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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A284855 Array read by antidiagonals: T(n,k) = number of necklaces with n beads and k colors that are the same when turned over.

Original entry on oeis.org

1, 2, 1, 3, 3, 1, 4, 6, 4, 1, 5, 10, 9, 6, 1, 6, 15, 16, 18, 8, 1, 7, 21, 25, 40, 27, 12, 1, 8, 28, 36, 75, 64, 54, 16, 1, 9, 36, 49, 126, 125, 160, 81, 24, 1, 10, 45, 64, 196, 216, 375, 256, 162, 32, 1, 11, 55, 81, 288, 343, 756, 625, 640, 243, 48, 1
Offset: 1

Views

Author

Andrew Howroyd, Apr 04 2017

Keywords

Comments

Number of periodic palindromes of length n using a maximum of k different symbols.
From Petros Hadjicostas, Sep 02 2018: (Start)
According to Christian Bower's theory of transforms, we have boxes of different sizes and different colors. The size of a box is determined by the number of balls it can hold. In this case, we assume all balls are the same and are unlabeled. Assume also that the number of possible colors a box with m balls can have is given by c(m). We place the boxes on a circle at equal distances from each other. Two configurations of boxes on the circle are considered equivalent if one can be obtained from the other by rotation. We are interested about circular configurations of boxes that are circular palindromes (i.e., necklaces with boxes that are the same when turned over). Let b(n) be the number of circularly palindromic configurations of boxes on a circle when the total number of balls in the boxes is n (and each box contains at least one ball).
Bower calls the sequence (b(n): n >= 1), the CPAL ("circular palindrome") transform of the input sequence (c(m): m >= 1). If the g.f. of the input sequence (c(m): m >= 1) is C(x) = Sum_{m>=1} c(m)*x^m, then the g.f. of the output sequence (b(n): n >= 1) is B(x) = Sum_{n >= 1} b(n)*x^n = (1 + C(x))^2/(2*(1 - C(x^2))) - 1/2.
In the current sequence, each box holds only one ball but can have one of k colors. Hence, c(1) = k but c(m) = 0 for m >= 2. Thus, C(x) = k*x. Then, for fixed k, the output sequence is (b(n): n >= 1) = (T(n, k): n >= 1), where T(n, k) = number of necklaces with n beads and k colors that are the same when turned over. If we let B_k(x) = Sum_{n>=1} T(n, k)*x^n, then B_k(x) = (1 + k*x)^2/(2*(1 - k*x^2)) - 1/2. From this, we can easily prove the formulae below.
Note that T(n, k=2) - 1 is the total number of Sommerville symmetric cyclic compositions of n. See pp. 301-304 in his paper in the links below. To see why this is the case, we use MacMahon's method of representing a cyclic composition of n with a necklace of 2 colors (see p. 273 in Sommerville's paper where the two "colors" are an x and a dot . rather than B and W). Given a Sommerville symmetrical composition b_1 + ... + b_r of n (with b_i >= 1 for all i and 1 <= r <= n), create the following circularly palindromic necklace with n beads of 2 colors: Start with a B bead somewhere on the circle and place b_1 - 1 W beads to the right of it; place a B bead to the right of the W beads (if any) followed by b_2 - 1 W beads; and so on. At the end, place a B bead followed with b_r - 1 W beads. (If b_i = 1 for some i, then a B bead follows a B bead since there are 0 W beads between them.) We thus get a circularly palindromic necklace with n beads of two colors. (The only necklace we cannot get with this method is the one than has all n beads colored W.)
It is interesting that the representation of a necklace of length n, say s_1, s_2, ..., s_n, as a periodic sequence (..., s_{-2}, s_{-1}, s_0, s_1, s_2, ...) with the property s_i = s_{i+n} for all i, as was done by Marks R. Nester in Chapter 2 of his 1999 PhD thesis, was considered by Sommerville in his 1909 paper (in the very first paragraph of his paper). (End)

