Given $f(0) = 1c$, what is the nonrecursive $f(n)$ that satify the following equation $f(n)\frac{1}{2}f(n1)f(n)+f(n1) = 1$
for n = 1,2,3,...?
Given $f(0) = 1c$, what is the nonrecursive $f(n)$ that satify the following equation $f(n)\frac{1}{2}f(n1)f(n)+f(n1) = 1$ for n = 1,2,3,...? 


$$ f(n) = 2{\frac {1c+ \left( 1 \right) ^{n}c \left( 1 \right) ^{n}\sqrt {2}c+ \left( 1 \right) ^{n}\sqrt {2}c}{2\sqrt {2}\sqrt {2}c+2 \left( 1 \right) ^{n}+ \left( 1 \right) ^{n}\sqrt {2}+ \left( 1 \right) ^{n}\sqrt {2}c}} $$ which is a nontrivial pattern to spot! [I used a CAS] The thing to notice is that one can transform this firstorder recurrence to the constant coefficient 2nd order linear recurrence $$a(n+2)2a(n) = 0, a(0) = 1, a(1) = 1c$$ and then transform back. This can be done as the equation is of Riccati type. I guess I 'cheated' in that I traced through the CAS's process of solving to extract the above information from it. Though if you don't know what you're looking for, it can be rather difficult information to extract... 


We can rewrite the given recurrence as $(f(n)  2)(f(n1)  2) = 2$. That makes it clear that the sequence is 2periodic. 

