n= -log2(log2(√√...√2))
where log2 is the logarithm base 2, and there are n square roots.
Can you manage to do the same (that is, represent all positive integers) using three THREEs instead? And by using only TWO twos?
N=lim x-->3^(3-3) d/dx (x*x*..*x)
Of course, you can also do without NO numbers at all, and just write
N=d/dx (x+x+...+x)
with N x's in the sum...
blackjack
flooble's webmaster puzzle