It's because the 1000 will end up adding one to the 999 which becomes 1000 which adds one to 998.

Yes, that is funky.

  def long_div(x,y,N=10):
    """Given two integers x and y, return us the N digits of x/y."""
    #First work out the integer part
    x = int(x)
    divisor = y = int(y)
    quotients = [x/y]
    dividend = x % y
    for digit in range(N):
      dividend *= 10
      quotient = dividend/divisor
      dividend = dividend%divisor
      quotients.append(quotient)
  
    return quotients
  
  def pretty_print(quotients, G=3):
    """Pretty print quotients by grouping digits by 3."""
    strp = ''
    cnt = 1
    for n in quotients[1:]:
      strp += str(n)
      cnt +=1
      if cnt > G:
        print strp
        strp = ''
        cnt = 1
  
  
  quotients = long_div(1,998001,N=3000)
  pretty_print(quotients,G=3)

And even the suggested 1/9801 for two digits produces no 98

    ... 95 96 97 99 00 01 02 ...

And extrapolating it to one digit, 1/81 gives

    0.012345679012345679...

Again, the 100 overflows to the 99. This makes the 99 a 100 and then that overflows again and the 98 turns to a 99.

you could always do 1/99980001 - that would have a 0998

But it would lack a 9998.

then i suppose we need 1/9999800001 !

But it would lack... just kidding.

Another way to get the real consecutive numbers is to make a rest so we avoid that annoying increment at the end:

1/998001 - 1e-1000