A213934 -id:A213934 - cp's OEIS frontend

A212359 Partition array for the number of representative necklaces (only cyclic symmetry) with n beads, each available in n colors. Only the color type (signature) matters.

1, 1, 1, 1, 1, 2, 1, 1, 2, 3, 6, 1, 1, 2, 4, 6, 12, 24, 1, 1, 3, 4, 5, 10, 16, 20, 30, 60, 120, 1, 1, 3, 5, 6, 15, 20, 30, 30, 60, 90, 120, 180, 360, 720, 1, 1, 4, 7, 10, 7, 21, 35, 54, 70, 42, 105, 140, 210, 318, 210, 420, 630, 840, 1260, 2520, 5040

Offset: 1

Wolfdieter Lang, Jun 25 2012

The row lengths sequence is A000041(n), n>=1.
The partitions are ordered like in Abramowitz-Stegun (A-St order). For the reference see A036036, where also a link to a work by C. F. Hindenburg from 1779 is found where this order has been used.
A necklace with n beads (n-necklace) has here only the cyclic C_n symmetry group. This is in contrast to, e.g., the Harary-Palmer reference, p. 44, where a n-necklace has the symmetry group D_n, the dihedral group of degree n (order 2n), which allows, in addition to C_n operations, also a turnover (in 3-space) or a reflection (in 2-space).
The necklace number a(n,k) gives the number of n-necklaces, with up to n colors for each bead, belonging to the k-th partition of n in A-St order in the following way. Write this partition with nonincreasing parts (this is the reverse of the partition as given by A-St), e.g., [3,1^2], not [1^2,3], is written as [3,1,1], a partition of n=5. In general [p[1],p[2],...,p[m]], with p[1]>=p[2]>=...>=p[m]>=1, with m the number of parts. To each such partition of n corresponds an n-multiset obtained by 'exponentiation'. For the given example the 5-multiset is {1^3,2^1,3^1}={1,1,1,2,3}. In general {1^p[1],2^p[2],...,m^p[m]}. Such an n-multiset representative (of a repetition class defined by the exponents, sometimes called signature) encodes the n-necklace color monomial by c[1]^p[1]*c[2]^p[2]*...*c[m]^p[m]. For the example one has c[1]^3*c[2]*c[3]. The number of 5-necklaces with this color assignment is a(5,4) because [3,1,1] is the 4th partition of 5 in A-St order. The a(5,4)=4 non-equivalent 5-necklaces with this color assignment are cyclic(c[1]c[1]c[1]c[2]c[3]), cyclic(c[1]c[1]c[1]c[3]c[2]), cyclic(c[1]c[1]c[2]c[1]c[3]) and cyclic(c[1]c[1]c[3]c[1]c[2]).
Such a set of a(n,k) n-necklaces for the given color assignment stands for other sets of the same order where different colors from the repertoire {c[1],...,c[n]} are chosen. In the example, the partition [3,1,1]  with the representative multiset [1^3,2,3] stands for all-together 5*binomial(4,2) = 30 such sets, each leading to 4 possible non-equivalent 5-necklace arrangements. Thus one has, in total, 30*4=120 5-necklaces with color signature determined from the partition [3,1,1]. See the partition array A212360 for these numbers.
For the example n=4, k=1..5, see the Stanley reference, last line, where the numbers a(4,k) are, in A-St order, 1, 1, 2, 3, 6, summing to A072605(4).
a(n,k) is computed from the cycle index Z(C_n) for the cyclic group (see A212357 and the link given there) after the variables x_j have been replaced by the j-th power sum sum(c[i]^j,i=1..n), abbreviated as Z(C_n,c_n) with c_n:=sum(c[i],i=1..n), n>=1. The coefficient of the color assignment representative determined by the k-th partition of n in A-St order, as explained above, is a(n,k). See the Harary-Palmer reference, p. 36, Theorem (PET) with A=C_n and p. 36 eq. (2.2.10) for the cycle index polynomial Z(C_n). See the W. Lang link for more details.
The corresponding triangle with summed entries of row n which belong to partitions of n with the same number of parts is A213934. [Wolfdieter Lang, Jul 12 2012]

