2.1 KiB
id | challengeType | title |
---|---|---|
5900f3ac1000cf542c50febf | 5 | Problem 64: Odd period square roots |
Description
√N = a0 + 1
a1 + 1
a2 + 1
a3 + ...
For example, let us consider √23:
√23 = 4 + √23 — 4 = 4 + 1 = 4 + 1
1√23—4
1 + √23 – 37
If we continue we would get the following expansion:
√23 = 4 + 1
1 + 1
3 + 1
1 + 1
8 + ...
The process can be summarised as follows:
a0 = 4,
1√23—4 = √23+47 = 1 + √23—37 a1 = 1,
7√23—3 = 7(√23+3)14 = 3 + √23—32 a2 = 3,
2√23—3 = 2(√23+3)14 = 1 + √23—47 a3 = 1,
7√23—4 = 7(√23+4)7 = 8 + √23—4 a4 = 8,
1√23—4 = √23+47 = 1 + √23—37 a5 = 1,
7√23—3 = 7(√23+3)14 = 3 + √23—32 a6 = 3,
2√23—3 = 2(√23+3)14 = 1 + √23—47 a7 = 1,
7√23—4 = 7(√23+4)7 = 8 + √23—4
It can be seen that the sequence is repeating. For conciseness, we use the notation √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: √2=[1;(2)], period=1 √3=[1;(1,2)], period=2 √5=[2;(4)], period=1 √6=[2;(2,4)], period=2 √7=[2;(1,1,1,4)], period=4 √8=[2;(1,4)], period=2 √10=[3;(6)], period=1 √11=[3;(3,6)], period=2 √12= [3;(2,6)], period=2 √13=[3;(1,1,1,1,6)], period=5 Exactly four continued fractions, for N ≤ 13, have an odd period. How many continued fractions for N ≤ 10000 have an odd period?
Instructions
Tests
tests:
- text: <code>euler64()</code> should return 1322.
testString: assert.strictEqual(euler64(), 1322, '<code>euler64()</code> should return 1322.');
Challenge Seed
function euler64() {
// Good luck!
return true;
}
euler64();
Solution
// solution required