! Copyright (c) 2009 Guillaume Nargeot. ! See http://factorcode.org/license.txt for BSD license. USING: kernel math lists lists.lazy project-euler.common sequences ; IN: project-euler.065 ! http://projecteuler.net/index.php?section=problems&id=065 ! 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 ! -------- 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