RFE July 2026: Ferrites and Barlow packings

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Sean A. Irvine

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Jul 2, 2026, 4:23:28 PMJul 2
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Hi

Can someone please compute more terms for the following sequences concerning ferrites and Barlow packings?


These are the six remaining sequences with the obnoxious keyword combination "easy", "more". The hardest part will likely be obtaining access to corresponding paper by T. J. McLarnan.

Thank you to all those who responded to last month's question. We did not get a complete resolution, but progress was made.

Robert McKone

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Jul 3, 2026, 12:47:55 PMJul 3
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Hello Sean,

Attached is a short note showing:
A011947(n)=A045683(2n+1)

It gives a direct coding between the Barlow packing condition and the binary necklace condition.  This gives a simple formula for computing a(n).

Mathematica code:
a[n_] := Total[MoebiusMu[#]*2^(((2 n + 1)/# - 1)/2) & /@ Divisors[2 n + 1]];

Sorry for its rough format, I saw your email earlier today when at work and I was thinking about the layers and necklaces since, and only had time to write this up after midnight.

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a011947_a045683_proof.pdf

Anthony Neves

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Jul 3, 2026, 5:29:29 PMJul 3
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Greetings,

Please find the McLarnan paper attached below.

Note that A011947 is found in Table 4; A011958 in Table 5; and A011961A011962A011963, and A011964 in Table 7 with parameters N = 4, 6, 8, and 10, respectively.

- AMN

McLarnan-1981.pdf

brad klee

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Jul 4, 2026, 12:58:01 AMJul 4
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This isn't, imo, a good use case for LLMs but one to triage anyways, mainly 
with a focus on autonomous refereeing. 

The setup is that two LLM's testify regarding evidence they've collected and 
the payoff matrix is essentially the one from prisoner's dilemma. 

The best we can do is a minimum sentence for Harm.On.ica and Claude, 
which they did obtain for the data of Table 7. 

I don't know if this is actionable, but the reference implementation looks 
concise enough for a human reviewer in finite time. 

[  ] chaotic disks update :  :

M.F. Hasler also asked for more rigor on the transcendent digits claim, so 
we ran a burner to 50K finding a 4:1 wall:body collision ratio and a very 
strong linear signal over the 10K essential data: 


The question we're debating on youtube (lol) is whether these "first terms" 
will ultimately reach a revival with roughly symmetric negative slope guiding
bitwise complexity back toward its crystalline initial condition. 

My opinion or belief is also an Occam's razor argument that once the velocity 
vectors move off an octagonal star, the feedback looping of position and 
momentum can't be expected to reach a logistic turnaround. 

The wildest periodicity conjecture we've come up with is this: 

If the phase volume is essentially zero in the momentum space, and the 
few admissible momentum vectors form a strict D4 star, then we expect periodic 
trajectories in the algebraic position space of a square or maybe rectangular 
container. 

That makes conceptual sense, but it's even more difficult to prove than an 
increasingly complex hierarchy of special case crystalline initial conditions. 

What we're doing is not the same as polygon billiards, so I guess it's not 
also immediately relevant to this recent paper: 

Miranda, Ramos, "Classical Billiards can Compute"

I don't know if any techniques from Veech or Rauzy would be helpful, but 
I can be interested to look more in that direction as necessary. 


Thanks, 





--Brad

Andrei Zabolotskii

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Jul 5, 2026, 1:54:17 PMJul 5
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The author of the paper is alive and well. I got in touch with him after I tried and failed to replicate A011958. We'll see how it goes.

Andrei

суббота, 4 июля 2026 г. в 05:58:01 UTC+1, brad...@proton.me:

Daniel Okwor

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Aug 4, 2026, 7:22:54 PM (24 hours ago) Aug 4
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Following up on this one in case A011958 is still open.

For what it is worth, A011961 through A011964 all picked up terms from R. J. Mathar on 19 July and none of them still carry the "more" keyword, so A011947 and A011958 look like the only two left of the original six.

The short version on A011958 is that McLarnan's Table 5 values are right, and the printed formula is what cannot be implemented as written. That would account for both Andrei's failed replication and the half-integers Mathar flagged on the sequence in July.

I checked it two ways, neither of which uses the formula from the paper.

First I enumerated directly. I built the group from McLarnan's prose in the CdI2 section rather than from his equations, so translations by an even number of layers, inversion centers between close-packed layers (the reflections with cycle structure x1^2 x2^(N-1)), and the anti-identification. Then I ran over all 2^(2N) Hagg symbols, kept the ones where 3 divides Sum r_i, counted orbits, and Moebius inverted for exact layer-number. No formula anywhere in that. It reproduces a(1) through a(14) exactly.

Second I did an independent Burnside count on the same power group, using a small mod 3 DP over cycle lengths. That agrees with all 25 published terms including a(25) = 3753005281872, and it extends cheaply.

So the table is sound and the data currently in the OEIS is correct. I have not worked out which symbol in the printed I'(N) is off, and Andrei may already have that straight from McLarnan.

Twenty further terms.

a(26) = 14434625803920
a(27) = 55600017771599
a(28) = 214457171447220
a(29) = 828248296810990
a(30) = 3202559919944296
a(31) = 12397005783794016
a(32) = 48038396763471872
a(33) = 186330749650975598
a(34) = 723401731328271460
a(35) = 2810932436006173716
a(36) = 10931403907303050080
a(37) = 42543842210921313990
a(38) = 165697069621392018234
a(39) = 645793707150723179160
a(40) = 2518595457718310508776
a(41) = 9828665200485440255650
a(42) = 38378597448833927426442
a(43) = 149944287705606068180436
a(44) = 586145851937364327840300
a(45) = 2292481554238049762231644

a(28) is the last one that fits on the DATA line, so the rest would want a b-file. I can put that together along with a program unless someone is already on it.

On A011947 I verified Robert's identity independently. Enumerating close-packings by their stabilizer in the power group leaves exactly one class matching 1, 1, 3, 7, 14, 31 at N = 2, 6, 10, 14, 18, 22, and that class has a stabilizer of order 4 generated by an inversion, a mirror and a 6_3 axis, which is what his two necklace conditions generate. His balance step earns its keep as well, since w = s s' forces Sum r_i = 0 and the packing comes out hexagonal with layer-number exactly 4n+2. I am happy to file the extended terms with credit to him, but it is his result so it should be his call.

Happy to share code for any of it.

Daniel

Andrei Zabolotskii

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4:41 AM (14 hours ago) 4:41 AM
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Daniel, that's excellent. I think you can go ahead with updating A011958.
I don't have anything to add from Timothy McLarnan: he responded to me quickly but couldn't immediately help.

Andrei


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