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.

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A001868 Number of n-bead necklaces with 4 colors.

Original entry on oeis.org

1, 4, 10, 24, 70, 208, 700, 2344, 8230, 29144, 104968, 381304, 1398500, 5162224, 19175140, 71582944, 268439590, 1010580544, 3817763740, 14467258264, 54975633976, 209430787824, 799645010860, 3059510616424, 11728124734500, 45035996273872, 173215372864600, 667199944815064
Offset: 0

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Author

Keywords

Comments

From Richard L. Ollerton, May 07 2021: (Start)
Here, as in A000031, turning over is not allowed.
(1/n) * Dirichlet convolution of phi(n) and 4^n, n>0. (End)

References

  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 162.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 7.112(a).

Crossrefs

Column 4 of A075195.
Cf. A054611.

Programs

  • Maple
    A001868 := proc(n) local d,s; if n = 0 then RETURN(1); else s := 0; for d in divisors(n) do s := s+phi(d)*4^(n/d); od; RETURN(s/n); fi; end;
  • Mathematica
    a[n_] := (1/n)*Total[ EulerPhi[#]*4^(n/#) &  /@ Divisors[n]]; a[0] = 1; Table[a[n], {n, 0, 24}] (* Jean-François Alcover, Oct 21 2011 *)
    mx=40;CoefficientList[Series[1-Sum[EulerPhi[i] Log[1-4*x^i]/i,{i,1,mx}],{x,0,mx}],x] (* Herbert Kociemba, Nov 01 2016 *)
    k=4; Prepend[Table[DivisorSum[n, EulerPhi[#] k^(n/#) &]/n, {n, 1, 30}], 1] (* Robert A. Russell, Sep 21 2018 *)
  • PARI
    a(n) = if (n, sumdiv(n, d, eulerphi(d)*4^(n/d))/n, 1); \\ Michel Marcus, Nov 01 2016

Formula

a(n) = (1/n)*Sum_{d|n} phi(d)*4^(n/d) = A054611(n)/n, n>0.
G.f.: 1 - Sum_{n>=1} phi(n)*log(1 - 4*x^n)/n. - Herbert Kociemba, Nov 01 2016
a(0) = 1; a(n) = (1/n) * Sum_{k=1..n} 4^gcd(n,k). - Ilya Gutkovskiy, Apr 17 2021
a(0) = 1; a(n) = (1/n)*Sum_{k=1..n} 4^(n/gcd(n,k))*phi(gcd(n,k))/phi(n/gcd(n,k)). - Richard L. Ollerton, May 07 2021

A333943 Numbers k such that the k-th composition in standard order is a reversed necklace.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 15, 16, 17, 18, 19, 21, 23, 31, 32, 33, 34, 35, 36, 37, 39, 41, 42, 43, 45, 47, 63, 64, 65, 66, 67, 68, 69, 71, 73, 74, 75, 77, 79, 81, 83, 85, 87, 91, 95, 127, 128, 129, 130, 131, 132, 133, 135, 136, 137, 138, 139, 141, 143
Offset: 1

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Author

Gus Wiseman, Apr 14 2020

Keywords

Comments

A necklace is a finite sequence that is lexicographically minimal among all of its cyclic rotations. Reversed necklaces are different from co-necklaces (A333764).
A composition of n is a finite sequence of positive integers summing to n. The k-th composition in standard order (row k of A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

Examples

			The sequence together with the corresponding reversed necklaces begins:
    1: (1)             32: (6)               69: (4,2,1)
    2: (2)             33: (5,1)             71: (4,1,1,1)
    3: (1,1)           34: (4,2)             73: (3,3,1)
    4: (3)             35: (4,1,1)           74: (3,2,2)
    5: (2,1)           36: (3,3)             75: (3,2,1,1)
    7: (1,1,1)         37: (3,2,1)           77: (3,1,2,1)
    8: (4)             39: (3,1,1,1)         79: (3,1,1,1,1)
    9: (3,1)           41: (2,3,1)           81: (2,4,1)
   10: (2,2)           42: (2,2,2)           83: (2,3,1,1)
   11: (2,1,1)         43: (2,2,1,1)         85: (2,2,2,1)
   15: (1,1,1,1)       45: (2,1,2,1)         87: (2,2,1,1,1)
   16: (5)             47: (2,1,1,1,1)       91: (2,1,2,1,1)
   17: (4,1)           63: (1,1,1,1,1,1)     95: (2,1,1,1,1,1)
   18: (3,2)           64: (7)              127: (1,1,1,1,1,1,1)
   19: (3,1,1)         65: (6,1)            128: (8)
   21: (2,2,1)         66: (5,2)            129: (7,1)
   23: (2,1,1,1)       67: (5,1,1)          130: (6,2)
   31: (1,1,1,1,1)     68: (4,3)            131: (6,1,1)
		

