Convergents of e
Problem Statement
The square root of 2 can be written as an infinite continued fraction. The infinite continued fraction can be written, √2 = [1;(2)], (2) indicates that 2 repeats ad infinitum. In a similar way, √23 = [4;(1,3,1,8)].
It turns out that the sequence of partial values of continued fractions for square roots provide the best rational approximations. Let us consider the convergents for √2.
Hence the sequence of the first ten convergents for √2 are: 1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, …
What is most surprising is that the important mathematical constant, e = [2; 1,2,1, 1,4,1, 1,6,1 , … , 1,2k,1, …].
The first ten terms in the sequence of convergents for e are: 2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, …
The sum of digits in the numerator of the 10th convergent is 1+4+5+7=17.
Find the sum of digits in the numerator of the 100th convergent of the continued fraction for e.
Approach
The solution involves:
- Generating the continued fraction coefficients for e: [2; 1,2,1, 1,4,1, 1,6,1, …]
- Computing convergents using the recurrence relation
- Using BigInteger to handle large numerators
- Extracting and summing the digits of the 100th numerator