Examples

			Table starts:
1  2   3    4    5     6     7      8      9     10 ...
1  3   6   10   15    21    28     36     45     55 ...
1  4   9   16   25    36    49     64     81    100 ...
1  6  18   40   75   126   196    288    405    550 ...
1  8  27   64  125   216   343    512    729   1000 ...
1 12  54  160  375   756  1372   2304   3645   5500 ...
1 16  81  256  625  1296  2401   4096   6561  10000 ...
1 24 162  640 1875  4536  9604  18432  32805  55000 ...
1 32 243 1024 3125  7776 16807  32768  59049 100000 ...
1 48 486 2560 9375 27216 67228 147456 295245 550000 ...
...
For n = 4 and k = 2, the palindromic necklaces are 0000, 0001, 0011, 0111, 0101, 1111 so T(4,2) = 6. Necklaces are only counted up to cyclic equivalence.
For n = 4 and k = 2, using MacMahon's bijection, with B = 0 and W = 1, the corresponding Sommerville symmetrical cyclic compositions of n = 4 are as follows: 1+1+1+1, 1+1+2, 1+3, 4, 2+2 (with none for 1111). If we let B = 1 and W = 0, we get the corresponding symmetrical cyclic compositions of n=4: (none for 0000) 4, 1+3, 1+1+2, 2+2, 1+1+1+1. (All these cyclic compositions must viewed on a circle.) - _Petros Hadjicostas_, Sep 02 2018
		

References

  • M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for the pdf file of Chap. 2]

Crossrefs

Programs

  • Mathematica
    a[n_, k_] := If[EvenQ[n], (k^(n/2) + k^(n/2 + 1))/2, k^((n+1)/2)];
    Table[a[n-k+1, k], {n, 1, 11}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Jun 05 2017, translated from PARI *)
  • PARI
    a(n,k) = if(n % 2 == 0, (k^(n/2) + k^(n/2+1))/2, k^((n+1)/2));
    for(n=1, 10, for(k=1, 10, print1( a(n,k),", ");); print(););

Formula

T(2*n, k) = (k^(n+1) + k^n) / 2.
T(2*n + 1, k) = k^(n+1).
T(n, k) = 2 * A081720(n, k) - A075195(n, k).
From Petros Hadjicostas, Sep 02 2018: (Start)
For fixed k >= 1, the k-th column (T(n, k): n >= 1) is the CPAL ("circular palindrome") transform of the sequence k, 0, 0, ...
G.f. of column k: Sum_{n>=1} T(n,k)*x^n = (1 + k*x)^2/(2*(1 - k*x^2)) - 1/2. (End)

A293496 Array read by antidiagonals: T(n,k) = number of chiral pairs of necklaces with n beads using a maximum of k 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, 38, 2, 0, 0, 0, 35, 105, 252, 270, 117, 6, 0, 0, 0, 56, 210, 672, 1130, 1044, 336, 14, 0, 0, 0, 84, 378, 1512, 3535, 5270, 3795, 976, 30, 0
Offset: 1

Views

Author

Andrew Howroyd, Oct 10 2017

Keywords

Comments

An orientable necklace when turned over does not leave it unchanged. Only one necklace in each pair is included in the count.
The number of chiral bracelets. An achiral bracelet is the same as its reverse, while a chiral bracelet is equivalent to its reverse. - Robert A. Russell, Sep 28 2018

Examples

			Array begins:
  ==========================================================
  n\k | 1  2    3     4      5       6        7        8
  ----+-----------------------------------------------------
   1  | 0  0    0     0      0       0        0        0 ...
   2  | 0  0    0     0      0       0        0        0 ...
   3  | 0  0    1     4     10      20       35       56 ...
   4  | 0  0    3    15     45     105      210      378 ...
   5  | 0  0   12    72    252     672     1512     3024 ...
   6  | 0  1   38   270   1130    3535     9156    20748 ...
   7  | 0  2  117  1044   5270   19350    57627   147752 ...
   8  | 0  6  336  3795  23520  102795   355656  1039626 ...
   9  | 0 14  976 14060 106960  556010  2233504  7440216 ...
  10  | 0 30 2724 51204 483756 3010098 14091000 53615016 ...
  ...
For T(3,4)=4, the chiral pairs are ABC-ACB, ABD-ADB, ACD-ADC, and BCD-BDC.
For T(4,3)=3, the chiral pairs are AABC-AACB, ABBC-ACBB, and ABCC-ACCB. - _Robert A. Russell_, Sep 28 2018
		