			n\k  1 2 3 4 5  6  7  8  9 10  11  12  13  14  15
1    1
2    1 1
3    1 1 2
4    1 1 2 3 6
5    1 1 2 4 6 12 24
6    1 1 3 4 5 10 16 20 30 60 120
7    1 1 3 5 6 15 20 30 30 60  90 120 180 360 720
...
See the link for the rows n=1..15 and the corresponding color polynomials for n=1..10.
a(4,5)=6 because the partition in question is 1^4, the corresponding color type representative multinomial is c[1]*c[2]*c[3]*c[4] (all four colors are involved), and there are the 6 C_4 non-equivalent 4-necklaces (we use here j for color c[j]): 1234, 1243, 1324, 1342, 1423 and 1432 (all taken as cyclically). For this partition there is only one color choice.
a(4,4)=3 because the partition is [2,1^2]=[2,1,1], the color representative monomial is c[1]^2*c[2]*c[3], and the arrangements are 1123, 1132  and  1213 (all taken cyclically). There are, in total, 4*binomial(3,2)=12 color multinomials of this signature (color type) in Z(C_4,c_4).

F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, NY, 1973, p. 36, (2.2.10).
R. Stanley, Enumerative combinatorics, Vol. 2, Cambridge University Press, Cambridge, 1999, p. 392, 7.24.3 Example.

Álvar Ibeas, First 25 rows, flattened
Wolfdieter Lang, Rows n=1..15 and color polynomials n=1..10.

Cf. A212357 for Z(C_n), A072605 for the row sums.

Cf. A000041 (row lengths), A036036, A185974, A212360, A213934, A318810.

a(n,k) is the number of necklace arrangements with n beads (respecting the cyclic C_n symmetry) with color assignment given by the multiset representative obtained uniquely from the k-th partition of n in A-St order. See the comment for more details and the A-St reference.
From Álvar Ibeas, Dec 12 2020: (Start)
Let L be the k-th partition of n in A-St and d be the gcd of its parts. Abusing the notation, we write a(n, L) for a(n, k) and accordingly for other partition arrays.
a(n, L) = n^(-1) * Sum_{v|d} phi(v) * A036038(n/v, L/v), where L/v is the partwise division of L by v.
a(n, L) = Sum_{v|d} A339677(L/v).
(End)
a(n,k) = A318810(A185974(n,k)). - Andrew Howroyd, Jan 23 2025

A213940 Triangle with entry a(n,m) giving the number of bracelets of n beads (dihedral D_n symmetry) with n colors available for each bead, but only m distinct fixed colors, say c[1],...,c[m], are present, with m from {1,...,n} and n>=1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 3, 2, 3, 1, 3, 6, 6, 12, 1, 7, 20, 26, 30, 60, 1, 8, 40, 93, 150, 180, 360, 1, 18, 106, 424, 633, 1050, 1260, 2520, 1, 22, 304, 1180, 3260, 5040, 8400, 10080, 20160, 1, 46, 731, 4844, 16212, 29244, 45360, 75600, 90720, 181440

Offset: 1

Wolfdieter Lang, Jul 20 2012

This triangle is obtained from the partition array A213939 by summing in row n, for n>=1, all entries related to partitions of n with the same number of parts m.
a(n,m) is the number of bracelets of n beads (dihedral D_n symmetry) corresponding to the representative color multinomials obtained from all partitions of n with m parts by 'exponentiation', hence only m from the available n colors are present. As a representative multinomial of each of the p(n,m)=A008284(n,m) such m-color classes we take the one where the considered m part partition of n, [p[1],...,p[m]], written in nonincreasing order, is distributed as exponents on the color indices like c[1]^p[1]*...*c[m]^p[m]. That is only the first m colors from the n available ones are involved.
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.
The row sums of this triangle coincide with the ones of array A213939, and they are given by A213943.
Number of n-length bracelets w over a k-ary alphabet {a1,a2,...,ak} such that #(w,a1) >= #(w,a2) >= ... >= #(w,ak) >= 1, where #(w,x) counts the letters x in word w (bracelet analog of A226874). - Andrew Howroyd, Sep 26 2017

