$\large 16, 7, 5, 6, 16, \mathbf{112}$
How? Let's go forth analyzing this slightly twisted series.
$16 = start$
$7 = 16 \times 2^{-1} - 2^0$
$5 = 7 \times 2^0 - 2^1$
$6 = 5 \times 2^1 - 2^2$
$16 = 6 \times 2^2 - 2^3$
$112 = 16 \times 2^3 - 2^4$
The trick lies in starting from $2^{-1}$.
-sudoankit
