Elegant way to solve $n\log_2(n) \le 10^6$ I'm studying Tomas Cormen Algorithms book and solve tasks listed after each chapter.
I'm curious about task 1-1. that is right after Chapter #1.
The question is: what is the best way to solve: $n\lg(n) \le 10^6$, $n \in \mathbb Z$, $\lg(n) = \log_2(n)$; ?
The simplest but longest one is substitution. Are there some elegant ways to solve this? Thank you!
Some explanations: $n$ - is what I should calculate - total quantity of input elements, $10^6$ - time in microseconds - total algorithm running time. I should figure out nmax.

If something cheesier than Newton's method is wanted, I recommend the fixed point iteration:
x := (10^6)/log_2(x) 

This is easily done with a calculator applet by storing constant (10^6) * log_10(2) and thereafter taking $\log_{10}(x)$, reciprocating, and then multiplying by the recalled constant.
Starting with obvious overestimate $x_0 = 10^6$, the fixed-point iterates alternate between under- and over-shooting:
x1 = 50,171.67
x2 = 64,042.69
x3 = 62,630.17
x4 = 62,756.63
x5 = 62,745.18
x6 = 62,746.21
x7 = 62,746.12

Thus the largest possible integer is $n = 62746$, as Michael already stated.