Crossrefs

Programs

  • Mathematica
    b[n_, k_] := (1/n)*DivisorSum[n, EulerPhi[#]*k^(n/#) &];
    c[n_, k_] := If[EvenQ[n], (k^(n/2) + k^(n/2 + 1))/2, k^((n + 1)/2)];
    T[, 1] = T[1, ] = 0; T[n_, k_] := (b[n, k] - c[n, k])/2;
    Table[T[n - k + 1, k], {n, 1, 11}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Oct 11 2017, translated from PARI *)
  • PARI
    \\ here b(n,k) is A075195 and c(n,k) is A284855
    b(n, k) = (1/n) * sumdiv(n, d, eulerphi(d)*k^(n/d));
    c(n, k) = if(n % 2 == 0, (k^(n/2) + k^(n/2+1))/2, k^((n+1)/2));
    T(n, k) = (b(n, k) - c(n, k)) / 2;

Formula

T(n,k) = (A075195(n,k) - A284855(n,k)) / 2.
From Robert A. Russell, Sep 28 2018: (Start)
T(n, k) = -(k^floor((n+1)/2) + k^ceiling((n+1)/2)) / 4 + (1/2n) * Sum_{d|n} phi(d) * k^(n/d)
G.f. for column k: -(kx/4)*(kx+x+2)/(1-kx^2) - Sum_{d>0} phi(d)*log(1-kx^d)/2d. (End)

A054625 Number of n-bead necklaces with 6 colors.

Original entry on oeis.org

1, 6, 21, 76, 336, 1560, 7826, 39996, 210126, 1119796, 6047412, 32981556, 181402676, 1004668776, 5597460306, 31345666736, 176319474366, 995685849696, 5642220380006, 32071565263716, 182807925027504, 1044616697187576, 5982804736593846
Offset: 0

Views

Author

N. J. A. Sloane, Apr 16 2000

Keywords

Examples

			G.f. = 1 + 6*x + 21*x^2 + 76*x^3 + 336*x^4 + 1650*x^5 + 7826*x^6 + 39996*x^7 + ...
		

Crossrefs

Column 6 of A075195.
Cf. A054613.

Programs

  • Maple
    with(combstruct):A:=[N,{N=Cycle(Union(Z$6))},unlabeled]: seq(count(A,size=n),n=0..22); # Zerinvary Lajos, Dec 05 2007
  • Mathematica
    f[n_] := Block[{d = Divisors@ n}, Total[EulerPhi[d]*6^(n/d)]/n]; f[0] = 1; Array[f, 23, 0] (* Robert G. Wilson v, Jan 01 2013 *)
    mx=40; CoefficientList[Series[1-Sum[EulerPhi[i] Log[1-6*x^i]/i, {i, 1, mx}], {x, 0, mx}], x] (* Herbert Kociemba, Nov 02 2016 *)

Formula

a(n) = (1/n)*Sum_{d|n} phi(d)*6^(n/d), n > 0.
G.f.: 1 - Sum_{n>=1} phi(n)*log(1 - 6*x^n)/n. - Herbert Kociemba, Nov 02 2016
a(0) = 1; a(n) = (1/n) * Sum_{k=1..n} 6^gcd(n,k). - Ilya Gutkovskiy, Apr 17 2021

Extensions

Edited by Christian G. Bower, Sep 07 2002
a(0) corrected by Herbert Kociemba, Nov 02 2016

A054630 T(n,k) = Sum_{d|k} phi(d)*n^(k/d)/k, triangle read by rows, T(n,k) for n >= 1 and 1 <= k <= n.

