cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 10 results.

A050534 Tritriangular numbers: a(n) = binomial(binomial(n,2),2) = n*(n+1)*(n-1)*(n-2)/8.

Original entry on oeis.org

0, 0, 0, 3, 15, 45, 105, 210, 378, 630, 990, 1485, 2145, 3003, 4095, 5460, 7140, 9180, 11628, 14535, 17955, 21945, 26565, 31878, 37950, 44850, 52650, 61425, 71253, 82215, 94395, 107880, 122760, 139128, 157080, 176715, 198135, 221445, 246753, 274170, 303810, 335790
Offset: 0

Views

Author

Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 29 1999

Keywords

Comments

"There are n straight lines in a plane, no two of which are parallel and no three of which are concurrent. Their points of intersection being joined, show that the number of new lines drawn is (1/8)n(n-1)(n-2)(n-3)." (Schmall, 1915).
Several different versions of this sequence are possible, beginning with either one, two or three 0's.
If Y is a 3-subset of an n-set X then, for n>=6, a(n-4) is the number of (n-6)-subsets of X which have exactly one element in common with Y. - Milan Janjic, Dec 28 2007
Number of distinct ways to select 2 pairs of objects from a set of n+1 objects, when order doesn't matter. For example, with n = 3 (4 objects), the 3 possibilities are (12)(34), (13)(24), and (14)(23). - Brian Parsonnet, Jan 03 2012
Partial sums of A027480. - J. M. Bergot, Jul 09 2013
For the set {1,2,...,n}, the sum of the 2 smallest elements of all subsets with 3 elements is a(n) (see Bulut et al. link). - Serhat Bulut, Jan 20 2015
a(n) is also the number of subgroups of S_{n+1} (the symmetric group on n+1 elements) that are isomorphic to D_4 (the dihedral group of order 8). - Geoffrey Critzer, Sep 13 2015
a(n) is the coefficient of x1^(n-3)*x2^2 in exponential Bell polynomial B_{n+1}(x1,x2,...) (number of ways to select 2 pairs among n+1 objects, see above), hence its link with A000292 and A001296 (see formula). - Cyril Damamme, Feb 26 2018
Also the number of 4-cycles in the complete graph K_{n+1}. - Eric W. Weisstein, Mar 13 2018
Number of chiral pairs of colorings of the 4 edges or vertices of a square using n or fewer colors. Each member of a chiral pair is a reflection, but not a rotation, of the other. - Robert A. Russell, Oct 20 2020

Examples

			For a(3)=3, the chiral pairs of square colorings are AABC-AACB, ABBC-ACBB, and ABCC-ACCB. - _Robert A. Russell_, Oct 20 2020
		

References

  • Arthur T. Benjamin and Jennifer Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 154.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, Problem 1, page 72.
  • Richard P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.5, case k=2.

Crossrefs

Cf. A000217, A000332, A033487, A107394, A034827, A210569, Second column of triangle A001498.
Cf. similar sequences listed in A241765.
Cf. (square colorings) A006528 (oriented), A002817 (unoriented), A002411 (achiral),
Row 2 of A325006 (orthoplex facets, orthotope vertices) and A337409 (orthotope edges, orthoplex ridges).
Row 4 of A293496 (cycles of n colors using k or fewer colors).

Programs

  • GAP
    List([0..40],n->3*Binomial(n+1,4)); # Muniru A Asiru, Mar 20 2018
  • Magma
    [3*Binomial(n+1, 4): n in [0..40]]; // Vincenzo Librandi, Feb 14 2015
    
  • Maple
    [seq(binomial(n+1,4)*3,n=0..40)]; # Zerinvary Lajos, Jul 18 2006
  • Mathematica
    Table[Binomial[Binomial[n, 2], 2], {n, 0, 50}] (* Stefan Steinerberger, Apr 08 2006 *)
    LinearRecurrence[{5, -10, 10, -5, 1}, {0, 0, 0, 3, 15}, 40] (* Harvey P. Dale, Dec 14 2011 *)
    (* Start from Eric W. Weisstein, Mar 13 2018 *)
    Binomial[Binomial[Range[0, 20], 2], 2]
    Nest[Binomial[#, 2] &, Range[0, 20], 2]
    Nest[PolygonalNumber[# - 1] &, Range[0, 20], 2]
    CoefficientList[Series[3 x^3/(1 - x)^5, {x, 0, 20}], x]
    (* End *)
  • PARI
    a(n)=n*(n+1)*(n-1)*(n-2)/8 \\ Charles R Greathouse IV, Nov 20 2012
    
