I wonder if there is some "surprising" function $f(\;)$ that, when input $2013$, produces $2014$? What I have in mind is more in line with the Lewis Carroll computation involving $137$ and $992$ that Gerry Myerson highlighted in this MO answer. $f(n)=n+1$ would not satisfy. :-)

$$2013 = 3 \cdot 11 \cdot 61$$ $$2014 = 2 \cdot 19 \cdot 53$$ Hmmm... Happy New Year!