Crossrefs

The non-reversed version is A065609.
The dual version is A328595.
Binary necklaces are A000031.
Necklace compositions are A008965.
Necklaces covering an initial interval are A019536.
Numbers whose prime signature is a necklace are A329138.
Length of co-Lyndon factorization of binary expansion is A329312.
Length of Lyndon factorization of reversed binary expansion is A329313.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Sum is A070939.
- Runs are counted by A124767.
- Rotational symmetries are counted by A138904.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Lyndon compositions are A275692.
- Co-Lyndon compositions are A326774.
- Aperiodic compositions are A328594.
- Length of Lyndon factorization is A329312.
- Rotational period is A333632.
- Co-necklaces are A333764.
- Length of co-Lyndon factorization is A334029.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    neckQ[q_]:=Array[OrderedQ[{q,RotateRight[q,#1]}]&,Length[q]-1,1,And];
    Select[Range[100],neckQ[Reverse[stc[#]]]&]

A179043 Number of n X n checkered tori.

Original entry on oeis.org

1, 2, 7, 64, 4156, 1342208, 1908897152, 11488774559744, 288230376353050816, 29850020237398264483840, 12676506002282327791964489728, 21970710674130840874443091905462272, 154866286100907105149651981766316633972736
Offset: 0

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Author

Rouben Rostamian (rostamian(AT)umbc.edu), Jun 25 2010

Keywords

Comments

Consider an n X n checkerboard whose tiles are assigned colors 0 and 1, at random. There are 2^(n^2) such checkerboards. We identify the opposite edges of each checkerboard, thus making it into a (topological) torus. There are a(n) such (distinct) tori. It is possible to show that a(n) >= 2^(n^2)/n^2 for all n.
Main diagonal of A184271.
Main diagonal of Table 3: The number a(m, n) of toroidal m X n binary arrays, allowing rotation of the rows and/or the columns but not reflection, for m, n = 1, 2, ..., 8, at page 5 of Ethier. - Jonathan Vos Post, Jan 14 2013
This is a 2-dimensional generalization of binary necklaces (A000031). - Gus Wiseman, Feb 04 2019

Examples

			From _Gus Wiseman_, Feb 04 2019: (Start)
Inequivalent representatives of the a(2) = 7 checkered tori:
  [0 0] [0 0] [0 0] [0 1] [0 1] [0 1] [1 1]
  [0 0] [0 1] [1 1] [0 1] [1 0] [1 1] [1 1]
(End)
		

Crossrefs

Cf. A184271 (n X k toroidal binary arrays).

Programs

  • Mathematica
    a[n_] := Sum[If[Mod[n, c] == 0, Sum[If[Mod[n, d] == 0, EulerPhi[c] EulerPhi[d] 2^(n^2/LCM[c, d]), 0], {d, 1, n}], 0], {c, 1, n}]/n ^2

Formula

a(n) = (1/n^2)*Sum_{ c divides n } Sum_{ d divides n } phi(c)*phi(d)*2^(n^2/lcm(c,d)), where phi is A000010 and lcm stands for least common multiple. - Stewart N. Ethier, Aug 24 2012

Extensions

Terms a(6) and a(7) from A184271
a(8)-a(12) from Stewart N. Ethier, Aug 24 2012
a(0)=1 prepended by Alois P. Heinz, Aug 20 2017

A000939 Number of inequivalent n-gons.