Original entry on oeis.org

1, 2, 3, 3, 6, 11, 4, 10, 24, 70, 5, 15, 45, 165, 629, 6, 21, 76, 336, 1560, 7826, 7, 28, 119, 616, 3367, 19684, 117655, 8, 36, 176, 1044, 6560, 43800, 299600, 2097684, 9, 45, 249, 1665, 11817, 88725, 683289, 5381685, 43046889, 10, 55, 340, 2530, 20008, 166870, 1428580, 12501280, 111111340, 1000010044
Offset: 1

Views

Author

N. J. A. Sloane, Apr 16 2000, revised Mar 21 2007

Keywords

Comments

T(n, k) is the number of n-ary necklaces of length k (see Ruskey, Savage and Wang). - Peter Luschny, Aug 12 2012, comment corrected at the suggestion of Petros Hadjicostas, Peter Luschny, Sep 10 2018
From Petros Hadjicostas, Sep 12 2018: (Start)
The programs by Peter Luschny below can generate all n-ary necklaces of length k (and all k-ary necklaces of length n) for any positive integer values of n and k, not just for 1 <= k <= n.
From the examples below, we see that the number of 4-ary necklaces of length 3 equals the number of 3-ary necklaces of length 4. The question is whether there are other pairs (n, k) of distinct positive integers such that the number of n-ary necklaces of length k equals the number of k-ary necklaces of length n.
(End)

Examples

			Triangle starts:
  1;
  2,  3;
  3,  6, 11;
  4, 10, 24, 70;
  5, 15, 45, 165,  629;
  6, 21, 76, 336, 1560, 7826;
The 24 necklaces over {0,1,2} of length 4 are:
  0000,0001,0002,0011,0012,0021,0022,0101,0102,0111,0112,0121,
  0122,0202,0211,0212,0221,0222,1111,1112,1122,1212,1222,2222.
The 24 necklaces over {0,1,2,3} of length 3 are:
  000,001,002,003,011,012,013,021,022,023,031,032,
  033,111,112,113,122,123,132,133,222,223,233,333.
		

References

  • D. E. Knuth, Generating All Tuples and Permutations. The Art of Computer Programming, Vol. 4, Fascicle 2, Addison-Wesley, 2005.

Crossrefs

Cf. A054631, A054618, A054619, A056665, A215474. Upper triangle of A075195.

Programs

  • Julia
    A054630(n::Int, k::Int) = div(sum(n^gcd(i,k) for i in 1:k), k)
    for n in 1:6
        println([A054630(n, k) for k in 1:n])
    end # Peter Luschny, Sep 10 2018
  • Maple
    T := (n,k) -> add(n^igcd(i,k), i=1..k)/k:
    seq(seq(T(n,k), k=1..n), n=1..10); # Peter Luschny, Sep 10 2018
  • Mathematica
    T[n_, k_] := 1/k Sum[EulerPhi[d] n^(k/d), {d, Divisors[k]}];
    Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 30 2018 *)
  • Sage
    def A054630(n,k): return (1/k)*add(euler_phi(d)*n^(k/d) for d in divisors(k))
    for n in (1..9):
        print([A054630(n,k) for k in (1..n)]) # Peter Luschny, Aug 12 2012
    

Formula

T(n,n) = A056665(n). - Peter Luschny, Aug 12 2012
T(n,k) = (1/k)*Sum_{i=1..k} n^gcd(i, k). - Peter Luschny, Sep 10 2018

A054624 Number of ways to color vertices of a 10-gon using <= n colors, allowing only rotations.

Original entry on oeis.org

0, 1, 108, 5934, 104968, 976887, 6047412, 28249228, 107377488, 348684381, 1000010044, 2593758618, 6191761368, 13785886387, 28925519364, 57665115096, 109951267744, 201599532153, 357046911756, 613106873542, 1024000320168, 1667988506415, 2655992794708
Offset: 0

Views

Author

N. J. A. Sloane, Apr 16 2000

Keywords

Crossrefs

Row 10 of A075195.

Formula

Sum_{d|10} phi(d)*n^(10/d)/10 = n*(n+1)*(n^8-n^7+n^6-n^5+n^4+4)/10.
G.f.: x*(100*x^8 +4783*x^7 +45547*x^6 +130963*x^5 +131119*x^4 +45469*x^3 +4801*x^2 +97*x+1) / (1-x)^11. - Colin Barker, Jun 12 2012

Extensions

Edited by Christian G. Bower, Sep 07 2002

A006565 Number of ways to color vertices of a hexagon using <= n colors, allowing only rotations.

