Fun with math: Dividing one by 998001 yields a surprising result

1---------

If x = 0.001 then the sum x^2 + 2x^3 + 3x^4…. in its decimal places will have all the three digit numbers except the the second last starting with 000,001 and then till 997,999.

As pointed out by someone below,the reason 998 is missing is because after 997998999 the next coefficient is 1000.This overflow 1 will carry over and mess up all the nines to the right until it hits the eight at which point it will make it a 9.Therefore the series will become 997999000001002003... as subsequent overflows keep messing up the last digit to the left.During the addition of the 2000th term a similar thing will happen and the series at that location will become ...997999001002003...... and it will keep losing a term from the beginning in the subsequent 998 repetitions.

More generally if x = 10^-n,n>0 then the sum x^2 + 2x^3 + 3x^4 will have all the n digit numbers starting with (n-zeroes),(n-1 zeroes 1)…except the second last (10^n-2).

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From http://en.wikipedia.org/wiki/Geometric_series

An infinite geometric series converges to a/(1-r) if and only if the absolute value of r is less than 1.

a + ar^2 + ar^3 .... = a/(1-r) if |r| [1]

3---------

1 + 2x + 3x^2.....

= (1) + (x+x) + (x^2+x^2+x^2)....

= (1+x+x^2...) + (x+x^2+x^3...) + ...

= 1/(1-x) + x/(1-x) + (x^2)/(1-x) + .... since x = (1+x+x^2....)/(1-x)

= 1/(1-x)^2 since x [2]

4---------

So the number 0.000001002003004……997999000001002003....

= (x^2 + 2x^3 + 3x^4…..) , x = 0.001

= (x^2) * (1 + 2x + 3x^2 …..)

= (x^2)/((1-x)^2)

= (0.001)^2/(0.999)^2

= ((1/999))^2

= 1/998001

More generally the sum which has all the n digit numbers except (10^n-2) in its decimal places is given by (10^n-1)^(-2).

  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)

There's a special place in hell for business owners who inflict Websense or any other method of censoring content on their own employees.

If you don't trust me to use it, don't give me a computer, or an internet connection.

Basically, the pattern is 1 over some number of 9s, followed by an 8, followed by the same number of 0s, followed by a 1.

So, 1/81, 1/9801, 1/998001, 1/99980001, 1/9999800001, etc.