All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers > Sequences
Fibonaccish ratio (Posted on 2015-12-07) Difficulty: 3 of 5
It is a very well known mathematical fact that the limiting ratio of consecutive terms of the Fibonacci sequence [F0=0, F1=1, Fn=Fn-1+Fn-2] is Ļ†=(1+āˆš5)/2 as nā†’āˆž.

Suppose we generalize the definition of the sequence to:
Fn=AFn-1+BFn-2.

Find an expression for the limiting ratio of consecutive terms (in terms of A and B.)

Find formulas for A and B to make the limiting ratio any whole number N.

No Solution Yet Submitted by Jer    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: solution | Comment 2 of 3 |
(In reply to solution by Charlie)

Nicely done, Charlie!


A few things, starting with B = ((2N - A)^2 - A^2) / 4

a) Not clear why you did not just substitute A =1, to get
B = ((2N - 1)^2 - 1^2) / 4 = (4N^2 - 4N)/4 = N(N-1)

b) I do not consider A = N, B = 0 to be cheating. :-)

c) Not all values of A and N work in the above formula.  Negative A and N do not give the expected result, I think.  And at first I thought A = 0, B = N^2 would work, but this does not converge unless f(0) = f(1).



  Posted by Steve Herman on 2015-12-07 19:41:56
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information