Original entry on oeis.org

0, 1, 14, 130, 700, 2635, 7826, 19684, 43800, 88725, 166870, 295526, 498004, 804895, 1255450, 1899080, 2796976, 4023849, 5669790, 7842250, 10668140, 14296051, 18898594, 24674860, 31853000, 40692925, 51489126, 64573614
Offset: 0

Views

Author

Keywords

References

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

Crossrefs

Programs

  • Maple
    A006565 := n-> (n^6+n^3+2*n^2+2*n)/6.
    A006565:=-(1+7*z+53*z**2+49*z**3+10*z**4)/(z-1)**7; [Conjectured by Simon Plouffe in his 1992 dissertation.]

Formula

n*(n+1)*(n^2+n+1)*(n^2-2*n+2)/6.

A054631 Triangle read by rows: row n (n >= 1) contains the numbers T(n,k) = Sum_{d|n} phi(d)*k^(n/d)/n, for k=1..n.

Original entry on oeis.org

1, 1, 3, 1, 4, 11, 1, 6, 24, 70, 1, 8, 51, 208, 629, 1, 14, 130, 700, 2635, 7826, 1, 20, 315, 2344, 11165, 39996, 117655, 1, 36, 834, 8230, 48915, 210126, 720916, 2097684, 1, 60, 2195, 29144, 217045, 1119796, 4483815, 14913200, 43046889
Offset: 1

Views

Author

N. J. A. Sloane, Apr 16 2000, revised Mar 21 2007

Keywords

Comments

T(n,k) is the number of n-bead necklaces with up to k different colored beads. - Yves-Loic Martin, Sep 29 2020

Examples

			1;
1,  3;                                   (A000217)
1,  4,  11;                              (A006527)
1,  6,  24,   70;                        (A006528)
1,  8,  51,  208,   629;                 (A054620)
1, 14, 130,  700,  2635,  7826;          (A006565)
1, 20, 315, 2344, 11165, 39996, 117655;  (A054621)
		

Crossrefs

Cf. A054630, A054618, A054619, A087854. Lower triangle of A075195.

Programs

  • Maple
    A054631 := proc(n,k) add( numtheory[phi](d)*k^(n/d),d=numtheory[divisors](n) ) ;  %/n ; end proc: # R. J. Mathar, Aug 30 2011
  • Mathematica
    Needs["Combinatorica`"]; Table[Table[NumberOfNecklaces[n, k, Cyclic], {k, 1, n}], {n, 1, 8}] //Grid (* Geoffrey Critzer, Oct 07 2012, after code by T. D. Noe in A027671 *)
    t[n_, k_] := Sum[EulerPhi[d]*k^(n/d)/n, {d, Divisors[n]}]; Table[t[n, k], {n, 1, 9}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jan 20 2014 *)
  • PARI
    T(n, k) = sumdiv(n, d, eulerphi(d)*k^(n/d))/n; \\ Seiichi Manyama, Mar 10 2021
    
  • PARI
    T(n, k) = sum(j=1, n, k^gcd(j, n))/n; \\ Seiichi Manyama, Mar 10 2021

Formula

T(n,k) = Sum_{j=1..k} binomial(k,j) * A087854(n, j). - Yves-Loic Martin, Sep 29 2020
T(n,k) = (1/n) * Sum_{j=1..n} k^gcd(j, n). - Seiichi Manyama, Mar 10 2021

A051137 Table T(n,k) read by antidiagonals: number of necklaces allowing turnovers (bracelets) with n beads of k colors.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 6, 10, 10, 5, 1, 1, 8, 21, 20, 15, 6, 1, 1, 13, 39, 55, 35, 21, 7, 1, 1, 18, 92, 136, 120, 56, 28, 8, 1, 1, 30, 198, 430, 377, 231, 84, 36, 9, 1, 1, 46, 498, 1300, 1505, 888, 406, 120, 45, 10, 1
Offset: 0

Views

Author

Keywords

Comments

Unlike A075195 and A284855, antidiagonals go from bottom-left to top-right.

