From Hugh Noland
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ

CB> You know, instead of just posting code for these things, it sure
  > would be nice to see the algorithm behind them!
 
  O.K., here's the e-spigot algorithm:
 
  1.  Initialize:  Let the first digit be 2 and initialize an array A
      of length n+1 to (1,1,...,1).
 
  2.  Repeat n-1 times:
      Multiply by 10:  Multiply each entry of A by 10.
      Take the fractional part:  Starting from the right, reduce
       the i-th entry of A modulo i+1, carrying the quotient one
       place left and adding it to the (i-1)-th entry of A.
      Output the next digit:  The final quotient is the next digit
       of e.
 
 Here is an example for n = 3:
 
           2    3    4    5 -- modulus
 
     2
           1    1    1    1 -- A
          10   10   10   10 -- 10*A
     7     4    3    2   -- -- (10*A+q)/(i+1) = q
        \  | \  | \  | \  |
          14   13   12   10 -- 10*A+q
           |    |    |    |
           0    1    0    0 -- (10*A+q) mod (i+1) = A
           0   10    0    0 -- 10*A
     1     3    0    0   -- -- q
        \  | \  | \  | \  |
           3   10    0    0 -- 10*A+q
 
 The first n (= 3) digits of e occur in the first column.
 The algorithm for pi is a good bit more complicated.  If you really
 want to see it, you'll need to go to the article mentioned.  Maybe
 your public library could have the U of Arkansas library mail a copy;
 or maybe you could get access to the article via the Internet.