDivsr*NDivsr=N then print N,NDivsr
  200   next

finds only 1 and 9, where 1 has only 1 divisor, itself, and 9 has 3 divisors: 1, 3 and 9.

The program did check all numbers up to 1,000,000.

What's needed is a proof that above some value there, the number of divisors will always be below the square root of the number.

Posted by Charlie on 2007-07-13 11:14:19