To compute a constant of calculus (2019)
Disappointing, did not see the extremely easy and quite efficient method of calculating e through continued fractions.<p><pre><code>  S =[2,1,2] 
  T =0

  for P in range(200):
    S += [1,1,(4 + 2*P)]

  for x in range(len(S)-1):
    T = 1 &#x2F; (T + S[-(x+1)])

  print(S[0] + T) # value of e</code></pre>