  • PARI
    x='x+O('x^100); concat([0, 0, 0], Vec(3*x^3/(1-x)^5)) \\ Altug Alkan, Nov 01 2015
    
  • Sage
    [(binomial(binomial(n,2),2)) for n in range(0, 39)] # Zerinvary Lajos, Nov 30 2009
    

Formula

a(n) = 3*binomial(n+1, 4) = 3*A000332(n+1).
From Vladeta Jovovic, May 03 2002: (Start)
Recurrence: a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
G.f.: 3*x^3 / (1-x)^5. (End)
a(n+1) = T(T(n)) - T(n); a(n+2) = T(T(n)+n) where T is A000217. - Jon Perry, Jun 11 2003
a(n+1) = T(n)^2 - T(T(n)) where T is A000217. - Jon Perry, Jul 23 2003
a(n) = T(T(n-1)-1) where T is A000217. - Jon E. Schoenfield, Dec 14 2014
a(n) = 3*C(n, 4) + 3*C(n, 3), for n>3.
From Alexander Adamchuk, Apr 11 2006: (Start)
a(n) = (1/2)*Sum_{k=1..n} k*(k-1)*(k-2).
a(n) = A033487(n-2)/2, n>1.
a(n) = C(n-1,2)*C(n+1,2)/2, n>2. (End)
a(n) = A052762(n+1)/8. - Zerinvary Lajos, Apr 26 2007
a(n) = (4x^4 - 4x^3 - x^2 + x)/2 where x = floor(n/2)*(-1)^n for n >= 0. - William A. Tedeschi, Aug 24 2010
E.g.f.: x^3*exp(x)*(4+x)/8. - Robert Israel, Nov 01 2015
a(n) = Sum_{k=1..n} Sum_{i=1..k} (n-i-1)*(n-k). - Wesley Ivan Hurt, Sep 12 2017
a(n) = A001296(n-1) - A000292(n-1). - Cyril Damamme, Feb 26 2018
Sum_{n>=3} 1/a(n) = 4/9. - Vaclav Kotesovec, May 01 2018
a(n) = A006528(n) - A002817(n) = (A006528(n) - A002411(n)) / 2 = A002817(n) - A002411(n). - Robert A. Russell, Oct 20 2020
Sum_{n>=3} (-1)^(n+1)/a(n) = 32*log(2)/3 - 64/9. - Amiram Eldar, Jan 09 2022
a(n) = Sum_{k=1..2} (-1)^(k+1)*binomial(n,2-k)*binomial(n,2+k). - Gerry Martens, Oct 09 2022

Extensions

Additional comments from Antreas P. Hatzipolakis, May 03 2002

A059076 Number of pairs of orientable necklaces with n beads and two colors; i.e., turning the necklace over does not leave it unchanged.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 2, 6, 14, 30, 62, 128, 252, 495, 968, 1866, 3600, 6917, 13286, 25476, 48916, 93837, 180314, 346554, 666996, 1284570, 2477342, 4781502, 9240012, 17871708, 34604066, 67060746, 130085052, 252548760, 490722344
Offset: 0

Views

Author

Henry Bottomley, Dec 22 2000

Keywords

Comments

Number of chiral bracelets with n beads and two colors.

Examples

			For n=6, the only chiral pair is AABABB-AABBAB.  For n=7, the two chiral pairs are AAABABB-AAABBAB and AABABBB-AABBBAB. - _Robert A. Russell_, Sep 24 2018
		

Crossrefs

Column 2 of A293496.
Cf. A059053.
Column 2 of A305541.
Equals (A000031 - A164090) / 2.
a(n) = (A052823(n) - A027383(n-2)) / 2.