Original entry on oeis.org

1, 1, 1, 2, 4, 14, 54, 332, 2246, 18264, 164950, 1664354, 18423144, 222406776, 2905943328, 40865005494, 615376173184, 9880209206458, 168483518571798, 3041127561315224, 57926238289970076, 1161157777643184900, 24434798429947993054, 538583682082245127336
Offset: 1

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Author

Keywords

Comments

Here two n-gons are said to be equivalent if they differ in starting point, orientation, or by a rotation (but not by a reflection - for that see A000940).
Number of cycle necklaces on n vertices, defined as equivalence classes of (labeled, undirected) Hamiltonian cycles under rotation of the vertices. The path version is A275527. - Gus Wiseman, Mar 02 2019

Examples

			Possibilities for n-gons without distinguished vertex can be encoded as permutation classes of vertices, two permutations being equivalent if they can be obtained from each other by circular rotation, translation mod n or complement to n+1.
n=3: 123.
n=4: 1234, 1243.
n=5: 12345, 12354, 12453, 13524.
n=6: 123456, 123465, 123564, 123645, 123654, 124365, 124635, 124653, 125364, 125463, 125634, 126435, 126453, 135264.
		

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A000940. Bisections give A094154, A094155.
For star polygons see A231091.

Programs

  • Maple
    with(numtheory):
    # for n odd:
    Ed:= proc(n) local t1, d; t1:=0; for d from 1 to n do
           if n mod d = 0 then t1:= t1+phi(n/d)^2*d!*(n/d)^d fi od:
           t1/(2*n^2)
         end:
    # for n even:
    Ee:= proc(n) local t1, d; t1:= 2^(n/2)*(n/2)*(n/2)!; for d
           from 1 to n do if n mod d = 0 then t1:= t1+
           phi(n/d)^2*d!*(n/d)^d; fi od: t1/(2*n^2)
         end:
    A000939:= n-> if n mod 2 = 0 then ceil(Ee(n)) else ceil(Ed(n)); fi:
    seq(A000939(n), n=1..25);
  • Mathematica
    a[n_] := (t = If[OddQ[n], 0, 2^(n/2)*(n/2)*(n/2)!]; Do[If[Mod[n, d]==0, t = t+EulerPhi[n/d]^2*d!*(n/d)^d], {d, 1, n}]; t/(2*n^2)); a[1] := 1; a[2] := 1; Print[a /@ Range[1, 450]] (* Jean-François Alcover, May 19 2011, after Maple prog. *)
    rotgra[g_,m_]:=Sort[Sort/@(g/.k_Integer:>If[k==m,1,k+1])];
    Table[Length[Select[Union[Sort[Sort/@Partition[#,2,1,1]]&/@Permutations[Range[n]]],#==First[Sort[Table[Nest[rotgra[#,n]&,#,j],{j,n}]]]&]],{n,8}] (* Gus Wiseman, Mar 02 2019 *)
  • PARI
    a(n)={if(n<3, n>=0, (if(n%2, 0, (n/2-1)!*2^(n/2-2)) + sumdiv(n, d, eulerphi(n/d)^2 * d! * (n/d)^d)/n^2)/2)} \\ Andrew Howroyd, Aug 17 2019

Formula

For formula see Maple lines.
a(2*n + 1) = A002619(2*n + 1)/2 for n > 0; a(2*n) = (A002619(2*n) + A002866(n-1))/2 for n > 1. - Andrew Howroyd, Aug 17 2019
a(n) ~ sqrt(2*Pi)/2 * n^(n-3/2) / e^n. - Ludovic Schwob, Nov 03 2022

Extensions

More terms from Pab Ter (pabrlos(AT)yahoo.com), May 05 2004
Added a(1) = 1 and a(2) = 1 by Gus Wiseman, Mar 02 2019

A008610 Molien series of 4-dimensional representation of cyclic group of order 4 over GF(2) (not Cohen-Macaulay).

Original entry on oeis.org

1, 1, 3, 5, 10, 14, 22, 30, 43, 55, 73, 91, 116, 140, 172, 204, 245, 285, 335, 385, 446, 506, 578, 650, 735, 819, 917, 1015, 1128, 1240, 1368, 1496, 1641, 1785, 1947, 2109, 2290, 2470, 2670, 2870, 3091, 3311, 3553, 3795, 4060, 4324, 4612, 4900, 5213, 5525, 5863
Offset: 0

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Author

Keywords

Comments

a(n) is the number of necklaces with 4 black beads and n white beads.
Also nonnegative integer 2 X 2 matrices with sum of elements equal to n, up to rotational symmetry.
The g.f. is Z(C_4,x), the 4-variate cycle index polynomial for the cyclic group C_4, with substitution x[i]->1/(1-x^i), i=1,...,4. Therefore by Polya enumeration a(n) is the number of cyclically inequivalent 4-necklaces whose 4 beads are labeled with nonnegative integers such that the sum of labels is n, for n=0,1,2,... See A102190 for Z(C_4,x). - Wolfdieter Lang, Feb 15 2005

Examples

			There are 10 inequivalent nonnegative integer 2 X 2 matrices with sum of elements equal to 4, up to rotational symmetry:
[0 0] [0 0] [0 0] [0 0] [0 1] [0 1] [0 1] [0 2] [0 2] [1 1]
[0 4] [1 3] [2 2] [3 1] [1 2] [2 1] [3 0] [1 1] [2 0] [1 1].
		

