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

Home > Numbers > Sequences
Pseudo-FIBO (Posted on 2014-04-14) Difficulty: 4 of 5

a. If the pseudo-Fibonacci numbers are defined by u(1) = 1 , u(2) = 4, u(n)= u(n-1)+u(n-2) show that u(1) = 1, u(2) = 4, and u(4) = 9 are the only squares in the series.

b. How many ordered integer pairs (a,b) both non-negative (a<b) exist, such that a pseudo-Fibo series based upon any of those pairs (i.e. u(1)=a, u(2)=b... etc) will contain 520 as a member generated by that pair ?

No Solution Yet Submitted by Ady TZIDON    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Striking a blow for equality | Comment 2 of 5 |
I know it's not my problem, but if it were I would have specified a <= b, given that the real fibonacci series starts with a = b = 1.

If Part (b) is changed so that a <= b, then there are four more solutions:

(40,40)
(65,65)
(104,104)
(260,260)

  Posted by Steve Herman on 2014-04-14 20:49:14
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 (3)
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