Talk:6-simplex honeycomb

Counts

I get 17 separate tilings here.

6D (17 - here is the spaces to the next node)
     7  61  52  511 43 421 4111
     331 322 3211 3121 31111
     2221 22111 21211 211111 1111111

7D  agrees with A00029 = 30
 7    8  71  62 611 53 521 5111
 6    44  431 422 4211 4121 41111
 8    332 3311 3131 3221 3212 32111 31211 311111
 8    2222 22211 22121 221111 212111 211211
      2111111 11111111
29

8D agrees with A00029 = 46
 7     9  81  72  711  63  621  6111
 6     54  531 522  5211  5121  51111
 9     441 432 4311 4131 4221 4212 42111 41211 411111
10     333 3321 3231  33111  31311  3222  32211  32121  32112 31221
 4     321111  312111 311211 3111111
 4     22221  222111  221211  212121
 5     2211111  2121111 2112111 21111111 111111111
45

This method is horribly unreliable. The case for 9D gives me here 49, although sloane's A000029 suggets 77 here, so I am missing some 28 examples. The calculations for 10D, 12D, 14D agree exactly with sloane --Wendy.krieger (talk) 10:37, 13 May 2011 (UTC)[reply]

Yes, I had 17 listed, but as a typo wrote 16. And second excellent! I see at [1], A000029, has the unique cyclic binary permutations, and count one larger than honeycombs (which skip the zero ring case). Knowing the count is more important than getting the complete list. I'm glad you have a good generating process for even dimensions! Tom Ruen (talk) 20:29, 13 May 2011 (UTC)[reply]

Walikhanpti3 37.111.171.138 (talk) 19:37, 30 June 2024 (UTC)[reply]