Programs

  • Mathematica
    nn=35;Table[CoefficientList[Series[CycleIndex[CyclicGroup[n],s]-CycleIndex[DihedralGroup[n],s]/.Table[s[i]->2,{i,1,n}],{x,0,nn}],x],{n,1,nn}]//Flatten  (* Geoffrey Critzer, Mar 26 2013 *)
    mx=40; CoefficientList[Series[(1-Sum[ EulerPhi[n]*Log[1-2*x^n]/n, {n, mx}]-(1+x)^2/(1-2*x^2))/2, {x, 0, mx}], x] (* Herbert Kociemba, Nov 02 2016 *)
    terms = 36; a29[0] = 1; a29[n_] := (1/4)*(Mod[n, 2] + 3)*2^Quotient[n, 2] + DivisorSum[n, EulerPhi[#]*2^(n/#) & ]/(2*n); Array[a29, 36, 0] - LinearRecurrence[{0, 2}, {1, 2, 3}, 36] (* Jean-François Alcover, Nov 05 2017 *)
    k = 2; Prepend[Table[DivisorSum[n, EulerPhi[#] k^(n/#) &]/(2n)(k^Floor[(n+1)/2] + k^Ceiling[(n+1)/2])/4, {n, 1, 30}], 0] (* Robert A. Russell, Sep 24 2018 *)

Formula

a(n) = A000031(n) - A000029(n) = A000029(n) - A029744(n) = (A000031(n) - A029744(n))/2 = A008965(n) - A091696(n)
G.f.: (1 - Sum_{n>=1} phi(n)*log(1 - 2*x^n)/n - (1 + x)^2/(1 - 2*x^2))/2. - Herbert Kociemba, Nov 02 2016
For n > 0, a(n) = -(k^floor((n + 1)/2) + k^ceiling((n + 1)/2))/4 + (1/(2*n))* Sum_{d|n} phi(d)*k^(n/d), where k = 2 is the maximum number of colors. - Robert A. Russell, Sep 24 2018

A293500 Number of orientable strings of length n using a maximum of k colors, array read by descending antidiagonals, T(n,k) for n >= 1 and k >= 1.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 3, 2, 0, 0, 6, 9, 6, 0, 0, 10, 24, 36, 12, 0, 0, 15, 50, 120, 108, 28, 0, 0, 21, 90, 300, 480, 351, 56, 0, 0, 28, 147, 630, 1500, 2016, 1053, 120, 0, 0, 36, 224, 1176, 3780, 7750, 8064, 3240, 240, 0, 0, 45, 324, 2016, 8232, 23220, 38750, 32640, 9720, 496, 0
Offset: 1

Views

Author

Andrew Howroyd, Oct 10 2017

Keywords

Comments

Reversing the string does not leave it unchanged. Only one string from each pair is counted.
Equivalently, the number of nonequivalent strings up to reversal that are not palindromes.
Except for the first term, column k is the "BHK" (reversible, identity, unlabeled) transform of k,0,0,0,... [Corrected by Petros Hadjicostas, Jul 01 2018]
From Petros Hadjicostas, Jul 01 2018: (Start)
Consider the input sequence (c_k(n): n >= 1) with g.f. C_k(x) = Sum_{n>=1} c_k(n)*x^n. Let a_k(n) = BHK(c_k(n): n >= 1) be the output sequence under Bower's BHK transform. It can be proved that the g.f. of BHK(c_k(n): n >= 1) is A_k(x) = (C_k(x)^2 - C_k(x^2))/(2*(1-C_k(x))*(1-C_k(x^2))) + C_k(x). (See the comments for sequences A032096, A032097, and A032098.)
For column k of this two-dimensional array, the input sequence is defined by c_k(1) = k and c_k(n) = 0 for n >= 1. Thus, C_k(x) = k*x, and hence the g.f. of column k (with the term C_k(x) = k*x excluded) is (C_k(x)^2 - C_k(x^2))/(2*(1-C_k(x))*(1-C_k(x^2))) = (1/2)*(k - 1)*k*x^2/((k*x^2 - 1)*(k*x - 1)), from which we can easily prove Howroyd's formula.
(End)
Comment from Bahman Ahmadi, Aug 05 2019: (Start)
We give an alternative definition for the square array A(n,k) = T(n,k) with n >= 2 and k >= 0. A(n,k) is the number of inequivalent "distinguishing colorings" of the path on n vertices using at most k colors. The rows are indexed by n, the number of vertices of the path, and the columns are indexed by k, the number of permissible colors.
A vertex-coloring of a graph G is called "distinguishing" if it is only preserved by the identity automorphism of G. This notion is considered in the context of "symmetry breaking" of simple (finite or infinite) graphs. Two vertex-colorings of a graph are called "equivalent" if there is an automorphism of the graph which preserves the colors of the vertices. Given a graph G, we use the notation Phi_k(G) to denote the number of inequivalent distinguishing colorings of G with at most k colors. This sequence gives A(n,k) = Phi_k(P_n), i.e., the number of inequivalent distinguishing colorings of the path P_n on n vertices with at most k colors.
For n=3, we can color the vertices of P_3 with at most 2 colors in 3 ways such that all the colorings distinguish the graph (i.e., no non-identity automorphism of the path P_3 preserves the coloring) and that all the three colorings are inequivalent.
We have Phi_k(P_n) = binomial(k,2)*k^(n-2) + k*Phi_k(P_(n-2)) for n >= 4; Phi_k(P_2) = binomial(k,2); Phi_k(P_3) = k*binomial(k,2).
(End)

