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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    
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Some Thoughts Part a; possible solution Comment 5 of 5 |

U(n) as above defined = F(n)+3*F(n-1), where F(n) is the nth Fibonacci number*. [1] 
Clearly, any two consecutive Fibonacci numbers are co-prime. [2]

Taking [1] and [2] together, only these cases are possible:

F(n)=1, F(n-1)=0; U(n) =1  
F(n)=F(n-1)=1; U(n) =4  
F(n)=3, F(n-1)=2; 3+2*3=3*3=9  

in fact, for n>4, U(n), modF(n)= L(n-1), an interesting result.

*Many other formulations in terms of the Fibonacci and Lucas numbers are also possible but this may be the most convenient.


Edited on April 19, 2014, 9:05 am
  Posted by broll on 2014-04-15 14:10:49

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