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id	challengeType	title	forumTopicId
5900f3ad1000cf542c50fec0	5	Problem 65: Convergents of e	302177

Description

The square root of 2 can be written as an infinite continued fraction.

\sqrt{2} = 1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + ...}}}}

The infinite continued fraction can be written, \sqrt{2} = [1; (2)] indicates that 2 repeats ad infinitum. In a similar way, \sqrt{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 \sqrt{2}.

$1 + \dfrac{1}{2} = \dfrac{3}{2}\\ 1 + \dfrac{1}{2 + \dfrac{1}{2}} = \dfrac{7}{5}\\ 1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2}}} = \dfrac{17}{12}\\ 1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2}}}} = \dfrac{41}{29}$

Hence the sequence of the first ten convergents for \sqrt{2} are:

1, \dfrac{3}{2}, \dfrac{7}{5}, \dfrac{17}{12}, \dfrac{41}{29}, \dfrac{99}{70}, \dfrac{239}{169}, \dfrac{577}{408}, \dfrac{1393}{985}, \dfrac{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, \dfrac{8}{3}, \dfrac{11}{4}, \dfrac{19}{7}, \dfrac{87}{32}, \dfrac{106}{39}, \dfrac{193}{71}, \dfrac{1264}{465}, \dfrac{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.

Instructions

Tests

tests:
  - text: <code>convergentsOfE()</code> should return a number.
    testString: assert(typeof convergentsOfE() === 'number');
  - text: <code>convergentsOfE()</code> should return 272.
    testString: assert.strictEqual(convergentsOfE(), 272);

Challenge Seed

function convergentsOfE() {

  return true;
}

convergentsOfE();

Solution

// solution required

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