I think the right approach would be to observe that R<sup>n</sup> is the leading term in ![j(n\tau)](http://latex.mathoverflow.net/png?j%28n%5Ctau%29), for ![\tau](http://latex.mathoverflow.net/png?%5Ctau) as above.  Then there is a modular polynomial ![\Phi\sb n](http://latex.mathoverflow.net/png?%5CPhi%5Fn) which satisfies ![\Phi\sb n(j(x),j(nx))=0](http://latex.mathoverflow.net/png?%5CPhi%5Fn%28j%28x%29%2Cj%28nx%29%29%3D0).  For small n you could presumably just solve this for j(n![\tau](http://latex.mathoverflow.net/png?%5Ctau)) in terms of j(![\tau](http://latex.mathoverflow.net/png?%5Ctau)), especially since modular curves are rational for small n.  The conclusion would be that j(n![\tau](http://latex.mathoverflow.net/png?%5Ctau)) is, again for small n, an integer.