 Complete quotient

For a simple continued fraction, the complete quotients satisfy ${\theta }_0=x$, $a_n=\lfloor {\theta }_n\rfloor$ and ${\theta }_{n+1}=1/({\theta }_n-a_n)$ whenever the denominator is nonzero. Thus ${\theta }_n=[a_n;a_{n+1},\dots ]$.