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065.factor
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065.factor
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! Copyright (c) 2009 Guillaume Nargeot.
! See https://factorcode.org/license.txt for BSD license.
USING: kernel math lists lists.lazy project-euler.common
sequences ;
IN: project-euler.065
! https://projecteuler.net/problem=65
! DESCRIPTION
! -----------
! The square root of 2 can be written as an infinite continued
! fraction.
! 1
! √2 = 1 + -------------------------
! 1
! 2 + ---------------------
! 1
! 2 + -----------------
! 1
! 2 + -------------
! 2 + ...
! The infinite continued fraction can be written, √2 = [1;(2)],
! (2) indicates that 2 repeats ad infinitum. In a similar way,
! √23 = [4;(1,3,1,8)].
! It turns out that the sequence of partial values of continued
! fractions for square roots provide the best rational
! approximations. Let us consider the convergents for √2.
! 1 3 1 7 1 17 1 41
! 1 + - = - ; 1 + ----- = - ; 1 + --------- = -- ; 1 + ------------- = --
! 2 2 1 5 1 12 1 29
! 2 + - 2 + ----- 2 + ---------
! 2 1 1
! 2 + - 2 + -----
! 2 1
! 2 + -
! 2
! Hence the sequence of the first ten convergents for √2 are: 1,
! 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985,
! 3363/2378, ...
! What is most surprising is that the important mathematical
! constant, e = [2; 1,2,1, 1,4,1, 1,6,1 , ... , 1,2k,1, ...].
! The first ten terms in the sequence of convergents for e are:
! 2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465,
! 1457/536, ...
! The sum of digits in the numerator of the 10th convergent is
! 1+4+5+7=17.
! Find the sum of digits in the numerator of the 100th
! convergent of the continued fraction for e.
! SOLUTION
! --------
<PRIVATE
: (e-frac) ( -- seq )
2 lfrom [
dup 3 mod zero? [ 3 / 2 * ] [ drop 1 ] if
] lmap-lazy ;
: e-frac ( n -- n )
1 - (e-frac) ltake list>array reverse 0
[ + recip ] reduce 2 + ;
PRIVATE>
: euler065 ( -- answer )
100 e-frac numerator number>digits sum ;
! [ euler065 ] 100 ave-time
! 4 ms ave run time - 0.33 SD (100 trials)
SOLUTION: euler065