References

  • D. J. Benson, Polynomial Invariants of Finite Groups, Cambridge, 1993, p. 104.
  • E. V. McLaughlin, Numbers of factorizations in non-unique factorial domains, Senior Thesis, Allegeny College, Meadville, PA, April 2004.

Crossrefs

Row n=2 of A343874.
Column k=4 of A037306 and A047996.

Programs

  • GAP
    a:=[1,1,3,5,10,14,22,30];; for n in [9..50] do a[n]:=2*a[n-1]-2*a[n-3] +2*a[n-4]-2*a[n-5]+2*a[n-7]-a[n-1]; od; a; # G. C. Greubel, Jan 31 2020
  • Magma
    R:=PowerSeriesRing(Integers(), 50); Coefficients(R!( (1+2*x^3+x^4)/((1-x)*(1-x^2)^2*(1-x^4)) )); // G. C. Greubel, Jan 31 2020
    
  • Maple
    1/(1-x)/(1-x^2)^2/(1-x^4)*(1+2*x^3+x^4);
    seq(coeff(series(%, x, n+1), x, n), n=0..40);
  • Mathematica
    k = 4; Table[Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n, {n, k, 30}] (* Robert A. Russell, Sep 27 2004 *)
    LinearRecurrence[{2,0,-2,2,-2,0,2,-1}, {1,1,3,5,10,14,22,30}, 50] (* G. C. Greubel, Jan 31 2020 *)
  • PARI
    a(n)=if(n,([0,1,0,0,0,0,0,0; 0,0,1,0,0,0,0,0; 0,0,0,1,0,0,0,0; 0,0,0,0,1,0,0,0; 0,0,0,0,0,1,0,0; 0,0,0,0,0,0,1,0; 0,0,0,0,0,0,0,1; -1,2,0,-2,2,-2,0,2]^n*[1;1;3;5;10;14;22;30])[1,1],1) \\ Charles R Greathouse IV, Oct 22 2015
    
  • PARI
    my(x='x+O('x^50)); Vec((1+2*x^3+x^4)/((1-x)*(1-x^2)^2*(1-x^4))) \\ G. C. Greubel, Jan 31 2020
    
  • Sage
    def A008610_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P( (1+2*x^3+x^4)/((1-x)*(1-x^2)^2*(1-x^4)) ).list()
    A008610_list(50) # G. C. Greubel, Jan 31 2020
    

Formula

G.f.: (1+2*x^3+x^4)/((1-x)*(1-x^2)^2*(1-x^4)) = (1-x+x^2+x^3)/((1-x)^2*(1-x^2)*(1-x^4)).
a(n) = (1/48)*(2*n^3 + 3*(-1)^n*(n + 4) + 12*n^2 + 25*n + 24 + 12*cos(n*Pi/2)). - Ralf Stephan, Apr 29 2014
G.f.: (1/4)*(1/(1-x)^4 + 1/(1-x^2)^2 + 2/(1-x^4)). - Herbert Kociemba, Oct 22 2016
a(n) = -A032801(-n), per formulae of Colin Barker (A032801) and R. Stephan (above). Also, a(n) - A032801(n+4) = (1+(-1)^signum(n mod 4))/2, i.e., (1,0,0,0,1,0,0,0,...) repeating, (offset 0). - Gregory Gerard Wojnar, Jul 09 2022

Extensions

Comment and example from Vladeta Jovovic, May 18 2000

A008646 Molien series for cyclic group of order 5.