			n\m  1  2   3    4     5     6     7     8     9     10 ...
1    1
2    1  1
3    1  1   1
4    1  3   2    3
5    1  3   6    6    12
6    1  7  20   26    30    60
7    1  8  40   93   150   180   360
8    1 18 106  424   633  1050  1260  2520
9    1 22 304 1180  3260  5040  8400 10080 20160
10   1 46 731 4844 16212 29244 45360 75600 90720 181440
...
a(5,3) = 2 + 4 = 6, from A213939(5,4) + A213939(5,5), because k(5,3,1) = 4 and p(5,3) = 2.
a(2,1) = 1 because the partition [2] of n=2 with part number m=1 corresponds to the representative color multinomial (here monomial) c[1]^2 = c[1]*c[1], and there is one such representative bracelet. There is another bracelet color monomial in this class of n=2 colors where only m=1 color is active: c[2]*c[2]. See the triangle entry A213941(2,1)=2. The same holds for the necklace case.
a(3,1) = 1 from the color monomial representative c[1]^3. This class has 2 other members: c[2]^3 and c[3]^3. See A213941(3,1)=3. The same holds for the necklace case.
Like in the necklace case one has in general a(n,1)=1 and A213941(n,1) = n from the partition [n] providing the color signature and a representative c[1]^n.
a(3,2) = 1 from the representative color multinomial c[1]^2*c[2] (from the m=2 partition [2,1] of n=3) leading to just one representative bracelet (and necklace) cyclic(112) (when one uses j for color c[j]). The whole class consists of A213941(3,2)=6 bracelets (or necklaces): cyclic(112), cyclic(113), cyclic(221), cyclic(223), cyclic(331) and cyclic(332).
a(3,3) = 1. The representative color multinomial is c[1]*c[2]*c[3] (from the m=3 partition [1,1,1]). There is only one bracelet cyclic(1,2,3) which constitutes already the whole class (A213941(3,3)=1). The necklace cyclic(1,3,2) becomes equivalent under D_3.
a(4,2) = 3 from two representative color multinomials c[1]^3*c[2] and c[1]^2*c[2]^2 (from the two m=2 partitions of n=4: [3,1] and [2,2]). The first one has one representative bracelet, namely cyclic(1112), the second one leads to the two representative bracelets: cyclic(1122) and cyclic(1212). Together these are the 3 bracelets counted by a(4,2). The first color class c[.]^3*c[.] consists of 4*3=12 bracelets, when all 4 colors are used. The second one consists of 2*6=12 bracelets. Together they sum up to the 24 bracelets counted by A213941(4,2). In this example the necklace case does not differ from the bracelet one.

Andrew Howroyd, Table of n, a(n) for n = 1..1275

Columns k=2..5 are A213942, A214307, A214309, A214311.
Cf. A008284, A213939, A213941, A213943 (row sums), A226874, A152176.
Cf. A213934 (cyclic symmetry).

Cyc(v)={my(s=vecsum(v)); sumdiv(gcd(v), d, eulerphi(d)*(s/d)!/prod(i=1, #v, (v[i]/d)!))/s}
CPal(v)={my(odds=#select(t->t%2,v), s=vecsum(v));  if(odds>2, 0, ((s-odds)/2)!/prod(i=1, #v, (v[i]\2)!))}
T(n,k)={my(t=0); forpart(p=n, t+=Cyc(Vec(p))+CPal(Vec(p)), [1,n], [k,k]); t/2}
for(n=1, 10, for(k=1,n, print1(T(n,k), ", ")); print); \\ Andrew Howroyd, Sep 26 2017

\\ faster version; here U is A226874 as vector of polynomials.
U(n)={Vec(serlaplace(prod(k=1, n, 1/(1-y*x^k/k!) + O(x*x^n))))}
T(n)={my(t=U(n)); vector(n, n, vector(n, k, ((1/n)*sumdiv(n, d, eulerphi(n/d) * polcoeff(t[d+1], k)) + if(n%2, sum(d=0, (n-1)/2, binomial((n-1)/2, d)*polcoeff(t[d+1], (k-1))), polcoeff(t[n/2+1], k) + sum(d=0, n/2-1, binomial(n/2-1, d)*(2^d + if(d%2, 0, binomial(d, d/2)))*polcoeff(t[n/2-d], k-2))/2))/2))}
{ my(t=T(10)); for(n=1, #t, print(t[n])) } \\ Andrew Howroyd, Dec 22 2017

a(n,m) = Sum_{j=1..p(n,m)}A213939(n,k(n,m,1)+j-1), with k(n,m,1) the position where in the list of partitions of n in A-St order the first with m parts appears, and p(n,m) is the number of partitions of n with m parts shown in the array A008284. E.g., n=5, m=3: k(5,3,1)=4, p(5,3)=2.

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A212359 Partition array for the number of representative necklaces (only cyclic symmetry) with n beads, each available in n colors. Only the color type (signature) matters.

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