Define a(n) as the number of "necklace partitions" of n with elements not exceeding 3. If my calculations are right, then OEIS does not contain this sequence. I have: [1]: a(1) = 1 [11, 2]: a(2) = 2 [111, 12, 3]: a(3) = 3 [1111, 112, 13, 22]: a(4) = 4 [11111, 1112, 113, 122, 23]: a(5) = 5 [111111, 11112, 1113, 1122, 1212, 123, 222, 33]: a(6) = 8 [1111111, 111112, 11113, 11122, 11212, 1123, 1213, 1222, 133, 223]: a(7) = 10 [11111111, 1111112, 111113, 111122, 111212, 11123, 112112, 11213, 11222, 1133, 12122, 1232, 1313, 2222, 233]: a(8) = 15 [111111111, 11111112, 1111113, 1111122, 1111212, 111123, 1112112, 111213, 111222, 11133, 112113, 112122, 11232, 11313, 121212, 12132, 12213, 12222, 1233, 1323, 2223, 333]: a(9) = 22 And OEIS returns no hits for 1,2,3,4,5,8,10,15,22. Perhaps I've miscounted, and I would appreciate confirmation before we submit a new sequence. On Wed, Dec 7, 2011 at 12:31 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Another math-fun member, using a brief Mathematica program, also found 88. It looks as though that's the correct answer.
(Though, I haven't yet compared the two sets of output, the other of which is in terms of 1's, 2's, and 3's.)
--Dan
Tom Duff wrote:
<< On Tue, 6 Dec 2011, math-fun-request@mailman.xmission.com wrote:
Date: Tue, 6 Dec 2011 16:17:33 -0800 (GMT-08:00) From: Dan Asimov <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Counting problem Message-ID: < 10736092.1323217054054.JavaMail.root@elwamui-norfolk.atl.sa.earthlink.net>
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Ha -- and in my spare moments today I counted 87 (by hand).
So, can we split the difference and agree on 86 ?
--Dan
Allan wrote:
<< As luck would have it, this afternoon I had a boring staff meeting in which I enumerated all of them ... I think. And the answer is 85 ... I think.
I found 88. Here they are. Where is my mistake? (Each string has 3 for a 30 degree angle, 6 for 60, 9 for 90. These are the lexicographically smallest representatives of their equivalence classes.)
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