A032198
"CIK" (necklace, indistinct, unlabeled) transform of 1,2,3,4,...
Original entry on oeis.org
1, 3, 6, 13, 25, 58, 121, 283, 646, 1527, 3601, 8678, 20881, 50823, 124054, 304573, 750121, 1855098, 4600201, 11442085, 28527446, 71292603, 178526881, 447919418, 1125750145, 2833906683, 7144450566, 18036423973
Offset: 1
From _Petros Hadjicostas_, Jan 07 2018: (Start)
We give some examples to illustrate the theory of C. G. Bower about transforms given in the weblink above. We assume we have boxes of different sizes and colors that we place on a circle to form a necklace. Two boxes of the same size and same color are considered identical (indistinct and unlabeled). We do, however, change the roles of the sequences (a(n): n>=1) and (b(n): n>=1) that appear in the weblink above. We assume (a(n): n>=1) = CIK((b(n): n>=1)).
Since b(1) = 1, b(2) = 2, b(3) = 3, etc., a box that can hold 1 ball only can be of 1 color only, a box that can hold 2 balls only can be one of 2 colors only, a box that can hold 3 balls can be one of 3 colors, and so on.
To prove that a(3) = 6, we consider three cases. In the first case, we have a single box that can hold 3 balls, and thus we have 3 possibilities for the 3 colors the box can be. In the second case, we have a box that can hold 2 balls and a box that can hold 1 ball. Here, we have 2 x 1 = 2 possibilities. In the third case, we have 3 identical boxes, each of which can hold 1 ball. This gives rise to 1 possibility. Hence, a(3) = 3 + 2 + 1 = 6.
To prove that a(4) = 13, we consider 5 cases: a box with 4 balls (4 possibilities), one box with 3 balls and one box with 1 ball (3 possibilities), two identical boxes each with 2 balls (3 possibilities), one box with 2 balls and two identical boxes each with 1 ball (2 possibilities), and four identical boxes each with 1 ball (1 possibility). Thus, a(4) = 4 + 3 + 3 + 2 + 1 = 13.
(End)
- Vaclav Kotesovec, Table of n, a(n) for n = 1..1000
- A. K. Agarwal, n-colour compositions, Indian J. Pure Appl. Math. 31 (11) (2000), 1421-1427.
- C. G. Bower, Transforms (2).
- Joshua P. Bowman, Compositions with an Odd Number of Parts, and Other Congruences, J. Int. Seq (2024) Vol. 27, Art. 24.3.6. See p. 31.
- P. Flajolet and M. Soria, The Cycle Construction, SIAM J. Discr. Math., vol. 4 (1), 1991, pp. 58-60.
- P. Flajolet and M. Soria, The Cycle Construction. [pdf file]
- Meghann Moriah Gibson, Combinatorics of compositions, Master of Science, Georgia Southern University, 2017.
- Meghann Moriah Gibson, Daniel Gray, and Hua Wang, Combinatorics of n-color compositions, Discrete Mathematics 341 (2018), 3209-3226.
- Index entries for sequences related to necklaces
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nmax = 30;
f[x_] = Sum[n*x^n, {n, 1, nmax}];
gf = Sum[(EulerPhi[n]/n)*Log[1/(1 - f[x^n])] + O[x]^nmax, {n, 1, nmax}]/x;
CoefficientList[gf, x] (* Jean-François Alcover, Jul 29 2018, after Joerg Arndt *)
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N = 66; x = 'x + O('x^N);
f(x)=sum(n=1, N, n*x^n );
gf = sum(n=1, N, eulerphi(n)/n*log(1/(1-f(x^n))) );
v = Vec(gf)
/* Joerg Arndt, Jan 21 2013 */
A308747
Number of achiral m-color cyclic compositions of n (that is, number of cyclic compositions of n with reflection symmetry where each part of size m can be colored with one of m colors).
Original entry on oeis.org
1, 3, 6, 13, 23, 44, 73, 131, 210, 365, 575, 984, 1537, 2611, 4062, 6877, 10679, 18052, 28009, 47315, 73386, 123933, 192191, 324528, 503233, 849699, 1317558, 2224621, 3449495, 5824220, 9030985, 15248099, 23643522, 39920141, 61899647, 104512392, 162055489, 273617107
Offset: 1
We have a(1) = 1 because we only have one symmetric cyclic composition of n = 1, namely 1_1 (and a part of size 1 can be colored with only one color).
We have a(2) = 3 because we have the following colored achiral cyclic compositions of n = 2: 2_1, 2_2, 1_1 + 1_1.
We have a(3) = 6 because we have the following colored achiral cyclic compositions of n = 3: 3_1, 3_2, 3_3, 1_1 + 2_1, 1_1 + 2_2, 1_1 + 1_1 + 1_1.
We have a(4) = 13 because we have the following colored achiral cyclic compositions of n = 4: 4_1, 4_2, 4_3, 4_4, 1_1 + 3_1, 1_1 + 3_2, 1_1 + 3_3, 2_1 + 2_1, 2_1 + 2_2, 2_2 + 2_2, 1_1 + 2_1 + 1_1, 1_1 + 2_2 + 1_1, 1_1 + 1_1 + 1_1 + 1_1.
