2.8 KiB
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 10\\,000
have an odd period?
--hints--
oddPeriodSqrts()
should return a number.
assert(typeof oddPeriodSqrts() === 'number');
oddPeriodSqrts()
should return 1322.
assert.strictEqual(oddPeriodSqrts(), 1322);
--seed--
--seed-contents--
function oddPeriodSqrts() {
return true;
}
oddPeriodSqrts();
--solutions--
// solution required