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+# Solution comment: Note the sequence of numerators, n_i%0A#      2, 3, 8, 11, 19, 87, 106, 193, 1264, 1457,...%0A# With the sequence of a_i:%0A#            2,  1,  1,  4,   1,   1,   6,     1,...%0A# For n_i, i %3E= 2, we have%0A#            n_i = a_i * n_%7Bi-1%7D + n_%7Bi-2%7D%0A%0A# Start with n_1 and n_2, compute n_3... n_100 (98 iterations)%0Anp, n = 2, 3%0Afor i in range(3, 100+1):%0A    a_i = 2 * (i//3) if i %25 3 == 0 else 1%0A    np, n = n, a_i * n + np%0A%0Adig_sum = 0%0Awhile n:%0A    dig_sum += n %25 10%0A    n //= 10%0A%0Aprint('Sum of the digits in the numerator of the 100th convergent%5Cn'%0A      'of the continued fraction of e:', dig_sum)%0A%0A

Add solution to 65.