We have a(5) = 23 because we have the following colored achiral cyclic compositions of n = 5:
(i) with one part: 5_1, 5_2, 5_3, 5_4, 5_5;
(ii) with two parts: 1_1 + 4_1, 1_1 + 4_2, 1_1 + 4_3, 1_1 + 4_4, 2_1 + 3_1, 2_1 + 3_2, 2_1 + 3_3, 2_2 + 3_1, 2_2 + 3_2, 2_2 + 3_3;
(iii) with three parts: 1_1 + 3_1 + 1_1, 1_1 + 3_2 + 1_1, 1_1 + 3_3 + 1_1, 2_1 + 1_1 + 2_1, 2_2 + 1_1 + 2_2;
(iv) with four parts: 1_1 + 1_1 + 2_1 + 1_1, 1_1 + 1_1 + 2_2 + 1_1 (here, the axis of symmetry passes through one of the 1's and through 2);
(v) with five parts: 1_1 + 1_1 + 1_1 + 1_1 + 1_1.
- A. K. Agarwal, n-colour compositions, Indian J. Pure Appl. Math. 31 (11) (2000), 1421-1427.
- Christian G. Bower, Transforms (2).
- Petros Hadjicostas, Generalized colored circular palindromic compositions, Moscow Journal of Combinatorics and Number Theory, 9(2) (2020), 173-186.
- Meghann Moriah Gibson, Combinatorics of compositions, Master of Science, Georgia Southern University, 2017.
- Meghann Moriah Gibson, Daniel Gray, and Hua Wang, Combinatorics of n-color compositions, Discrete Mathematics 341 (2018), 3209-3226.
- D. M. Y. Sommerville, On certain periodic properties of cyclic compositions of numbers, Proc. London Math. Soc. S2-7(1) (1909), 263-313.
A307415
Total number of parts in all symmetric m-color cyclic compositions of n (that is, the total number of parts in all achiral cyclic compositions of n where a part with size m can be colored with one of m colors).
Original entry on oeis.org
1, 4, 10, 26, 53, 116, 215, 434, 766, 1480, 2539, 4776, 8045, 14864, 24722, 45094, 74305, 134236, 219619, 393790, 640646, 1141844, 1849175, 3279696, 5291353, 9346396, 15031450, 26458994, 42438221, 74479940, 119182319, 208629386, 333170830, 581904544, 927617347, 1616924664
Offset: 1
We have a(1) = 1 because we only have one symmetric cyclic composition of n = 1, namely 1_1, which has 1 part.
We have a(2) = 4 because we have the following colored achiral cyclic compositions of n = 2: 2_1, 2_2, 1_1 + 1_1; hence, a(2) = 1 + 1 + 2 = 4.
We have a(3) = 10 because we have the following colored achiral cyclic compositions of n = 3: 3_1, 3_2, 3_3, 1_1 + 2_1, 1_1 + 2_2, 1_1 + 1_1 + 1_1; hence, a(3) = 1 + 1 + 1 + 2 + 2 + 3 = 10.
We have a(4) = 26 because we have the following colored achiral cyclic compositions of n = 4: 4_1, 4_2, 4_3, 4_4, 1_1 + 3_1, 1_1 + 3_2, 1_1 + 3_3, 2_1 + 2_1, 2_1 + 2_2, 2_2 + 2_2, 1_1 + 2_1 + 1_1, 1_1 + 2_2 + 1_1, 1_1 + 1_1 + 1_1 + 1_1; hence, a(4) = 1 + 1 + 1 + 1 + 2 + 2 + 2 + 2 + 2 + 2 + 3 + 3 + 4 = 26.
We have a(5) = 53 because we have the following colored achiral cyclic compositions of n = 5:
(i) With one part: 5_1, 5_2, 5_3, 5_4, 5_5; i.e., a total of 5 parts.
(ii) With two parts: 1_1 + 4_1, 1_1 + 4_2, 1_1 + 4_3, 1_1 + 4_4, 2_1 + 3_1, 2_1 + 3_2, 2_1 + 3_3, 2_2 + 3_1, 2_2 + 3_2, 2_2 + 3_3; i.e., a total of 20 parts.
(iii) With three parts: 1_1 + 3_1 + 1_1, 1_1 + 3_2 + 1_1, 1_1 + 3_3 + 1_1, 2_1 + 1_1 + 2_1, 2_2 + 1_1 + 2_2; i.e., a total of 15 parts.
(iv) With four parts: 1_1 + 1_1 + 2_1 + 1_1, 1_1 + 1_1 + 2_2 + 1_1 (here, the axis of symmetry passes through one of the 1's and through 2); i.e., a total of 8 parts.
(v) With five parts: 1_1 + 1_1 + 1_1 + 1_1 + 1_1; i.e., a total of 5 parts.
Thus, a(5) = 5 + 20 + 15 + 8 + 5 = 53.
- A. K. Agarwal, n-colour compositions, Indian J. Pure Appl. Math. 31(11) (2000), 1421-1427.
- C. G. Bower, Transforms (2).
