[seqfan] Re: a question about arrangements of pennies

[William Keith]

> There are several ways to enforce rigidity.  Perhaps
> the best is to add a further constraint to the first
> problem.  Form a second graph with nodes = triangles,
> edges = two triangles that share an edge.  Then this second graph
> must be connected.

Counterexample!

   ****
  *****
**    **
  *     *
**    **
  *****
   ****

If that doesn't come across well graphically, picture a "mostly 
hexagon" with 4 in the first row, 5 in the second, the third row being 
2-blank-2, the fourth row being two pennies, just one in the middle of 
each pair, and the bottom half a vertical reflection.

This is mechanically rigid -- I think that's the intuition you sought 
to capture -- but the associated graph as you describe would be two 
disconnected arcs.

(Which is not to say that your sequence isn't also interesting; it 
would be counting such graphs with a weight of 3 per node, minus 2 per 
edge, plus a somewhat more complicated summand accounting for the 
possibility of cycles and part-cycles of various sizes.  It would be a 
"stronger" subset of rigid.)

William Keith