Calculated $f(2004)$ where the function $f:N \rightarrow Q^*$ It is considered that the function $f:N \rightarrow Q^*$ which has properties:
$a) f(7) = 4;$
$b) f(8013) = 8015;$
$c) f(n+2).f(n) = 1+ f(n+1).$
Calculated $f(2004)$.
All my attempts to find the required number had no result. Can anyone help me? Thanks so much!

Using (c) to express a simple recurrence $f(n+2)=\frac{f(n+1)+1}{f(n)}$, it's easy to demonstrate the sequence operates a cycle of length 5  (provided zeros are avoided):
$$ f(1)=x $$
$$f(2)=y$$
$$ f(3) = \frac{y+1}{x}$$
$$ f(4) = \frac{x+y+1}{xy}$$
$$f(5) = \frac{x(xy+x+y+1)}{xy(y+1)} = \frac{(x+1)(y+1)}{y(y+1)} = \frac{x+1}{y}$$
$$f(6) = \frac{xy(y+x+1)}{y(x+y+1)} = x$$
$$f(7) = \frac{y(x+1)}{(x+1)} = y$$
Therefore $f(2)= f(7)=4$ and $f(3)=f(8013)= 8015$ 
So $f(2004)=f(4) = \frac{8016}{4} = 2004$
