[Pehmyt]

I'm not sure if I understand what you're saying, but naturally the formulas are not supposed to work for regular 2d6 (except for results below 6), the distribution is different as soon as we start rerolling.

In some approximate sense k represents the number of 6's rolled, at least for large values, but not exactly. See my example for getting DRN=7: there we have 7 = 5*1+1+1, so k=1 and r=1 and the P(DRN=7) = ((6-1)*5*1+1)/6^3 = 26/216 = 13/108, but among the 6 different ways of getting this there are 4 without any 6's and 2 with 1.

So k and r don't really depend on how you got your result, they are computed simply from the desired number X as
r = (X-1) % 5 (remainder)
k = (X-1)/5 (integer division, remainder dropped)

It might be possible to divide the result in two parts, each representing probabilities of getting the asked DRN with a given number of 6's. I have not looked into that, but if it is possible, it might lead to a more intuitive proof. Mine is using induction on k, and does not look very nice. On the other hand it works for any N-sided dice (N>1).