Original entry on oeis.org

1, 1, 3, 7, 14, 26, 42, 66, 99, 143, 201, 273, 364, 476, 612, 776, 969, 1197, 1463, 1771, 2126, 2530, 2990, 3510, 4095, 4751, 5481, 6293, 7192, 8184, 9276, 10472, 11781, 13209, 14763, 16451, 18278, 20254, 22386, 24682, 27151, 29799, 32637, 35673
Offset: 0

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Author

Keywords

Comments

a(n) is the number of necklaces with 5 black beads and n white beads.
The g.f. is Z(C_5,x), the 5-variate cycle index polynomial for the cyclic group C_5, with substitution x[i]->1/(1-x^i), i=1,...,5. Therefore by Polya enumeration a(n) is the number of cyclically inequivalent 5-necklaces whose 5 beads are labeled with nonnegative integers such that the sum of labels is n, for n=0,1,2,... See A102190 for Z(C_5,x). - Wolfdieter Lang, Feb 15 2005

References

  • B. Sturmfels, Algorithms in Invariant Theory, Springer, '93, p. 65.

Crossrefs

Programs

  • Magma
    [Ceiling((n+4)*(n+3)*(n+2)*(n+1)/120): n in [0..50]]; // Vincenzo Librandi, Jun 11 2013
    
  • Maple
    seq(coeff(series((1+x^2+3*x^3+4*x^4+6*x^5+4*x^6+3*x^7+x^8+x^10)/((1-x)* (1-x^2)*(1-x^3)*(1- x^4)*(1-x^5)), x, n+1), x, n), n = 0..50); # corrected by G. C. Greubel, Sep 06 2019
    seq(ceil(binomial(n,4)/5), n=4..41); # Zerinvary Lajos, Jan 12 2009
  • Mathematica
    k = 5; Table[Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n, {n, k, 50}] (* Robert A. Russell, Sep 27 2004 *)
    CoefficientList[Series[(1 +x^2 +3*x^3 +4*x^4 +6*x^5 +4*x^6 +3*x^7 +x^8 +x^10)/((1-x)*(1-x^2)*(1-x^3)*(1- x^4)*(1-x^5)), {x,0,50}], x] (* Vincenzo Librandi, Jun 11 2013 *)
    LinearRecurrence[{4,-6,4,-1,1,-4,6,-4,1}, {1,1,3,7,14,26,42,66,99}, 50] (* Harvey P. Dale, Jan 11 2017 *)
  • PARI
    a(n)=ceil((n+4)*(n+3)*(n+2)*(n+1)/120)
    
  • PARI
    Vec((1-3*x+5*x^2-3*x^3+x^4)/((1-x)^4*(1-x^5)) + O(x^50)) \\ Altug Alkan, Oct 31 2015
    
  • Sage
    [ceil(binomial(n+5,5)/(n+5)) for n in (0..50)] # G. C. Greubel, Sep 06 2019

Formula

G.f.: (1 +x^2 +3*x^3 +4*x^4 +6*x^5 +4*x^6 +3*x^7 +x^8 +x^10)/((1-x)*(1-x^2)*(1-x^3)*(1- x^4)*(1-x^5)).
a(-5-n) = a(n) for all integers.
a(n) = ceiling( binomial(n+5, 5) / (n+5) ).
G.f.: (1 -3*x +5*x^2 -3*x^3 +x^4)/((1-x)^4*(1-x^5)). - Michael Somos, Dec 04 2001
a(n) = (n^4 +10*n^3 +35*n^2 +50*n +24*(3 -2*(-1)^(2^(n-5*floor(n/5)) )))/120. - Luce ETIENNE, Oct 31 2015
G.f.: (4/(1-x^5) + 1/(1-x)^5)/5. - Herbert Kociemba, Oct 15 2016

A138904 Number of rotational symmetries in the binary expansion of a number.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 2, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 0

