M Theory Lesson 246

The inverses of $C^n$ (that is, powers of the circulant $C=(0,1,1)$) also behave very nicely. A little algebra shows that for $C^n=(x,y,y)$, $C^{-n}=(a,b,b)$, where

$a=\frac{x+y}{x^2-2y^2+xy}$
$b=\frac{1-ax}{2y}$

In the special cases of interest, $(x,y)=(n,n+1)$ or $(n+1,n)$, we find (respectively) that

$C^{-n}=(\frac{-(2n+1)}{3n+2},\frac{n+1}{3n+2},\frac{n+1}{3n+2})$ or
$C^{-n}=(\frac{2n+1}{3n+1},\frac{-n}{3n+1},\frac{-n}{3n+1})$

and $\mid a\mid +\mid b\mid =1$ in all cases. For example, the inverse of $(2,3,3)$ is the circulant $(1/8)(-5,3,3)$. These forms for the inverse of a positive circulant hold even when $n$ is not an ordinal. For the more general case of a positive circulant of the form $(n,n+d,n+d)$, the sum $\mid a\mid +\mid b\mid =1/d$.