Time for some more mathfitti: Let's build numbers using just +, *, and 1's. (And parentheses if you must.) So 7 = 1 + (1+1)*(1+1+1) and 18=(1+1)*(1+1+1)*(1+1+1) . For n>1 let c(n) count the smallest number of ones necessary to form n. It is easy to show 3lg n <= c(n) < 5 lg n, where lg is log base 3. Is c(n) <= 4 lg n for all n>1? So far the extreme case seems to be n=1439.
The answer is nicely expressed in terms of lg. The calculations are done via program, not using base 3 arithmetic. See OEIS for links to a nice version in C.
@JonBeardsley One of my friends was learning LaTeX. He was very fond of using custom commands. One day, I created a python program that overwrites all newcommands and makes them into various nonsense things. I then aliased latex to it. Poor guy was completely confused :P
in case you didn't know.... $\mathbf{D}f(\mathbf{x})\mathbf{v}&=\mathrm{grad}f(\mathbf{x})\cdot\mathbf{v}=\nabla f(\mathbf{x})\cdot\mathbf{v}\\&=\left[\frac{df}{dx}(\mathbf{x})\right]v_1+\left[\frac{df}{dy}(\mathbf{x})\right]v_2+\left[\frac{df}{dz}(\mathbf{x})\right]v_3$