U(n) as above defined = F(n)+3*F(n1), where F(n) is the nth Fibonacci number*. [1]
Clearly, any two consecutive Fibonacci numbers are coprime. [2]
Taking [1] and [2] together, only these cases are possible:
F(n)=1, F(n1)=0; U(n) =1
F(n)=F(n1)=1; U(n) =4
F(n)=3, F(n1)=2; 3+2*3=3*3=9
in fact, for n>4, U(n), modF(n)= L(n1), 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 20140415 14:10:49 