Examples

			Table begins with T[0,1]:
1  1    1     1      1       1        1        1         1         1
1  2    3     4      5       6        7        8         9        10
1  3    6    10     15      21       28       36        45        55
1  4   10    20     35      56       84      120       165       220
1  6   21    55    120     231      406      666      1035      1540
1  8   39   136    377     888     1855     3536      6273     10504
1 13   92   430   1505    4291    10528    23052     46185     86185
1 18  198  1300   5895   20646    60028   151848    344925    719290
1 30  498  4435  25395  107331   365260  1058058   2707245   6278140
1 46 1219 15084 110085  563786  2250311  7472984  21552969  55605670
1 78 3210 53764 493131 3037314 14158228 53762472 174489813 500280022
		

Crossrefs

Columns 2-6 are A000029, A027671, A032275, A032276, and A056341.
Rows 2-7 are A000217, A000292, A002817, A060446, A027670, and A060532.
Cf. A000031.
T(n,k) = (A075195(n,k) + A284855(n,k)) / 2.

Programs

  • Mathematica
    b[n_, k_] := DivisorSum[n, EulerPhi[#]*k^(n/#) &] / n;
    c[n_, k_] := If[EvenQ[n], (k^(n/2) + k^(n/2+1))/2, k^((n+1)/2)];
    T[0, ] = 1; T[n, k_] := (b[n, k] + c[n, k])/2;
    Table[T[n, k-n], {k, 1, 11}, {n, k-1, 0, -1}] // Flatten
    (* Robert A. Russell, Sep 21 2018 after Jean-François Alcover *)

Formula

T(n, k) = (k^floor((n+1)/2) + k^ceiling((n+1)/2)) / 4 + (1/(2*n)) * Sum_{d divides n} phi(d) * k^(n/d). - Robert A. Russell, Sep 21 2018
G.f. for column k: (kx/4)*(kx+x+2)/(1-kx^2) - Sum_{d>0} phi(d)*log(1-kx^d)/2d. - Robert A. Russell, Sep 28 2018
T(n, k) = (k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/(2*n))*Sum_{i=1..n} k^gcd(n,i). (See A075195 formulas.) - Richard L. Ollerton, May 04 2021

A285522 Array read by antidiagonals: T(m,n) = number of circulant digraphs up to Cayley isomorphism on n vertices with edges colored according to step value using a maximum of m-1 colors.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 6, 6, 4, 1, 1, 6, 18, 10, 5, 1, 1, 20, 24, 40, 15, 6, 1, 1, 14, 135, 70, 75, 21, 7, 1, 1, 48, 130, 544, 165, 126, 28, 8, 1, 1, 52, 648, 700, 1625, 336, 196, 36, 9, 1, 1, 140, 1137, 4480, 2635, 3996, 616, 288, 45, 10, 1
Offset: 1

Views

Author

Andrew Howroyd, Apr 20 2017

Keywords

Comments

For the base case of m=2 the sequence counts circulant digraphs up to Cayley isomorphism. Two circulant graphs are Cayley isomorphic if there is a d, which is necessarily prime to n, that transforms through multiplication modulo n the step values of one graph into those of the other. For squarefree n this is the only way that two circulant graphs can be isomorphic. (See Liskovets reference for a proof.)
Alternatively, the number of mappings with domain {1..n-1} and codomain {1..m} up to equivalence. Mappings A and B are equivalent if there is a d, prime to n, such that A(i) = B(i*d mod n) for i in {1..n-1}. This sequence differs from A132191 only in that sequence also includes 0 in the domain which introduces an extra factor of m into the results since zero multiplied by anything is zero.
All column sequences are polynomials of order n-1 and these are the cycle index polynomials.
This sequence is also related to A075195(n, m) which counts necklaces and A285548(m, n) which is the sequence described in the Titsworth reference. In particular, A075195 is the analogous array with equivalence determined through the additive group instead of by multiplication whereas A285548 allows for both addition and multiplication.

