Riemann R-function is defined as: $$R (x)=\sum _{n=1}^{\infty } \frac{\mu (n) \text{li}\left(x^{1/n}\right)}{n}$$
I know how it appears in Riemann explicit formula for prime counting function $\pi (x)$ and that it is (with some minor terms) the so called best continuous estimator for $\pi (x)$
But what it really represents?
Look instead at
$$Li(x) = \int_2^x\frac{1_{t > 2}}{\log t}dt, \qquad R(x) =\sum_{n=1}^\infty \frac{\mu(n)}{n} Li(x^{1/n})$$
The Riemann explicit formula is $$\psi(x)= \sum_{p^k \le x} \log p = x - \sum_\rho \frac{x^\rho}{\rho} - \log 2\pi -\sum_{k =1}^\infty \frac{x^{-2k}}{-2k}$$
(for $x > 1$)
Which gives $$\Pi(x) = \sum_{p^k \le x} \frac{1}{k} = \int_{2-\epsilon}^x \frac{\psi'(x)}{\log x} dx = Li(x) - \sum_\rho Li(x^\rho) -\sum_{k =1}^\infty Li(x^{-2k})$$
And hence $$\pi(x) = \sum_{p \le x} 1 = \sum_{n=1}^\infty \frac{\mu(n)}{n} \Pi(x^{1/n})= R(x) - \sum_\rho R(x^\rho) -\sum_{k =1}^\infty R(x^{-2k})$$
Thus $R(x)$ is the main term in the explicit formula for $\pi(x)$.