[seqfan] Brick sequences
franktaw at netscape.net
franktaw at netscape.net
Sat Oct 24 13:44:25 CEST 2009
I came up with an idea for a sequence. Actually, it's several
sequences, some of which I'm not sure how to define mathematically.
Consider a connected structure of bricks, idealized as congruent
rectangles. The bricks on the bottom are at integer offsets; those on
the next row up are at half-integer offsets, the next row is again at
integer offsets, etc. The structure has to stand up.
For the simplest case, we require that each brick be supported by two
bricks under it. If we regard mirror images as the same, this sequence
starts, from n=1 (hand-calculated):
(I suppose a(0) = 1, but I'm not including it here.)
For example, the two structures for n=4 are:
(Imagine the lines on the = are farther apart than most fonts will make
If mirror images are different, the sequence starts:
I'm pretty sure that neither of these is in the OEIS: there are
sequences matching either, but none with even a vaguely related
Note that, for these two sequences, we could just as well use hexagons
as bricks. This won't work for the next one.
But now things get more interesting. The structure:
is not stable; the center of gravity of the top brick is right over the
edge of the brick below it. A small perturbation will send it
is stable; the weight of the top brick holds the center of gravity for
the weight supported by the bricks under it to be over the base brick.
On the other hand, consider:
Now the two top bricks are braced against each other. I think that, by
the definition I want to use, this is actually unstable; perturb the
top two bricks a bit to the right, and the left one will start to tilt,
pushing the right one further right until the left brick falls.
At this point I'm not sure how to proceed. How do you determine
whether a given structure is stable? If you measure the center gravity
of the weight supported by a brick, it must be over another brick; this
is certainly a necessary and sufficient condition. But how do you
distribute the weight for a brick that has two bricks under it? My
second thought (the first thought was clearly wrong) is that the weight
should be distributed according to the position of the center of
gravity of the weight supported by the brick; if this is 1/4 of the way
across the brick, distribute the left-most 3/4 of the weight to the
left and the remaining 1/4 to right. I'm pretty dubious that this is
correct in general, however.
Once this question is settled, there will again be two sequences: one
for mirror images considered the same, and the other considering them
Franklin T. Adams-Watters
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