Examples

			Table starts:
\n  1 2  3   4    5    6    7     8      9      10
m\ ---------------------------------------------------
1 | 1 1  1   1    1    1    1     1      1       1 ...
2 | 1 2  3   6    6   20   14    48     52     140 ...
3 | 1 3  6  18   24  135  130   648   1137    4995 ...
4 | 1 4 10  40   70  544  700  4480  11056   65824 ...
5 | 1 5 15  75  165 1625 2635 20625  65425  489125 ...
6 | 1 6 21 126  336 3996 7826 72576 280596 2521476 ...
...
Case n=10:
Only 1, 3, 7, 9 are prime to 10.
Multiplication modulo 10 is described by the following multiplication table.
  1, 2, 3, 4, 5, 6, 7, 8, 9  => (1)(2)(3)(4)(5)(6)(7)(8)(9) => m^9
  3, 6, 9, 2, 5, 8, 1, 4, 7  => (1397)(2684)(5)             => m^3
  7, 4, 1, 8, 5, 2, 9, 6, 3  => (1793)(2486)(5)             => m^3
  9, 8, 7, 6, 5, 4, 3, 2, 1  => (19)(28)(37)(46)(5)         => m^5
Each row of the multiplication table can be viewed as a permutation and together these form a commutative group on 4 elements. In this case the group is isomorphic to the cyclic group C_4. Each permutation can be represented in cycle notation. (shown above to the right of the corresponding multiplication table row). In order to count the equivalence classes using Polya's enumeration theorem only the number of cycles in each permutation is needed.
This gives the cycle index polynomial (1/4)*(m^9 + m^5 + 2*m^3). Putting m = 1..4 gives 1, 140, 4995, 65824.
		

Crossrefs

Programs

  • Mathematica
    A132191[m_, n_] := (1/EulerPhi[n])*Sum[If[GCD[k, n] == 1, m^DivisorSum[n, EulerPhi[#] / MultiplicativeOrder[k, #] &], 0], {k, 1, n}];
    T[m_, n_] := A132191[m, n]/m;
    Table[T[m - n + 1, n], {m, 1, 11}, {n, m, 1, -1}] // Flatten (* Jean-François Alcover, Jun 06 2017 *)
  • PARI
    a(n,x)=sum(k=1, n, if(gcd(k, n)==1, x^(sumdiv(n, d, eulerphi(d)/znorder(Mod(k, d)))-1), 0))/eulerphi(n);
    for(m=1, 6, for(n=1, 10, print1( a(n,m), ", ") ); print(); );

Formula

T(m, n) = A132191(m, n) / m.

A054627 Number of n-bead necklaces with 8 colors.

Original entry on oeis.org

1, 8, 36, 176, 1044, 6560, 43800, 299600, 2097684, 14913200, 107377488, 780903152, 5726645688, 42288908768, 314146329192, 2345624810432, 17592187093524, 132458812569728, 1000799924679192, 7585009898729264, 57646075284033552, 439208192231379680
Offset: 0

Views

Author

N. J. A. Sloane, Apr 16 2000

Keywords

Examples

			G.f. = 1 + 8*x + 36*x^2 + 176*x^3 + 1044*x^4 + 6560*x^5 + 43800*x^6 + ...
		

Crossrefs

Column 8 of A075195.
Column k=1 of A184294.
Cf. A054615.

Programs

  • Maple
    with(combstruct):A:=[N,{N=Cycle(Union(Z$8))},unlabeled]: seq(count(A,size=n),n=0..20); # Zerinvary Lajos, Dec 05 2007
  • Mathematica
    mx=40; CoefficientList[Series[1-Sum[EulerPhi[i] Log[1-8*x^i]/i, {i, 1, mx}], {x, 0, mx}], x] (* Herbert Kociemba, Nov 02 2016 *)
    k=8; Prepend[Table[DivisorSum[n, EulerPhi[#] k^(n/#) &]/n, {n, 1, 30}], 1] (* Robert A. Russell, Sep 21 2018 *)
  • PARI
    a(n)=if(n==0, 1, 1/n*sumdiv(n, d, eulerphi(d)*8^(n/d))); \\ Altug Alkan, Sep 21 2018

Formula

a(n) = (1/n)*Sum_{d|n} phi(d)*8^(n/d), n > 0.
G.f.: 1 - Sum_{n>=1} phi(n)*log(1 - 8*x^n)/n. - Herbert Kociemba, Nov 02 2016
a(0) = 1; a(n) = (1/n) * Sum_{k=1..n} 8^gcd(n,k). - Ilya Gutkovskiy, Apr 17 2021

Extensions

Edited by Christian G. Bower, Sep 07 2002
a(0) corrected by Herbert Kociemba, Nov 02 2016
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