Comments on: Multiples in the Fibonacci series http://www.joaoff.com/2008/05/09/multiples-in-the-fibonacci-series/ Programming, Algorithms, and Mathematics Tue, 06 Jan 2009 15:36:58 +0000 http://wordpress.org/?v=2.6.1 By: jff http://www.joaoff.com/2008/05/09/multiples-in-the-fibonacci-series/#comment-880 jff Sun, 11 May 2008 20:26:00 +0000 http://www.joaoff.com/?p=64#comment-880 Yes, it's just the (n*k)th Fibonacci number :-) Thanks for your comments! Yes, it’s just the (n*k)th Fibonacci number :-) Thanks for your comments!

]]> By: Hugo http://www.joaoff.com/2008/05/09/multiples-in-the-fibonacci-series/#comment-879 Hugo Sun, 11 May 2008 19:59:38 +0000 http://www.joaoff.com/?p=64#comment-879 That's right, I disregarded fib.(n*k) represents a number, thanks for the help. Congratulations! Your work it's really interesting. That’s right, I disregarded fib.(n*k) represents a number, thanks for the help. Congratulations! Your work it’s really interesting.

]]> By: jff http://www.joaoff.com/2008/05/09/multiples-in-the-fibonacci-series/#comment-878 jff Sun, 11 May 2008 19:08:11 +0000 http://www.joaoff.com/?p=64#comment-878 Hi Hugo, If fib.k divides fib.(n*k), then the greatest common divisor of fib.k and fib.(n*k) is fib.k (it divides both, and is the largest divisor of itself). Also, if fib.k is the greatest common divisor of itself and fib.(n*k), then it clearly divides fib.(n*k). Do you agree? Joao Hi Hugo,

If fib.k divides fib.(n*k), then the greatest common divisor of fib.k and fib.(n*k) is fib.k (it divides both, and is the largest divisor of itself).

Also, if fib.k is the greatest common divisor of itself and fib.(n*k), then it clearly divides fib.(n*k).

Do you agree?

Joao

]]> By: Hugo http://www.joaoff.com/2008/05/09/multiples-in-the-fibonacci-series/#comment-877 Hugo Sun, 11 May 2008 18:52:45 +0000 http://www.joaoff.com/?p=64#comment-877 Can you help me understand your second equality? Can you help me understand your second equality?

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