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 Pseudo-FIBO (Posted on 2014-04-14)

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.)
 re: 'generated'........... att: All solvers | Comment 4 of 5 |
(In reply to 'generated' by Jer)

re Jer    Your solution would include all series with positive integers each below 520, and a<b.  They would include inter alia 520  as a number generated by you, contradicting the text.

re. Steve.   a=b generates a regular Fibonacci series (a=b=1) ,each number being multiplied by 520/f  f being a  Fibonacci number dividing 520 (520/2,520/5,520/8,520/13). I just excluded them.

Both modifications would  fit an altered  text, creating another valid puzzle.

re: All solvers-
1. One does have to list all solutions, a counting is possible like
for u(3 )=520   there are 259 solutions: u(1)=260-k, u(2)=260+k,  etc
2.  The 1st part of the puzzle is still open.

 Posted by Ady TZIDON on 2014-04-15 11:19:27

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