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