tag:blogger.com,1999:blog-3844331164714701431.post8324931816444046888..comments2012-11-27T19:10:14.221+05:30Comments on Puzzles for humans: The even fibonacciAnonymoushttp://www.blogger.com/profile/06503740596277603928noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-3844331164714701431.post-39692991479559969152012-07-22T23:25:42.402+05:302012-07-22T23:25:42.402+05:30if F[n] is even and if any of F[n-1] and F[n+1] ar...if F[n] is even and if any of F[n-1] and F[n+1] are even then whole sequence will be even integers which contradicts the base case f[1] = 1.Vinod Reddy Ghttps://www.blogger.com/profile/17259494234235824937noreply@blogger.comtag:blogger.com,1999:blog-3844331164714701431.post-21403662513280319352012-07-04T02:16:26.112+05:302012-07-04T02:16:26.112+05:30What he meant was that F[n-1] + F[n+1] will be eve...What he meant was that F[n-1] + F[n+1] will be even if F[n] is even. Is it so vinod?Anonymoushttps://www.blogger.com/profile/06503740596277603928noreply@blogger.comtag:blogger.com,1999:blog-3844331164714701431.post-2541891989186894862012-07-04T02:12:19.940+05:302012-07-04T02:12:19.940+05:30"if F[n] is even then both F[n-1] and F[n+1] ..."if F[n] is even then both F[n-1] and F[n+1] should be odd leaking F[2*n] as a multiple of 4." How can you claim that?Aakashhttps://www.blogger.com/profile/08446859313227705373noreply@blogger.comtag:blogger.com,1999:blog-3844331164714701431.post-33276681357047857092012-06-29T10:52:19.546+05:302012-06-29T10:52:19.546+05:30F[n] = 1*F[n-1]+1*F[n-2]
=> F[n] = F[2]*F[n-1]+...F[n] = 1*F[n-1]+1*F[n-2]<br />=> F[n] = F[2]*F[n-1]+F[1]*F[n-2]<br />=> F[n] = F[3]*F[n-2]+F[2]*F[n-3]<br />in general F[n] = F[k+1]*F[n-k]+F[k]*F[n-k-1] for any k < n;<br />now put n = 2*n and k = n in above equation <br />we get F[2*n] = F[n]*(F[n-1]+F[n+1])<br />now in above equation if F[n] is even then both F[n-1] and F[n+1] should be odd leaking F[2*n] as a multiple of 4.<br />if F[n] is odd then one of F[n-1] and F[n+1] should be odd and the other even<br />leaving F[2*n] odd.<br />From above results F[2n] cannot be of the form 4k+2Vinod Reddy Ghttps://www.blogger.com/profile/17259494234235824937noreply@blogger.comtag:blogger.com,1999:blog-3844331164714701431.post-77938726731212749432012-06-29T01:10:20.701+05:302012-06-29T01:10:20.701+05:30My Solution :
First, note that F[n](mod 4) depends...My Solution :<br />First, note that F[n](mod 4) depends only on F[n-1](mod 4) and F[n-2](mod 4).<br />F[0] = 0 = 0 mod 4<br />F[1] = 1 = 1 mod 4<br />F[2] = 1 = 1 mod 4<br />F[3] = 2 = 2 mod 4<br />F[4] = 3 = 3 mod 4<br />F[5] = 5 = 1 mod 4<br />F[6] = 8 = 0 mod 4<br />F[7] = 13 = 1 mod 4<br /><br />Note that (F[6], F[7]) == (F[0], F[1]) (mod 4). Thus the sequence repeats with period of size 6. Thus, as is clear from the sequence, the only candidates for F[2n] (mod 4) are 0, 1, and 3. Q.E.D.Aakashhttps://www.blogger.com/profile/08446859313227705373noreply@blogger.com