Yan Cui

I help clients go faster for less using serverless technologies.

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#### Problem

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 10^{th}convergent is 1+4+5+7=17.

Find the sum of digits in the numerator of the 100^{th}convergent of the continued fraction for e.

#### Solution

If you look at the convergents of *?2* and the numerators in the convergents of *e*, you’ll see a pattern emerging:

1 + 2 * 1 = 3

2 + 3 * 2 = 8

3 + 8 * 1 = 11

8 + 11 * 1 = 19

11 + 19 * 4 = 87

If you look at the sequence of numerators in the convergents of *e* ( 2, 3, 8, 11, … ) and the sequence of numbers in the convergents of *?2 *( 1, 2, 1, 1, 4, 1, … ), given the current numerator ( *n *) and the previous numerator ( *n-1* ) in the sequence and the corresponding number in the convergents of *?2* ( *i* )the next numerator ( *n+1* ) can be calculated using the formula:

( n+1) = ( n-1 ) + n * i

Once we have this formula to work with, the rest is simple, the solution runs in 7 milliseconds on my machine.

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