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