Examples

			Array begins:
======================================================
n\k| 1   2    3     4      5      6       7       8
---|--------------------------------------------------
1  | 0   0    0     0      0      0       0       0...
2  | 0   1    3     6     10     15      21      28...
3  | 0   2    9    24     50     90     147     224...
4  | 0   6   36   120    300    630    1176    2016...
5  | 0  12  108   480   1500   3780    8232   16128...
6  | 0  28  351  2016   7750  23220   58653  130816...
7  | 0  56 1053  8064  38750 139320  410571 1046528...
8  | 0 120 3240 32640 195000 839160 2881200 8386560...
...
For T(4,2)=6, the chiral pairs are AAAB-BAAA, AABA-ABAA, AABB-BBAA, ABAB-BABA, ABBB-BBBA, and BABB-BBAB.
		

Crossrefs

Columns 2-5 for n > 1 are A032085, A032086, A032087, A032088.
Column 6 is A320524.
Rows 2-6 are A161680, A006002(n-1), A083374, A321672, A085744.
Cf. A003992 (oriented), A277504 (unoriented), A321391 (achiral).

Programs

  • Mathematica
    Table[Function[n, (k^n - k^(Ceiling[n/2]))/2][m - k + 1], {m, 11}, {k, m, 1, -1}] // Flatten (* Michael De Vlieger, Oct 11 2017 *)
  • PARI
    T(n,k) = (k^n - k^(ceil(n/2)))/2;

Formula

T(n,k) = (k^n - k^(ceiling(n/2)))/2.
G.f. for column k: (1/2)*(k - 1)*k*x^2/((k*x^2 - 1)*(k*x - 1)). - Petros Hadjicostas, Jul 07 2018
From Robert A. Russell, Nov 16 2018: (Start)
T(n,k) = (A003992(k,n) - A321391(n,k)) / 2.
T(n,k) = = A003992(k,n) - A277504(n,k) = A277504(n,k) - A321391(n,k).
G.f. for row n: (Sum_{j=0..n} S2(n,j)*j!*x^j/(1-x)^(j+1) - Sum_{j=0..ceiling(n/2)} S2(ceiling(n/2),j)*j!*x^j/(1-x)^(j+1)) / 2, where S2 is the Stirling subset number A008277.
G.f. for row n>1: x * Sum_{k=1..n-1} A145883(n,k) * x^k / (1-x)^(n+1).
E.g.f. for row n: (Sum_{k=0..n} S2(n,k)*x^k - Sum_{k=0..ceiling(n/2)} S2(ceiling(n/2),k)*x^k) * exp(x) / 2, where S2 is the Stirling subset number A008277.
T(0,k) = T(1,k) = 0; T(2,k) = binomial(k,2); for n>2, T(n,k) = k*(T(n-3,k)+T(n-2,k)-k*T(n-1,k)).
For k>n, T(n,k) = Sum_{j=1..n+1} -binomial(j-n-2,j) * T(n,k-j). (End)

A305541 Triangle read by rows: T(n,k) is the number of chiral pairs of color loops of length n with exactly k different colors.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 3, 3, 0, 0, 12, 24, 12, 0, 1, 35, 124, 150, 60, 0, 2, 111, 588, 1200, 1080, 360, 0, 6, 318, 2487, 7845, 11970, 8820, 2520, 0, 14, 934, 10240, 46280, 105840, 129360, 80640, 20160, 0, 30, 2634, 40488, 254676, 821592, 1481760, 1512000, 816480, 181440, 0, 62, 7503, 158220, 1344900, 5873760, 14658840, 21772800, 19051200, 9072000, 1814400
Offset: 1

Views

Author

Robert A. Russell, Jun 04 2018

Keywords

Comments

In other words, the number of n-bead bracelets with beads of exactly k different colors that when turned over are different from themselves. - Andrew Howroyd, Sep 13 2019

