5 * 9^3 = 3645, so any number of digits above 4 will reduce the number of digits at each iteration. So we need look only up to 4-digit numbers.

The program below reports the first number where the procedure leads to a given number. The results were:

1 1

2 371

3 153

4 133 55 250 133

7 370

16 217 352 160 217

47 407

49 1099 1459 919 1459

55 55 250 133 55

79 352 160 217 352

136 153 136 244 136

160 160 217 352 160

163 163 244 136 244

250 250 133 55 250

478 478 919 1459 919

There are indeed only 5 fixed points, where the result stays constant: 1, 371, 153, 370 and 407. I would not say, however, that there are only "five possible choices", as one does not know ahead of time that one will hit a fixed point.

Another possibility is that you cycle ... 133 55 250 133 ...

another cycle is ... 217 352 160 217 ...

or back and forth between 919 and 1459 

or between 244 and 136.

So there are 5 fixed points possible, and 4 cycles possible.

DefDbl A-Z

Dim crlf$

Private Sub Form_Load()

 Form1.Visible = True

 

 

 Text1.Text = ""

 crlf = Chr$(13) + Chr$(10)

 

 For n = 1 To 9999

  nn = n

 Do

   ppprev = pprev

   pprev = prev

   prev = nn

   nn = socd(nn)

   DoEvents

  Loop Until nn = prev Or nn = pprev Or nn = ppprev

    s$ = Str(nn)

    If InStr(allNN$, s) = 0 Then

      Text1.Text = Text1.Text & n

      If nn <> prev Then Text1.Text = Text1.Text & Str(ppprev) & Str(pprev) & Str(prev)

      Text1.Text = Text1.Text & s & crlf

      allNN = allNN + s

    End If

 Next

 

 Text1.Text = Text1.Text & crlf & " done"

  

End Sub

Function socd(n)

  s$ = LTrim(Str(n))

  tot = 0

  For i = 1 To Len(s$)

   d = Val(Mid(s$, i, 1))

   tot = tot + d * d * d

  Next

  socd = tot

End Function