Project Euler Problem 64

http://projecteuler.net/problem=64

Problem 64

27 February 2004

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

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 – 3 7

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+4 7	= 1 +	23—3 7
a1 = 1,		7 23—3	=	7(23+3) 14	= 3 +	23—3 2
a2 = 3,		2 23—3	=	2(23+3) 14	= 1 +	23—4 7
a3 = 1,		7 23—4	=	7(23+4) 7	= 8 +	23—4
a4 = 8,		1 23—4	=	23+4 7	= 1 +	23—3 7
a5 = 1,		7 23—3	=	7(23+3) 14	= 3 +	23—3 2
a6 = 3,		2 23—3	=	2(23+3) 14	= 1 +	23—4 7
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?

Answer:

1322

* Racket

#lang racket
(define (period-root n)
  (define limit (floor (sqrt n)))
  (if (= n (sqr limit))
      0
      (let ([period 0]
            [d 1]
            [m 0]
            [a limit])
        (let loop ()         
          (set! m (- (* d a) m))
          (set! d (quotient (- n (sqr m)) d))
          (set! a (quotient (+ limit m) d))
          (set! period (add1 period))
          (unless (= a (* 2 limit)) (loop)))
        period)))

(for/sum ([n (in-range 2 10001)])
  (modulo (period-root n) 2))