Examples

			Triangle T(n,k) begins:
  0;
  0,  0;
  0,  0,    1;
  0,  0,    3,     3;
  0,  0,   12,    24,     12;
  0,  1,   35,   124,    150,     60;
  0,  2,  111,   588,   1200,   1080,     360;
  0,  6,  318,  2487,   7845,  11970,    8820,    2520;
  0, 14,  934, 10240,  46280, 105840,  129360,   80640,  20160;
  0, 30, 2634, 40488, 254676, 821592, 1481760, 1512000, 816480, 181440;
  ...
For T(4,3)=3, the chiral pairs are AABC-AACB, ABBC-ACBB, and ABCC-ACCB.
For T(4,4)=3, the chiral pairs are ABCD-ADCB, ABDC-ACDB, and ACBD-ADBC.
		

Crossrefs

Columns 2-6 are A059076, A305542, A305543, A305544, and A305545.
Row sums are A326895.

Programs

  • Mathematica
    Table[(k!/(2n)) DivisorSum[n, EulerPhi[#] StirlingS2[n/#, k] &] - (k!/4) (StirlingS2[Floor[(n+1)/2], k] + StirlingS2[Ceiling[(n+1)/2], k]), {n, 1, 15}, {k, 1, n}] // Flatten
  • PARI
    T(n,k) = {-k!*(stirling((n+1)\2,k,2) + stirling(n\2+1,k,2))/4 + k!*sumdiv(n,d, eulerphi(d)*stirling(n/d,k,2))/(2*n)} \\ Andrew Howroyd, Sep 13 2019

Formula

T(n,k) = -(k!/4)*(S2(floor((n+1)/2),k) + S2(ceiling((n+1)/2),k)) + (k!/(2 n))*Sum_{d|n} phi(d)*S2(n/d,k), where S2(n,k) is the Stirling subset number A008277.
T(n,k) = A087854(n,k) - A273891(n,k).
T(n,k) = (A087854(n,k) - A305540(n,k)) / 2.
T(n, k) = Sum_{i=0..k} (-1)^(k-i)*binomial(k,i)*A293496(n, i). - Andrew Howroyd, Sep 13 2019

A309528 The number of non-equivalent distinguishing colorings of the cycle on n vertices with at most k colors (k>=1). The cycle graph is defined for n>=3; extended to n=1,2 using the closed form. Square array read by descending antidiagonals: the rows are indexed by n, the number of vertices of the cycle and the columns are indexed by k, the number of permissible colors.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 4, 3, 0, 0, 0, 0, 10, 15, 12, 1, 0, 0, 0, 20, 45, 72, 37, 2, 0, 0, 0, 35, 105, 252, 266, 117, 6, 0, 0, 0, 56, 210, 672, 1120, 1044, 333, 14, 0, 0, 0, 84, 378, 1512, 3515, 5270, 3788, 975, 30, 0, 0, 0, 120, 630, 3024, 9121, 19350, 23475, 14056, 2712, 62, 0
Offset: 1

Views

Author

Bahman Ahmadi, Aug 06 2019

Keywords

Comments

A vertex-coloring of a graph G is called distinguishing if it is only preserved by the identity automorphism of G. This notion is considered in the subject of symmetry breaking of simple (finite or infinite) graphs. Two vertex-colorings of a graph are called equivalent if there is an automorphism of the graph which preserves the colors of the vertices. Given a graph G, we use the notation Phi_k(G) to denote the number of non-equivalent distinguishing colorings of G with at most k colors. The sequence here, displays A(n,k)=Phi_k(C_n), i.e., the number of non-equivalent distinguishing colorings of the cycle C_n on n vertices with at most k colors.

