--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;
}