- Petros Hadjicostas, Generalized colored circular palindromic compositions, Moscow Journal of Combinatorics and Number Theory, 9(2) (2020), 173-186.
- Meghann Moriah Gibson, Combinatorics of compositions, Master of Science, Georgia Southern University, 2017.
- Meghann Moriah Gibson, Daniel Gray, and Hua Wang, Combinatorics of n-color compositions, Discrete Mathematics 341 (2018), 3209-3226.
- D. M. Y. Sommerville, On certain periodic properties of cyclic compositions of numbers, Proc. London Math. Soc. S2-7(1) (1909), 263-313.
A032278
Number of ways to partition n elements into pie slices each with at least 2 elements allowing the pie to be turned over.
Original entry on oeis.org
0, 1, 1, 2, 2, 4, 4, 7, 8, 13, 15, 25, 30, 48, 63, 98, 132, 208, 290, 454, 656, 1021, 1509, 2358, 3544, 5535, 8441, 13200, 20318, 31835, 49352, 77435, 120710, 189673, 296853, 467159, 733362, 1155646, 1818593, 2869377, 4524080
Offset: 1
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A032278_list := proc(n)
local ele ;
ele := [0,seq(1,i=1..30)] ;
DIK(ele) ; # defined in A032287
end proc:
A032278_list(50) ; # R. J. Mathar, Feb 14 2025
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seq(n)={Vec(x^2/((1-x)*(1-x^2-x^4)) + sum(d=1, n, eulerphi(d)/d*log((1-x^d)/(1-x^d-x^(2*d)) + O(x*x^n))), -n)/2} \\ Andrew Howroyd, Jun 20 2018
A308817
Total number of parts in all m-color dihedral compositions of n (that is, the total number of parts in all dihedral compositions of n where a part of size m may be colored with one of m colors).
Original entry on oeis.org
1, 4, 10, 26, 56, 138, 299, 726, 1686, 4158, 10130, 25678, 64725, 166538, 428456, 1112226, 2888604, 7533750, 19653903, 51367462, 134277878, 351284164, 919080550, 2405427698, 6295780309, 16480373968, 43141303978, 112939105716, 295664584064, 774042041090, 2026429360115, 5305210333758
Offset: 1
We have a(1) = 1 because 1_1 is the only m-color dihedral composition of n = 1 and the total number of parts is 1.
We have a(2) = 4 because 2_1, 2_2, 1_1 + 1_1 are all the m-color dihedral compositions of 2 and the total number of parts is 1 + 1 + 2 = 4.
We have a(3) = 10 because 3_1, 3_2, 3_3, 1_1 + 2_1, 1_1 + 2_2, 1_1 + 1_1 + 1_1 are all the m-color dihedral compositions of n = 3 and the total number of parts is 1 + 1 + 1 + 2 + 2 + 3 = 10.
We have a(4) = 26 because 4_1, 4_2, 4_3, 4_4, 1_1 + 3_1, 1_1 + 3_2, 1_1 + 3_3, 2_1 + 2_1, 2_1 + 2_2, 2_2 + 2_2, 1_1 + 2_1 + 1_1, 1_1 + 2_2 + 1_1, 1_1 + 1_1 + 1_1 + 1_1 are all the m-color dihedral compositions of n = 4 and the total number of parts is 1 + 1 + 1 + 1 + 2 + 2 + 2 + 2 + 2 + 2 + 3 + 3 + 4 = 26.
Note that a(n) = A307415(n) = A308723(n) for n = 1, 2, 3, 4 because all cyclic compositions of n when 1 <= n <= 4 are symmetric as well (and thus are dihedral).
For n = 5, we have A307415(5) = 53 <> 59 = A308723(5), in which case, a(n) = (53 + 59)/2 = 56. For example, 1_1 + 2_1 + 3_1 is a dihedral composition of n = 5, but it is not symmetric, so it corresponds to two (inequivalent) cyclic compositions: 1_1 + 2_1 + 3_1 and 3_1 + 2_1 + 1_1. Similarly, 1_1 + 2_1 + 2_2 is a dihedral composition of n = 5, but it is not symmetric, so it corresponds to two (inequivalent) cyclic compositions: 1_1 + 2_1 + 2_2 and 2_2 + 2_1 + 1_1.
- A. K. Agarwal, n-colour compositions, Indian J. Pure Appl. Math. 31 (11) (2000), 1421-1427.
- C. G. Bower, Transforms (2).
- Petros Hadjicostas, Generalized colored circular palindromic compositions, Moscow Journal of Combinatorics and Number Theory, 9(2) (2020), 173-186.
- Meghann Moriah Gibson, Combinatorics of compositions, Master of Science, Georgia Southern University, 2017.
- Meghann Moriah Gibson, Daniel Gray, and Hua Wang, Combinatorics of n-color compositions, Discrete Mathematics 341 (2018), 3209-3226.
- Arnold Knopfmacher and Neville Robbins, Some properties of dihedral compositions, Util. Math. 92 (2013), 207-220.
- D. M. Y. Sommerville, On certain periodic properties of cyclic compositions of numbers, Proc. London Math. Soc. S2-7(1) (1909), 263-313.
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