OFFSET
1,2
COMMENTS
This triangle is obtained from the array A212360 by summing in the row number n, for n>=1, all entries related to partitions of n with the same number of parts m.
a(n,m) is the total number of necklaces of n beads (C_n symmetry) corresponding to all the color multinomials obtained from all p(n,m)=A008284(n,m) partitions of n with m parts, written in nonincreasing form, by 'exponentiation'. Therefore only m from the available n colors are present, and a(n,m) gives the number of necklaces with n beads with only m of the n available colors present, for m from 1,2,...,n, and n>=1. All of the possible color assignments are counted.
See the comments on A212359 for the Abramowitz-Stegun (A-St) order of partitions, and the 'exponentiation' to obtain multisets, used to encode color multinomials, from partitions.
FORMULA
EXAMPLE
n\m 1 2 3 4 5 8 7 8 ...
1 1
2 2 1
3 3 6 2
4 4 24 36 6
5 5 60 300 240 24
6 6 180 1820 3900 1800 120
7 7 378 9030 42000 50400 15120 72
8 8 952 40824 357420 882000 670320 141120 5040
...
Row n=9: 9 2088 169512 2610720 11677680 17781120 9313920 1451520 40320.
Row n=10: 10 4770 673560 17193960 128598624 345144240 355622400 136080000 16329600 362880.
a(2,2)=1 from the color monomial c[1]^1*c[2]^1= c[1]*c[2] (from the m=2 partition [1,1] of n=2). The necklace in question is cyclic(12) (we use j for color c[j] in these examples).
a(3,1) = 3 from the color monomials c[1]^3, c[2]^3 and c[3]^1. The three necklaces are cyclic(111), cyclic(222) and cyclic(333).
In general a(n,1)=n from the partition [n] providing the color signature (exponent), and the n color choices.
a(3,2) = 6 from the color signature c[.]^2 c[.]^1, (from the m=2 partition [2,1] of n=3), and there are 6 choices for the color indices. The 6 necklaces are cyclic(112), cyclic(113), cyclic(221), cyclic(223), cyclic(331) and cyclic(332).
a(3,3) = 2. The color multinomial is c[1]*c[2]*c[3] (from the m=3 partition [1,1,1]). All three available colors are used. There are two non-equivalent necklaces: cyclic(1,2,3) and cyclic(1,3,2).
a(4,2) = 24 from two color signatures c[.]^3 c[.] and c[.]^2 c[.]^2 (from the two m=2 partitions of n=4: [3,1] and [2,2]). The first one produces 4*3=12 necklaces, namely 1112, 1113, 1114, 2221, 2223, 2224, 3331, 3332, 3334, 4441, 4442 and 4443 all taken cyclically. The second color signature leads to another 2*6=12 necklaces: 1122, 1133, 1144, 2233, 2244, 3344, 1212, 1313, 1414, 2323, 2424 and 3434, all taken cyclically. Together they provide the 24 necklaces counted by a(4,2).
CROSSREFS
KEYWORD
AUTHOR
Wolfdieter Lang, Jun 27 2012
STATUS
approved