A sequence from Buckminster Fuller's Synergetics based on constructions of great circles.

[Alex Howe]

I've been investigating a sequence implicitly defined by Buckminster Fuller's Synergetics (although I've cleaned up the construction and dropped the proto-sacred geometry angle). It's defined by an iterated construction of great circles on a sphere. Fuller computed the first 3-5 terms (depending on how you count them), but to my knowledge, no one has tried to extend it further.

The way I prefer to construct it, you start with the icosahedral symmetry group. This group can be represented by reflections across 15 great circles like these, so a(1)=15:

These 15 great circles intersect at 62 points. If you rotate the sphere around each of these points, you can draw 31 great circles along their equators (including the original 15), so we have a(2)=31:

Fuller actually called these the "primary great circles," and he constructed them by rotating an icosahedron around its vertices, edge midpoints, and face centers, but you can equally construct them just from the 15 mirror planes.

You then repeat the process. These 31 great circles intersect at 242 points, so they can be used to draw 121 great circles (again including the previous steps), and a(3)=121. Fuller called the new circles "secondary great circles":

There's no indication that Fuller or anyone else continued the construction to "tertiary great circles," but if you count up the number of intersections in each fundamental triangle, you find that a(4)=2431.

This sequence obviously grows very fast, since a(n+1) is on the order of a(n)^2. I wrote a Python script that verified a(2), a(3), and a(4) and computed a(5)=1286101. I believe it could compute a(6) in about 500 core-hours of runtime, but I'm hoping someone can check my work before I invest that kind of time/money.

Questions:

1. Has this construction been studied in any other context? (It's definitely not in the OEIS.)

2. Might there be a formula that can compute the terms directly rather than finding every intersection point?

3. Would it be more appropriate to use Fuller's definition where a(1)=31, or my definition where a(1)=15? Or alternatively, the 15 mirror planes can be generated by a subset of 6, so you could have a(1)=6 instead. (If you try to push it back another step, you're back to 15.)

4. Is anyone interested in checking my script and my computation of a(5)?

Finally, note that similar sequences can be constructed for other symmetry groups. Fuller himself explored octahedral symmetry, but he applied a different construction in the first step, so there's no prior literature there.

[D. S. McNeil]

Pretty!

FWIW I can reproduce your values up to a(5), taking as you recommend a(1)=15, which seems more consistent to me than either of the others.

Doug

[Alex Howe]

Great, thanks. Did you do it by computing all the intersection points up to symmetry, or is there another method I'm not aware of?

Alex

[D. S. McNeil]

I counted too, but used H_3 orbit reps.  You take pairs of them, double cosets from the stabilizers, compute the cross product, and canonicalize to get another orbit rep.

Pretty quick to get a(5), about 8 min or so, but it feels like a(6) would take a while!

Doug

[Alex Howe]

My abstract algebra is not strong enough to follow all of that. I just pre-computed all the circle centers in a(4), numbered them so I could track duplicates (long story), and computed all of their cross products up to symmetry (21540 rather than 1.3 million). It only takes a few seconds, but as I said, a(6) is so much bigger that it would need ~500 core-hours.

On a separate note, I took a look at the corresponding sequences for other symmetry groups in 3 dimensions, and I found a sequence that may generate the octahedral case already in the OEIS. It's defined more generally ("starting with four points in general position in the real projective plane"), but it agrees with my results for octahedral symmetry with the subsequence 9, 13, 25, 97, 1741. (I'd need to rewrite some code to check the next term.)

However, this does not work for icosahedral symmetry. The corresponding sequence for six points is A244024, but it overcounts icosahedral symmetry with 6, 15, 51, 870, 342051.

Tetrahedral symmetry is identical to octahedral, except a(1)=6 instead of 9.

The dihedral and cyclic groups are trivial. For a(k>1) they always give n+1 for even n and 2n+1 for odd n.

Alex

[Alex Howe]

And I somehow deleted the number. The first sequence I highlighted was A140468.

[Marc LeBrun]

> Alex Howe

> The dihedral and cyclic groups are trivial. For a(k>1) they always give n+1 for even n and 2n+1 for odd n.

This might make a great addition to the comments at A114753.
(Trivial sequences need love too...)