Extended comment. Robin's criterion, equivalent to RH, is fairly widely known. First, however, his adviser, J.L. Nicolas, came up with [THIS as pdf][1]. The description of this on wikipedia is poor.

First, in a procedure invented by Ramanujan for his "superior highly composite numbers," it is easy to show that the smallest value of $\frac{\phi(n)}{n^\delta}$ for $0 < \delta < 1$ occurs when $n = n_\delta$ is a primorial, the product of consecutive primes beginning with 2.   

Let's see, Rosser and Schoenfeld gave some effective bounds, the way I am writing this comes out
$$ \frac{\phi(n)}{n} > \frac{1}{e^\gamma \log \log n + \frac{3}{\log \log n}}  $$

So the reasonable question comes, we know we get surprisingly small values of $\frac{\phi(n)}{n}$ when $n$ is a primorial. In that case, is it possible to replace the $3$ by a $0,$ giving
 $$ \frac{\phi(n)}{n} > \frac{1}{e^\gamma \log \log n } ? $$

And here is the answer: If RH is true, we can drop the 3. On the other hand, if RH is false, the inequality (no 3) is true for infinitely many primorials and false for infinitely many primorials. So, we have a statement equivalent to RH. 

Next, it is much easier to compute this comparison than Robin's. All you do is let $P$ be a primorial, and calculate 
$$ \frac{e^\gamma \log \log P \phi(P)}{P}  $$
which can be update fairly nicely as each $P$ is multiplied by the next prime. For all known primorials, this quantity strictly increases with $P.$ Since RH says it is below 1, we see the ratio increasing to 1, very pretty. i wrote out my own C++ program. There is, however, an amusing catch. [Michael Planat and colleagues][2] showed that Cramer's conjecture on prime gaps would be violated if the sequence increased forever. 

Enough for now, let me see if I can find the C++ program and post some early output. 
 
 


  [1]: http://math.univ-lyon1.fr/~nicolas/petitsphi83.pdf
  [2]: http://arxiv.org/abs/1012.3613