Views

Author

Max Sills, Apr 03 2008, Apr 04 2008

Keywords

Comments

Mersenne numbers of form (2^n - 1) have n rotational symmetries.
For prime length binary expansions these are the only nontrivial symmetries.
For composite length expansions it seems that when the number of symmetries is nontrivial it is equal to a factor of the length. We're working on an explicit formula.
Discovered in the context of random circulant matrices, examining if there's a correlation between degrees of freedom and number of symmetries in the first row.
When combined with A138954, these two sequences should give a full account of the number of redundant rows in a circulant square matrix with at most two distinct values, where a(n) is the encoding of the first row of the matrix into binary such that value a = 1 and value b = 0.
Discovered on the night of Apr 02, 2008 by Maxwell Sills and Gary Doran.
Conjecture: For binary expansions of length n, there are d(n) distinct values that will show up as symmetries, where d is the divisor function. The symmetry values will be precisely the divisors of n.
Example: for binary expansions of length 12, one sees that d(12) = 6 distinct values show up as symmetries (1, 2, 3, 4, 6, 12).
Conjecture: For numbers whose binary expansion has length n which has proper divisors which are all coprime: There will be only one number of length n with n symmetries. That number is 2^n - 1. For each proper divisor d (excluding 1), you can generate all numbers of length n that have n/d symmetries like so: (2^0 + 2^d + 2^2d ... 2^(n-d)) * a, where 2^(d-1) <= a < (2^d) - 1. The rest of the expansions of length n will have only the trivial symmetry.
Also the number of rotational symmetries of the n-th composition in standard order (graded reverse-lexicographic). This composition (row n of A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of n, prepending 0, taking first differences, and reversing again. - Gus Wiseman, Apr 19 2020
From Gus Wiseman, Apr 19 2020: (Start)
Aperiodic compositions are counted by A000740.
Aperiodic binary words are counted by A027375.
The orderless period of prime indices is A052409.
Numbers whose binary expansion is periodic are A121016.
Periodic compositions are counted by A178472.
Period of binary expansion is A302291.
Compositions by sum and number of distinct rotations are A333941.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Necklaces are A065609.
- Sum is A070939.
- Runs are counted by A124767.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Lyndon compositions are A275692.
- Co-Lyndon compositions are A326774.
- Aperiodic compositions are A328594.
- Reversed co-necklaces are A328595.
- Rotational period is A333632.
- Co-necklaces are A333764.
- Reversed necklaces are A333943.
(End).

Examples

			a(10) = 2 because the binary expansion of 10 is 1010 and it has two rotational symmetries (including identity).
		

Crossrefs

Programs

  • Mathematica
    Table[IntegerLength[n,2]/Length[Union[Array[RotateRight[IntegerDigits[n,2],#]&,IntegerLength[n,2]]]],{n,100}] (* Gus Wiseman, Apr 19 2020 *)

Formula

a(n) = A070939(n)/A302291(n) = A000120(n)/A333632(n). - Gus Wiseman, Apr 19 2020

A053635 a(n) = Sum_{d|n} phi(d)*2^(n/d).

Original entry on oeis.org

0, 2, 6, 12, 24, 40, 84, 140, 288, 540, 1080, 2068, 4224, 8216, 16548, 32880, 65856, 131104, 262836, 524324, 1049760, 2097480, 4196412, 8388652, 16782048, 33554600, 67117128, 134218836, 268452240, 536870968, 1073777040, 2147483708, 4295033472, 8589938808
Offset: 0

Views

Author

N. J. A. Sloane, Mar 23 2000

Keywords

Comments

Dirichlet convolution of phi(n) and 2^n. - Richard L. Ollerton, May 06 2021

Crossrefs

Column k=2 of A185651.

Programs

  • Magma
    [0] cat  [&+[EulerPhi(d)*2^(n div d): d in Divisors(n)]: n in [1..40]]; // Vincenzo Librandi, Jul 20 2019
  • Maple
    with(numtheory); A053685:=n->add( phi(n/d)*2^d, d in divisors(n)); # N. J. A. Sloane, Nov 21 2009
  • Mathematica
    a[0] = 0; a[n_] := Sum[EulerPhi[d] 2^(n/d), {d, Divisors[n]}];
    Table[a[n], {n, 0, 31}] (* Jean-François Alcover, Aug 30 2018 *)
  • PARI
    a(n) = if (n, sumdiv(n, d, eulerphi(d)*2^(n/d)), 0); \\ Michel Marcus, Sep 20 2017
    

Formula

a(n) = n * A000031(n).
a(n) = Sum_{k=1..n} 2^gcd(n,k). - Ilya Gutkovskiy, Apr 16 2021
a(n) = Sum_{k=1..n} 2^(n/gcd(n,k))*phi(gcd(n,k))/phi(n/gcd(n,k)). - Richard L. Ollerton, May 06 2021

A061417 Number of permutations up to cyclic rotations; permutation siteswap necklaces.

