1, 1, 2, 3, 4, 5, 8, 10, 16, 23, 36, 53, 88, 135, 225, 366, 616, 1025, 1759, 2987, 5174, 8929, 15578, 27158, 47703, ... and yes not in OEIS. Perhaps it would be more correct to call them "bracelet" partitions rather than necklace partitions since you are allowing these necklaces to be "turned over." The necklace version is A093305. As far as a program: consider the one I wrote in PARI for the OEIS https://oeis.org/transforms_pari.txt ? V=vector(25) %1 = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ? V[2]=V[3]=V[4]=1 %2 = 1 ? trv_polygon(V) %3 = [1, 1, 2, 3, 4, 5, 8, 10, 16, 23, 36, 53, 88, 135, 225, 366, 616, 1025, 1759, 2987, 5174, 8929, 15578, 27158, 47703] ------ Original Message ------ Received: Wed, 07 Dec 2011 01:13:39 PM PST From: Allan Wechsler <acwacw@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] math-fun Digest, Vol 106, Issue 11
Excellent, did you use a program or just debug my attempt by eye?
On Wed, Dec 7, 2011 at 3:26 PM, Veit Elser <ve10@cornell.edu> wrote:
For n=8 you left out 1223. I got a(8)=16 and a(9)=23. This sequence is not in OEIS.
Veit
On Dec 7, 2011, at 2:28 PM, Allan Wechsler wrote:
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>
Content-Type: text/plain; charset=UTF-8
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.)
It goes without saying that .
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