A geometric interpretation of recurrence relation r[n]==r[n-1]+1/r[n-1]

Drawing triangles defined by intersections of a set of concentric circles and two half lines starting at the center of the circles (Mathematica 10.2)

l1= HalfLine[{0,0},{1,1}];
l2= HalfLine[{0,0},{1,2}];
c1 = Circle[{0,0},1];
c2 = Circle[{0,0},2];
c3=Circle[{0,0},5/2];
c4=Circle[{0,0},29/10];
p1=RegionIntersection[l1,c1];
p2=RegionIntersection[l2,c1];
p3=RegionIntersection[l1,c2];
p4=RegionIntersection[l2,c2];
p5=RegionIntersection[l1,c3];
p6=RegionIntersection[l2,c3];
p7=RegionIntersection[l1,c4];
Graphics[
{
l1,l2,c1,c2,c3,c4,
Red,PointSize[0.02],p1,p2,p3,p4,p5,p6,p7,
Blue,
Triangle[Sequence@@@{p1,p2,p3}],
Triangle[Sequence@@@{p3,p4,p5}],
Triangle[Sequence@@@{p5,p6,p7}]
}
]

Deriving the recurrence equation

Area of triangle closest to the center:

Area@Triangle[{r[n]{Cos[ϕ1],Sin[ϕ1]},r[n+1]{Cos[ϕ1],Sin[ϕ1]},r[n]{Cos[ϕ2],Sin[ϕ2]}}]//FullSimplify

Area of second triangle:

Area@Triangle[{r[n+1]{Cos[ϕ1],Sin[ϕ1]},r[n+2]{Cos[ϕ1],Sin[ϕ1]},r[n+1]{Cos[ϕ2],Sin[ϕ2]}}]//FullSimplify

Requiring areas to be equal yields a recurrence equation:

r[n] (r[n]-r[1+n])==r[1+n] (r[1+n]-r[2+n])

which can be rewritten as:

Solve[r[n] (r[n]-r[1+n])==r[1+n] (r[1+n]-r[2+n])/.n->n-2,r[n]]//FullSimplify

RecurrenceTable[{r[n] (r[n]-r[1+n])==r[1+n] (r[1+n]-r[2+n]),r[1]==1,r[2]==2},r[n],{n,1,10}]

Proving equivalence with r[n]==r[n-1]+1/r[n-1] (assuming r[n]>0) :

r[n]==r[n-1]+1/r[n-1] =>
r[n+1]==r[n]+1/r[n]=>
r[n](r[n + 1] - r[n])==1

Also:

r[n]==r[n-1]+1/r[n-1] =>
r[n+2]==r[n+1]+1/r[n+1]=>
r[n+1](r[n + 2] - r[n+1])==1

Hence:

r[n](r[n + 1] - r[n])==r[n+1](r[n + 2] - r[n+1])

This is the relation derived above that causes the triangles to have equal area.

A nice illustration:

numRings=15;
numSectors=60;
rad=RecurrenceTable[{r[n] (r[n]-r[1+n])==r[1+n] (r[1+n]-r[2+n]),r[1]==1,r[2]==2},r[n],{n,1,numRings}]//N;
points=Flatten[Table[rad[[n]] {Cos[ϕ],Sin[ϕ]},{n,1,numRings},{ϕ,0,2π-2π/numSectors,2π/numSectors}],1];
triangles=Flatten[Table[{{numSectors(i+1)+j,numSectors i +Mod[1+j,numSectors,1],numSectors i+j},{numSectors(i+1)+j,numSectors(i+1)+Mod[1+j,numSectors,1],numSectors i +Mod[1+j,numSectors,1]}},{i,0,numRings-2},{j,1,numSectors}],2];
triangles=Table[{numSectors(i+1)+j,numSectors i +Mod[1+j,numSectors,1],numSectors i+j},{i,0,numRings-2},{j,1,numSectors}];
Graphics[GraphicsComplex[points,{FaceForm[LightBlue],EdgeForm[Darker@Red],Polygon/@triangles}]]