Sequence 1: Take each of the blue circles in turn as the starting point of an Arbelos chain.
Let be the outer circle, be the kth blue circle, and let them be tangent at .
To form the new (purplse) sequence, I need
1. The two radii
2. The line (vector) passing through the two centers
3. A point of tangency with another circle, , which is tangent to both .
The point satisfies 3. for every blue circle.

Notes:
1) Adjacent blue circles are tangent along the red circle (in the projection, this is the center line).
Call this circle Charles.
The reflection of a solution to the problem across Charles, is a solution to the problem.
2) Drag tangent point A1. Take the greatest circle in the chain (green) as the reference position for the sequence. (Alternately, I could constrain A1).
2)There is a smallest circle size, calculated before a sequence is formed.
3) The A_k can be merged without overlapping circles.
4) I have assumed the circles are anonymous (algebraically equivalent). My assumption is valid. It may also prove false for an apparently identical problem. Say I place n pictures in each of the n blue circles. A1 is the position of the first picture. Now consider Note 2. Measure will not be preserved: the first picture will cycle abruptly, snapping to the greatest circle.
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The Tangent Circle Problem: