3.7 KiB

Raw Blame History

id	title	challengeType	forumTopicId	dashedName
5900f3ac1000cf542c50febf	Problem 64: Odd period square roots	5	302176	problem-64-odd-period-square-roots

--description--

All square roots are periodic when written as continued fractions and can be written in the form:

\\displaystyle \\quad \\quad \\sqrt{N}=a_0+\\frac 1 {a_1+\\frac 1 {a_2+ \\frac 1 {a3+ \\dots}}}

For example, let us consider \\sqrt{23}::

\\quad \\quad \\sqrt{23}=4+\\sqrt{23}-4=4+\\frac 1 {\\frac 1 {\\sqrt{23}-4}}=4+\\frac 1 {1+\\frac{\\sqrt{23}-3}7}

If we continue we would get the following expansion:

\\displaystyle \\quad \\quad \\sqrt{23}=4+\\frac 1 {1+\\frac 1 {3+ \\frac 1 {1+\\frac 1 {8+ \\dots}}}}

The process can be summarized as follows:

\\quad \\quad a_0=4, \\frac 1 {\\sqrt{23}-4}=\\frac {\\sqrt{23}+4} 7=1+\\frac {\\sqrt{23}-3} 7

\\quad \\quad a_1=1, \\frac 7 {\\sqrt{23}-3}=\\frac {7(\\sqrt{23}+3)} {14}=3+\\frac {\\sqrt{23}-3} 2

\\quad \\quad a_2=3, \\frac 2 {\\sqrt{23}-3}=\\frac {2(\\sqrt{23}+3)} {14}=1+\\frac {\\sqrt{23}-4} 7

\\quad \\quad a_3=1, \\frac 7 {\\sqrt{23}-4}=\\frac {7(\\sqrt{23}+4)} 7=8+\\sqrt{23}-4

\\quad \\quad a_4=8, \\frac 1 {\\sqrt{23}-4}=\\frac {\\sqrt{23}+4} 7=1+\\frac {\\sqrt{23}-3} 7

\\quad \\quad a_5=1, \\frac 7 {\\sqrt{23}-3}=\\frac {7 (\\sqrt{23}+3)} {14}=3+\\frac {\\sqrt{23}-3} 2

\\quad \\quad a_6=3, \\frac 2 {\\sqrt{23}-3}=\\frac {2(\\sqrt{23}+3)} {14}=1+\\frac {\\sqrt{23}-4} 7

\\quad \\quad a_7=1, \\frac 7 {\\sqrt{23}-4}=\\frac {7(\\sqrt{23}+4)} {7}=8+\\sqrt{23}-4

It can be seen that the sequence is repeating. For conciseness, we use the notation \\sqrt{23}=\[4;(1,3,1,8)], to indicate that the block (1,3,1,8) repeats indefinitely.

The first ten continued fraction representations of (irrational) square roots are:

\\quad \\quad \\sqrt{2}=\[1;(2)], period = 1

\\quad \\quad \\sqrt{3}=\[1;(1,2)], period = 2

\\quad \\quad \\sqrt{5}=\[2;(4)], period = 1

\\quad \\quad \\sqrt{6}=\[2;(2,4)], period = 2

\\quad \\quad \\sqrt{7}=\[2;(1,1,1,4)], period = 4

\\quad \\quad \\sqrt{8}=\[2;(1,4)], period = 2

\\quad \\quad \\sqrt{10}=\[3;(6)], period = 1

\\quad \\quad \\sqrt{11}=\[3;(3,6)], period = 2

\\quad \\quad \\sqrt{12}=\[3;(2,6)], period = 2

\\quad \\quad \\sqrt{13}=\[3;(1,1,1,1,6)], period = 5

Exactly four continued fractions, for N \\le 13, have an odd period.

How many continued fractions for N \\le n have an odd period?

--hints--

oddPeriodSqrts(13) should return a number.

assert(typeof oddPeriodSqrts(13) === 'number');

oddPeriodSqrts(500) should return 83.

assert.strictEqual(oddPeriodSqrts(500), 83);

oddPeriodSqrts(1000) should return 152.

assert.strictEqual(oddPeriodSqrts(1000), 152);

oddPeriodSqrts(5000) should return 690.

assert.strictEqual(oddPeriodSqrts(5000), 690);

oddPeriodSqrts(10000) should return 1322.

assert.strictEqual(oddPeriodSqrts(10000), 1322);

--seed--

--seed-contents--

function oddPeriodSqrts(n) {

  return true;
}

oddPeriodSqrts(13);

--solutions--

function oddPeriodSqrts(n) {
  // Based on https://www.mathblog.dk/project-euler-continued-fractions-odd-period/
  function getPeriod(num) {
    let period = 0;
    let m = 0;
    let d = 1;
    let a = Math.floor(Math.sqrt(num));
    const a0 = a;
    while (2 * a0 !== a) {
      m = d * a - m;
      d = Math.floor((num - m ** 2) / d);
      a = Math.floor((Math.sqrt(num) + m) / d);
      period++;
    }
    return period;
  }

  function isPerfectSquare(num) {
    return Number.isInteger(Math.sqrt(num));
  }

  let counter = 0;
  for (let i = 2; i <= n; i++) {
    if (!isPerfectSquare(i)) {
      if (getPeriod(i) % 2 !== 0) {
        counter++;
      }
    }
  }
  return counter;
}

3.7 KiB Raw Blame History