Original entry on oeis.org

1, 2, 4, 10, 28, 136, 726, 5100, 40362, 363288, 3628810, 39921044, 479001612, 6227066928, 87178295296, 1307675013928, 20922789888016, 355687438476444, 6402373705728018, 121645100594641896, 2432902008177690360, 51090942175425331320, 1124000727777607680022
Offset: 1

Views

Author

Antti Karttunen, May 02 2001

Keywords

Comments

If permutations are converted to (i,p(i)) permutation arrays, then this automorphism is obtained by their "SW-NE diagonal toroidal shifts" (see Matthias Engelhardt's Java program in A006841), while the Maple procedure below converts each permutation to a siteswap pattern (used in juggling), rotates it by one digit and converts the resulting new (or same) siteswap pattern back to a permutation.
When the subset of permutations listed by A064640 are subjected to the same automorphism one gets A002995.
The number of conjugacy classes of the symmetric group of degree n when conjugating only with the cyclic permutation group of degree n. - Attila Egri-Nagy, Aug 15 2014
Also the number of equivalence classes of permutations of {1...n} under the action of rotation of vertices in the cycle decomposition. The corresponding action on words applies m -> m + 1 for m < n and n -> 1, and rotates once to the right. For example, (24531) first becomes (35142) under the application of cyclic rotation, and then is rotated right to give (23514). - Gus Wiseman, Mar 04 2019

Examples

			If I have a five-element permutation like 25431, in cycle notation (1 2 5)(3 4), I mark the numbers 1-5 clockwise onto a circle and draw directed edges from 1 to 2, from 2 to 5, from 5 to 1 and a double-way edge between 3 and 4. All the 5-element permutations that produce some rotation (discarding the labels of the nodes) of that chord diagram belong to the same equivalence class with 25431. The sequence gives the count of such equivalence classes.
		

Crossrefs

Cf. A006841, A060495. For other Maple procedures, see A060501 (Perm2SiteSwap2), A057502 (CountCycles), A057509 (rotateL), A060125 (PermRank3R and permul).
A061417[p] = A061860[p] = (p-1)!+(p-1) for all prime p's.
A064636 (derangements-the same automorphism).
A061417[n] = A064649[n]/n.
Cf. A000031, A000939, A002995, A008965, A060223, A064640, A086675 (digraphical necklaces), A179043, A192332, A275527 (path necklaces), A323858, A323859, A323870, A324513, A324514 (aperiodic permutations).

Programs

  • GAP
    List([1..10],n->Size( OrbitsDomain( CyclicGroup(IsPermGroup,n), SymmetricGroup( IsPermGroup,n),\^))); # Attila Egri-Nagy, Aug 15 2014
    
  • Haskell
    a061417 = sum . a047917_row  -- Reinhard Zumkeller, Mar 19 2014
    
  • Maple
    Algebraic formula: with(numtheory); SSRPCC := proc(n) local d,s; s := 0; for d in divisors(n) do s := s + phi(n/d)*((n/d)^d)*(d!); od; RETURN(s/n); end;
    Empirically: with(group); SiteSwapRotationPermutationCycleCounts := proc(upto_n) local b,u,n,a,r; a := []; for n from 1 to upto_n do b := []; u := n!; for r from 0 to u-1 do b := [op(b),1+PermRank3R(SiteSwap2Perm1(rotateL(Perm2SiteSwap2(PermUnrank3Rfix(n,r)))))]; od; a := [op(a),CountCycles(b)]; od; RETURN(a); end;
    PermUnrank3Rfixaux := proc(n,r,p) local s; if(0 = n) then RETURN(p); else s := floor(r/((n-1)!)); RETURN(PermUnrank3Rfixaux(n-1, r-(s*((n-1)!)), permul(p,[[n,n-s]]))); fi; end;
    PermUnrank3Rfix := (n,r) -> convert(PermUnrank3Rfixaux(n,r,[]),'permlist',n);
    SiteSwap2Perm1 := proc(s) local e,n,i,a; n := nops(s); a := []; for i from 1 to n do e := ((i+s[i]) mod n); if(0 = e) then e := n; fi; a := [op(a),e]; od; RETURN(convert(invperm(convert(a,'disjcyc')),'permlist',n)); end;
  • Mathematica
    a[n_] := (1/n)*Sum[ EulerPhi[n/d]*(n/d)^d*d!, {d, Divisors[n]}]; Table[a[n], {n, 1, 21}] (* Jean-François Alcover, Oct 09 2012, from formula *)
    Table[Length[Select[Permutations[Range[n]],#==First[Sort[NestList[RotateRight[#/.k_Integer:>If[k==n,1,k+1]]&,#,n-1]]]&]],{n,8}] (* Gus Wiseman, Mar 04 2019 *)
  • PARI
    a(n) = (1/n)*sumdiv(n, d, eulerphi(n/d)*(n/d)^d*d!); \\ Indranil Ghosh, Apr 10 2017
    
  • Python
    from sympy import divisors, factorial, totient
    def a(n):
        return sum(totient(n//d)*(n//d)**d*factorial(d) for d in divisors(n))//n
    print([a(n) for n in range(1, 22)]) # Indranil Ghosh, Apr 10 2017

Formula

a(n) = (1/n)*Sum_{d|n} phi(n/d)*((n/d)^d)*(d!).