Examples

			The table begins:
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...
0, 0, 1, 4, 10, 20, 35, 56, 84, 120, ...
0, 0, 3, 15, 45, 105, 210, 378, 630, 990, ...
0, 0, 12, 72, 252, 672, 1512, 3024, 5544, 9504, ...
0, 1, 37, 266, 1120, 3515, 9121, 20692, 42456, 80565, ...
0, 2, 117, 1044, 5270, 19350, 57627, 147752, 338364, 709290, ...
0, 6, 333, 3788, 23475, 102690, 355446, 1039248, 2673810, 6222150, ...
0, 14, 975, 14056, 106950, 555990, 2233469, 7440160, 21493836, 55505550, ...
0, 30, 2712, 51132, 483504, 3009426, 14089488, 53611992, 174189024, 499720518, ...
------
For n=4, we can color the vertices of the cycle C_4 with at most 3 colors, in 3 ways, such that all the colorings distinguish the graph (i.e., no non-identity automorphism of C_4 preserves the coloring) and that all the three colorings are non-equivalent. The color classes are as follows:
{ { 1 }, { 2 }, { 3, 4 } }
{ { 1 }, { 2, 3 }, { 4 } }
{ { 1, 2 }, { 3 }, { 4 } }
		

Crossrefs

Columns k=2..5 for n >= 3 are A032239, A032240, A032241, A032242.
Different from A293496.

Programs

  • PARI
    A(n,k)={sumdiv(n, d, moebius(n/d)*(k^d/n - if(d%2, k^((d+1)/2), (k+1)*k^(d/2)/2)))/2} \\ Andrew Howroyd, Aug 11 2019

Formula

A(n,k) = (A074650(n,k) - A284856(n,k))/2. - Andrew Howroyd, Aug 11 2019

A278639 Number of pairs of orientable necklaces with n beads and up to 3 colors; i.e., turning the necklace over does not leave it unchanged. The turned-over necklace is not included in the count.

Original entry on oeis.org

0, 0, 0, 1, 3, 12, 38, 117, 336, 976, 2724, 7689, 21455, 60228, 168714, 475037, 1338861, 3788400, 10742588, 30556305, 87112059, 248967564, 713032782, 2046325125, 5883428618, 16944975048, 48880471500, 141212377489, 408509453511, 1183275193908, 3431504760514
Offset: 0

Views

Author

Herbert Kociemba, Nov 24 2016

Keywords

Comments

Number of chiral bracelets of n beads using up to three different colors.

Examples

			Example: The 3 orientable necklaces with 4 beads and the colors A, B and C are AABC, BBAC and CCAB. The turned-over necklaces AACB, BBCA and CCBA are not included in the count.
For n=6, the three chiral pairs using just two colors are AABABB-AABBAB, AACACC-AACCAC, and BBCBCC-BBCCBC.  The other 35 use three colors. - _Robert A. Russell_, Sep 24 2018
		

Crossrefs

Column 3 of A293496.
Cf. A059076 (2 colors).
a(n) = (A001867(n) - A182751(n-1)) / 2.
Equals A001867 - A027671.
a(n) = A027671(n) - A182751(n-1).

Programs

  • Mathematica
    mx=40;f[x_,k_]:=(1-Sum[EulerPhi[n]*Log[1-k*x^n]/n,{n,1,mx}]-Sum[Binomial[k,i]*x^i,{i,0,2}]/(1-k*x^2))/2;CoefficientList[Series[f[x,3],{x,0,mx}],x]
    k=3; Prepend[Table[DivisorSum[n, EulerPhi[#] k^(n/#) &]/(2n) -(k^Floor[(n+1)/2] + k^Ceiling[(n+1)/2])/4, {n, 1, 30}], 0] (* Robert A. Russell, Sep 24 2018 *)

Formula

G.f.: k=3, (1 - Sum_{n>=1} phi(n)*log(1 - k*x^n)/n - Sum_{i=0..2} binomial(k,i)*x^i / (1 - k*x^2))/2.
For n > 0, a(n) = -(k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/2n)* Sum_{d|n} phi(d)*k^(n/d), where k=3 is the maximum number of colors. - Robert A. Russell, Sep 24 2018

A321791 Table read by descending antidiagonals: T(n,k) is the number of unoriented cycles (bracelets) of length n using up to k available colors.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 4, 6, 4, 1, 0, 1, 5, 10, 10, 6, 1, 0, 1, 6, 15, 20, 21, 8, 1, 0, 1, 7, 21, 35, 55, 39, 13, 1, 0, 1, 8, 28, 56, 120, 136, 92, 18, 1, 0, 1, 9, 36, 84, 231, 377, 430, 198, 30, 1, 0
Offset: 0

