Efficient Algorithm for Calculating (n^x)/d

[Atran]

Hi, I've recently found the following algorithm and I'm willing to share it:

For example, given an integer n=46, an exponent x=5, and d=13, we have 46⁵/13.
First, we set the equation, n = q*d + r = q*13 + r
We have, 46 = 3*13 + 7
Thus, [q=3, r=7, d=13]
And: 46⁵/13 = 3*46⁴ + 7*(3*46³ + 7*(3*46² + 7*(3*46¹ + 7*(46/13))))

Let [q, r, d] and we have (n^4)/d, then:
(n^4)/d = q*n^(4-1) + r*(q*n^(4-2) + r*(q*n^(4-3) + r*(n/d)))

I don't know if it's interesting to you or not, but what are your thoughts about it?

 

[Stephen Tashi]

Perhaps you'll get more comments if you write the expression in a less computationally efficient manner.

$\frac{(qd + r)^5}{d} = q(qd+r)^4 + rq(qd+r)^3 + r^2q(qd+r)^2 + r^3q(qd+r) + r^4\frac{(qd+r)}{d}$