A192332 For n >= 3, draw a regular n-sided polygon and its n(n-3)/2 diagonals, so there are n(n-1)/2 lines; a(n) is the number of ways to choose a subset of these lines (subsets differing by a rotation are regarded as identical). a(1)=1, a(2)=2 by convention.

Original entry on oeis.org

1, 2, 4, 22, 208, 5560, 299600, 33562696, 7635498336, 3518440564544, 3275345183542208, 6148914696963883712, 23248573454127484129024, 176848577040808821410837120, 2704321280486889389864215362560, 83076749736557243209409446411255936, 5124252113632955685095523500148980125696, 634332307869315502692705867068871886072665600
Offset: 1

Views

Author

N. J. A. Sloane, Jun 28 2011

Keywords

Comments

Suggested by A192314.
Also the number of graphical necklaces with n vertices. We define a graphical necklace to be a simple graph that is minimal among all n rotations of the vertices. Alternatively, it is an equivalence class of simple graphs under rotation of the vertices. These are a kind of partially labeled graphs. - Gus Wiseman, Mar 04 2019

Examples

			From _Gus Wiseman_, Mar 04 2019: (Start)
Inequivalent representatives of the a(1) = 1 through a(4) = 22 graphical necklace edge-sets:
  {}  {}      {}              {}
      {{12}}  {{12}}          {{12}}
              {{12}{13}}      {{13}}
              {{12}{13}{23}}  {{12}{13}}
                              {{12}{14}}
                              {{12}{24}}
                              {{12}{34}}
                              {{13}{24}}
                              {{12}{13}{14}}
                              {{12}{13}{23}}
                              {{12}{13}{24}}
                              {{12}{13}{34}}
                              {{12}{14}{23}}
                              {{12}{24}{34}}
                              {{12}{13}{14}{23}}
                              {{12}{13}{14}{24}}
                              {{12}{13}{14}{34}}
                              {{12}{13}{24}{34}}
                              {{12}{14}{23}{34}}
                              {{12}{13}{14}{23}{24}}
                              {{12}{13}{14}{23}{34}}
                              {{12}{13}{14}{23}{24}{34}}
(End)
		

Crossrefs

Cf. A192314, A191563 (orbits under dihedral group).
Cf. A000031, A000939 (cycle necklaces), A008965, A059966, A060223, A061417, A086675 (digraph version), A184271, A275527, A323858, A324461, A324463, A324464.

Programs

  • Maple
    with(numtheory);
    f:=proc(n) local t0, t1, d; t0:=0; t1:=divisors(n);
    for d in t1 do
    if d mod 2 = 0 then t0:=t0+phi(d)*2^(n^2/(2*d))
    else t0:=t0+phi(d)*2^(n*(n-1)/(2*d)); fi; od; t0/n; end;
    [seq(f(n), n=1..30)];
  • Mathematica
    Table[ 1/n* Plus @@ Map[Function[d, EulerPhi[d]*2^((n*(n - Mod[d, 2])/2)/d)], Divisors[n]], {n, 1, 20}]  (* Olivier Gérard, Aug 27 2011 *)
    rotgra[g_,m_]:=Sort[Sort/@(g/.k_Integer:>If[k==m,1,k+1])];
    Table[Length[Select[Subsets[Subsets[Range[n],{2}]],#=={}||#==First[Sort[Table[Nest[rotgra[#,n]&,#,j],{j,n}]]]&]],{n,0,5}] (* Gus Wiseman, Mar 04 2019 *)
  • PARI
    a(n) = sumdiv(n, d, if (d%2, eulerphi(d)*2^(n*(n-1)/(2*d)), eulerphi(d)*2^(n^2/(2*d))))/n; \\ Michel Marcus, Mar 08 2019

Formula

a(n) = (1/n)*(Sum_{d|n, d odd} phi(d)*2^(n*(n-1)/(2*d)) + Sum_{d|n, d even} phi(d)*2^(n^2/(2*d))).
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