Views

Author

Robert A. Russell, Dec 18 2018

Keywords

Examples

			Table begins with T(0,0):
  1 1  1    1     1      1       1        1        1         1         1 ...
  0 1  2    3     4      5       6        7        8         9        10 ...
  0 1  3    6    10     15      21       28       36        45        55 ...
  0 1  4   10    20     35      56       84      120       165       220 ...
  0 1  6   21    55    120     231      406      666      1035      1540 ...
  0 1  8   39   136    377     888     1855     3536      6273     10504 ...
  0 1 13   92   430   1505    4291    10528    23052     46185     86185 ...
  0 1 18  198  1300   5895   20646    60028   151848    344925    719290 ...
  0 1 30  498  4435  25395  107331   365260  1058058   2707245   6278140 ...
  0 1 46 1219 15084 110085  563786  2250311  7472984  21552969  55605670 ...
  0 1 78 3210 53764 493131 3037314 14158228 53762472 174489813 500280022 ...
For T(3,3)=10, the unoriented cycles are 9 achiral (AAA, AAB, AAC, ABB, ACC, BBB, BBC, BCC, CCC) and 1 chiral pair (ABC-ACB).
		

Crossrefs

Cf. A075195 (oriented), A293496(chiral), A284855 (achiral).
Cf. A051137 (ascending antidiagonals).
Columns 0-6 are A000007, A000012, A000029, A027671, A032275, A032276, and A056341.
Main diagonal gives A081721.

Programs

  • Mathematica
    Table[If[k>0, DivisorSum[k, EulerPhi[#](n-k)^(k/#)&]/(2k) + ((n-k)^Floor[(k+1)/2]+(n-k)^Ceiling[(k+1)/2])/4, 1], {n, 0, 12}, {k, 0, n}] // Flatten

Formula

T(n,k) = [n==0] + [n>0] * (k^floor((n+1)/2) + k^ceiling((n+1)/2)) / 4 + (1/(2*n)) * Sum_{d|n} phi(d) * k^(n/d).
T(n,k) = (A075195(n,k) + A284855(n,k)) / 2.
T(n,k) = A075195(n,k) - A293496(n,k) = A293496(n,k) + A284855(n,k).
Linear recurrence for row n: T(n,k) = Sum_{j=0..n} -binomial(j-n-1,j+1) * T(n,k-1-j) for k >= n + 1.
O.g.f. for column k >= 0: Sum_{n>=0} T(n,k)*x^n = 3/4 + (1 + k*x)^2/(4*(1 - k*x^2)) - (1/2) * Sum_{d >= 1} (phi(d)/d) * log(1 - k*x^d). - Petros Hadjicostas, Feb 07 2021

A278640 Number of pairs of orientable necklaces with n beads and up to 4 colors; i.e., turning the necklace over does not leave it unchanged. The turned-over necklace is not included in the count.

Original entry on oeis.org

0, 0, 0, 4, 15, 72, 270, 1044, 3795, 14060, 51204, 188604, 694130, 2572920, 9567090, 35758704, 134137875, 505159200, 1908554190, 7233104844, 27486506268, 104713296760, 399817262550, 1529746919604, 5864041395730, 22517964582504, 86607602546220, 333599838189804, 1286742419927070, 4969488707124120, 19215357085867800
Offset: 0

Views

Author

Herbert Kociemba, Nov 24 2016

Keywords

Comments

Number of chiral bracelets of n beads using up to four different colors.

Examples

			Example: For 3 beads and the colors A, B, C and D the 4 orientable necklaces are ABC, ABD, ACD and BCD. The turned-over necklaces ACB, ADB, ADC and BDC are not included in the count.
		

Crossrefs

Column 4 of A293496.
Cf. A059076 (2 colors), A278639 (3 colors).
Equals (A001868 - A056486) / 2 = A001868 - A032275 = A032275 - A056486.

Programs

  • Mathematica
    mx=40;f[x_,k_]:=(1-Sum[EulerPhi[n]*Log[1-k*x^n]/n,{n,1,mx}]-Sum[Binomial[k,i]*x^i,{i,0,2}]/(1-k*x^2))/2;CoefficientList[Series[f[x,4],{x,0,mx}],x]
    k=4; Prepend[Table[DivisorSum[n, EulerPhi[#] k^(n/#) &]/(2n) - (k^Floor[(n+1)/2] + k^Ceiling[(n+1)/2])/4, {n, 1, 30}], 0] (* Robert A. Russell, Sep 24 2018 *)

Formula

G.f.: k=4, (1 - Sum_{n>=1} phi(n)*log(1 - k*x^n)/n - Sum_{i=0..2} Binomial[k,i]*x^i / ( 1-k*x^2) )/2.
For n > 0, a(n) = -(k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/2n)* Sum_{d|n} phi(d)*k^(n/d), where k=4 is the maximum number of colors. - Robert A. Russell, Sep 24 2018

A278641 Number of pairs of orientable necklaces with n beads and up to 5 colors; i.e., turning the necklace over does not leave it unchanged. The turned-over necklace is not included in the count.

