rem - return true (-1) if n is perfect, otherwise 0
def fn.isperfect%(n%)
  sum% = 1 : rem 1 is a divisor of every number
  f1% = 2
  f2% = 1 :  rem dummy value to start
  while (f1% * f1%) <= n%
    if n% = (n% / f1%) * f1% then \
      sum% = sum% + f1%  : \
      f2% = n% / f1% : \
      sum% = sum% + f2%
    rem  don't double count sqrt of perfect square!
    if f1% = f2% then sum% = sum% - f2%
    f1% = f1% + 1
  wend
  fn.isperfect% = (sum% = n%)
  return
fend

print "Searching to 10000 for perfect numbers"
count% = 0
for i% = 2 to 10000
  if fn.isperfect%(i%) then \
    print i% : count% = count% + 1
next i%

print count%; "were found"

end