Prime numbers of the form $(1\times11\times111\times1111\times...)-(1+11+111+1111+...)$ Let
$$R(1) = 1-1,$$
$$R(2) = (1\times11) - (1+11),$$
$$R(3) = (1\times11\times111) - (1+11+111),$$ and so on...

$$R(4)=1355297\quad\text{(a prime number!)}$$

$R(4)$ is the only prime I found of such form up to $R(200)$. Are there anymore primes of such form?
Probabilistic answer: If $\phi_n\in {\Bbb N}$ is a sequence of 'random' integers going to infinity then the probability of $\phi_n$ being a prime is $\sim 1/\ln \phi_n$. When $\sum_{n\geq N} \frac{1}{\ln \phi_n}$ goes to zero fast with $N$ it is most 'likely' that there are no primes among these numbers for $N$ large. The words 'random' and 'likely' are obviously subject to interpretations.
I wanted to post this as a comment but it was too long.

I don't know if it will help but consider to rewrite your function in this way:

$$R(n) = (1 \times 11 \times 111 \times \ldots) - (1 + 11 + 111 + 1111 + \ldots)$$

$$R(n) = \left((10^0) \times (10^0 + 10^1) \times (10^0 + 10^1 + 10^2) \times \ldots\right) - \left((10^0) + (10^1 + 10^0) + (10^2 + 10^1 + 10^0) + \ldots \right)$$

Now with a bit of maths you can check that 

$$\left((10^0) \times (10^0 + 10^1) \times (10^0 + 10^1 + 10^2) \times \ldots\right) = \prod_{k = 1}^n \frac{10^k - 1}{9}$$

For what concerns the second part, it's more amusing. Indeed we have

$$\left((10^0) + (10^1 + 10^0) + (10^2 + 10^1 + 10^0) + \ldots \right)$$

But as we tend to $n$, we can easily check that there will be $n$ terms of $10^0$, $n-1$ terms of $10^1$, $n-2$ terms of $10^2$ and so on, which means

$$\left((10^0) + (10^1 + 10^0) + (10^2 + 10^1 + 10^0) + \ldots \right) = \sum_{k = 0}^{n-1} (n-k)10^k$$

Now with a bit of help (mathematica rules), the productory gives

$$\prod_{k = 1}^n \frac{10^k - 1}{9} = 9^{-k} {Q}_p (10, 10, k)$$

Where ${Q}_p (10, 10, k)$ is the so called q-Pochammer symbol, whose definition is

$${Q}_p (a, q, k) = \prod_{i = 0}^{k-1} (1 - aq^i)$$

Ref https://en.wikipedia.org/wiki/Q-Pochhammer_symbol

Whereas the sum gives

$$\sum_{k = 0}^{n-1} (n-k)10^k = \frac{1}{81}\left(10^{1+n} - 9n - 10\right)$$

Thus your function is

$$R(n) = 9^{-k} {Q}_p (10, 10, k) - \frac{1}{81}\left(10^{1+n} - 9n - 10\right)$$

As I said, I don't know if this helps, but it's a good way to check $R(n)$ quite fast.

Please If I made some mistake, make a comment. I liked this question and I immediately got involved in trying to find a suitable general form for $R(n)$.