Original entry on oeis.org

0, 0, 0, 10, 45, 252, 1130, 5270, 23520, 106960, 483756, 2211650, 10149805, 46911060, 217868310, 1017057518, 4767797895, 22438419120, 105960938380, 501928967930, 2384171386941, 11353241261180, 54185968572450, 259150507387910, 1241763071712930, 5960463867187752, 28656077411358180, 137973711706163210
Offset: 0

Views

Author

Herbert Kociemba, Nov 24 2016

Keywords

Comments

Number of chiral bracelets of n beads using up to five different colors.

Crossrefs

Column 5 of A293496.
Cf. A059076 (2 colors), A278639 (3 colors), A278640 (4 colors).
a(n) = (A001869(n) - A056487(n+1)) / 2 = A032276(n) - A056487(n+1).
Equals A001869 - A032276.

Programs

  • Mathematica
    mx=40;f[x_,k_]:=(1-Sum[EulerPhi[n]*Log[1-k*x^n]/n,{n,1,mx}]-Sum[Binomial[k,i]*x^i,{i,0,2}]/(1-k*x^2))/2;CoefficientList[Series[f[x,5],{x,0,mx}],x]
    k=5; Prepend[Table[DivisorSum[n, EulerPhi[#] k^(n/#) &]/(2n) - (k^Floor[(n+1)/2] + k^Ceiling[(n+1)/2])/4, {n, 1, 30}], 0] (* Robert A. Russell, Sep 24 2018 *)

Formula

G.f.: k=5, (1 - Sum_{n>=1} phi(n)*log(1 - k*x^n)/n - Sum_{i=0..2} Binomial[k,i]*x^i / ( 1-k*x^2) )/2.
For n>0, a(n) = -(k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/2n)* Sum_{d|n} phi(d)*k^(n/d), where k=5 is the maximum number of colors. - Robert A. Russell, Sep 24 2018

A278642 Number of pairs of orientable necklaces with n beads and up to 6 colors; i.e., turning the necklace over does not leave it unchanged. The turned-over necklace is not included in the count.

Original entry on oeis.org

0, 0, 0, 20, 105, 672, 3535, 19350, 102795, 556010, 3010098, 16467450, 90619690, 502194420, 2798240265, 15671993560, 88156797855, 497837886000, 2821092554035, 16035752398770, 91403856697944, 522308167195260, 2991401733402075, 17168047238861070, 98716274117752900, 568605754068247644, 3280417827002225910, 18953525314104758810
Offset: 0

Views

Author

Herbert Kociemba, Nov 24 2016

Keywords

Comments

Number of chiral bracelets of n beads using up to six different colors.

Crossrefs

Column 6 of A293496.
Cf. A059076 (2 colors), A278639 (3 colors), A278640 (4 colors), A278641 (5 colors).

Programs

  • Mathematica
    mx = 40; f[x_, k_] := (1 - Sum[EulerPhi[n] * Log[1 - k * x^n]/n,{n, mx}] - Sum[Binomial[k, i] * x^i, {i, 0, 2}]/(1 - k * x^2))/2; CoefficientList[Series[f[x, 6], {x, 0, mx}], x]
    k = 6; Prepend[Table[DivisorSum[n, EulerPhi[#] k^(n/#) &]/(2n) - (k^Floor[(n + 1)/2] + k^Ceiling[(n + 1)/2])/4, {n, 30}], 0] (* Robert A. Russell, Sep 24 2018 *)

Formula

Equals (A054625(n) - A056488(n)) / 2 = A054625(n) - A056341(n) = A056341(n) - A056488(n), for n >= 1.
G.f.: k = 6, (1 - Sum_{n >= 1} phi(n)*log(1 - k*x^n)/n - Sum_{i = 0..2} Binomial[k, i]*x^i / ( 1 - k*x^2) )/2.
For n > 0, a(n) = -(k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/2n)* Sum_{d|n} phi(d)*k^(n/d), where k = 6 is the maximum number of colors. - Robert A. Russell, Sep 24 2018
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