A208535 -id:A208535 - cp's OEIS frontend

A111492 Triangle read by rows: a(n,k) = (k-1)! * C(n,k).

1, 2, 1, 3, 3, 2, 4, 6, 8, 6, 5, 10, 20, 30, 24, 6, 15, 40, 90, 144, 120, 7, 21, 70, 210, 504, 840, 720, 8, 28, 112, 420, 1344, 3360, 5760, 5040, 9, 36, 168, 756, 3024, 10080, 25920, 45360, 40320, 10, 45, 240, 1260, 6048, 25200, 86400, 226800, 403200, 362880

Offset: 1

Ross La Haye, Nov 15 2005

For k > 1, a(n,k) = the number of permutations of the symmetric group S_n that are pure k-cycles.
Reverse signed array is A238363. For a relation to (Cauchy-Euler) derivatives of the Vandermonde determinant, see Chervov link. - Tom Copeland, Apr 10 2014
Dividing the k-th column of T by (k-1)! for each column generates A135278 (the f-vectors, or face-vectors for the n-simplices). Then ignoring the first column gives A104712, so T acting on the column vector (-0,d,-d^2/2!,d^3/3!,...) gives the Euler classes for hypersurfaces of degree d in CP^n. Cf. A104712 and Dugger link therein. - Tom Copeland, Apr 11 2014
With initial i,j,n=1, given the n X n Vandermonde matrix V_n(x_1,...,x_n) with elements a(i=row,j=column)=(x_j)^(i-1), its determinant |V_n|, and the column vector of n ones C=(1,1,...,1), the n-th row of the lower triangular matrix T is given by the column vector determined by (1/|V_n|) * V_n(:x_1*d/dx_1:,...,:x_n*d/dx_n:)|V_n| * C, where :x_j*d/dx_j:^n = (x_j)^n*(d/dx_j)^n. - Tom Copeland, May 20 2014
For some other combinatorial interpretations of the first three columns of T, see A208535 and the link to necklace polynomials therein. Because of the simple relation of the array to the Pascal triangle, it can easily be related to many other arrays, e.g., T(p,k)/(p*(k-1)!) with p prime gives the prime rows of A185158 and A051168 when the non-integers are rounded to 0. - Tom Copeland, Oct 23 2014

			a(3,3) = 2 because (3-1)!C(3,3) = 2.
1;
2 1;
3 3 2;
4 6 8 6;
5 10 20 30 24;
6 15 40 90 144 120;
7 21 70 210 504 840 720;
8 28 112 420 1344 3360 5760 5040;
9 36 168 756 3024 10080 25920 45360 40320;

A. Chervov, A sum involving derivatives of Vandermonde

/* As triangle: */ [[Factorial(k-1)*Binomial(n,k): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Oct 21 2014

Flatten[Table[(k - 1)!Binomial[n, k], {n, 10}, {k, n}]]

a(n, k) = (k-1)!C(n, k) = P(n, k)/k.
E.g.f. (by columns) = exp(x)((x^k)/k).
a(n, 1) = A000027(n);
a(n, 2) = A000217(n-1);
a(n, 3) = A007290(n);
a(n, 4) = A033487(n-3).
a(n, n) = A000142(n-1);
a(n, n-1) = A001048(n-1) for n > 1.
Sum[a(n, k), {k, 1, n}] = A002104(n);
Sum[a(n, k), {k, 2, n}] = A006231(n).
a(n,k) = sum(j=k..n-1, j!/(j-k)!) (cf. Chervov link). - Tom Copeland, Apr 10 2014
From Tom Copeland, Apr 28 2014: (Start)
E.g.f. by row: [(1+t)^n-1]/t.
E.g.f. of row e.g.f.s: {exp[(1+t)*x]-exp(x)}/t.
O.g.f. of row e.g.f.s: {1/[1-(1+t)*x] - 1/(1-x)}/t.
E.g.f. of row o.g.f.s: -exp(x) * log(1-t*x). (End)

A208536 Number of 5-bead necklaces of n colors not allowing reversal, with no adjacent beads having the same color.

Original entry on oeis.org

0, 0, 6, 48, 204, 624, 1554, 3360, 6552, 11808, 19998, 32208, 49764, 74256, 107562, 151872, 209712, 283968, 377910, 495216, 639996, 816816, 1030722, 1287264, 1592520, 1953120, 2376270, 2869776, 3442068, 4102224, 4859994, 5725824, 6710880

Offset: 1

R. H. Hardin, Feb 27 2012

This sequence would be better defined as a(n) = (n^5-n)/5 with offset 0, which is an integer by Fermat's little theorem. - N. J. A. Sloane, Nov 13 2023

			All solutions for n=3:
..1....1....1....1....1....1
..3....3....2....2....2....2
..1....2....1....3....3....1
..3....3....3....2....1....2
..2....2....2....3....3....3

R. H. Hardin, Table of n, a(n) for n = 1..210
Jack Jeffries, Differentiating by prime numbers, Notices Amer. Math. Soc., 70:11 (2023), 1772-1779.
Wikipedia, p-derivation.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Row 5 of A208535.
Also, row 5 (with different offset) of A074650. - Eric M. Schmidt, Dec 08 2017
Cf. A208537.

A208536[n_]:=((n-1)^5-(n-1))/5;Array[A208536,50] (* Paolo Xausa, Nov 14 2023 *)

Vec(6*x^3*(1 + x)^2 / (1 - x)^6 + O(x^40)) \\ Colin Barker, Nov 11 2017

Empirical: a(n) = (1/5)*n^5 - 1*n^4 + 2*n^3 - 2*n^2 + (4/5)*n.
Equivalently: a(n) = ((n-1)^5 - (n-1))/5. - M. F. Hasler, Mar 05 2016
Empirical formula confirmed by Petros Hadjicostas, Nov 05 2017 (see A208535).
a(n+2) = delta(-n) = -delta(n) for n >= 0, where delta is the p-derivation over the integers with respect to prime p = 5. - Danny Rorabaugh, Nov 10 2017
From Colin Barker, Nov 11 2017: (Start)
G.f.: 6*x^3*(1 + x)^2 / (1 - x)^6.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n>6.
(End)

A208537 Number of 7-bead necklaces of n colors not allowing reversal, with no adjacent beads having the same color.

Original entry on oeis.org

0, 0, 18, 312, 2340, 11160, 39990, 117648, 299592, 683280, 1428570, 2783880, 5118828, 8964072, 15059070, 24408480, 38347920, 58619808, 87460002, 127695960, 182857140, 257298360, 356336838, 486403632, 655210200, 871930800, 1147401450

Offset: 1

R. H. Hardin, Feb 27 2012

This sequence would be better defined as a(n) = (n^7-n)/7 with offset 0, which is an integer by Fermat's little theorem. - N. J. A. Sloane, Nov 13 2023
Row 7 of A208535.
Also, row 7 (with different offset) of A074650. - Eric M. Schmidt, Dec 08 2017

			All solutions for n=3:
..1...1...1...1...1...1...1...1...1...1...1...1...1...1...1...1...1...1
..2...2...2...3...2...2...2...2...2...3...3...3...2...2...2...2...2...2
..1...1...1...1...1...1...3...3...3...2...2...1...3...1...3...1...3...1
..2...3...2...3...3...3...2...1...2...3...1...3...2...2...1...3...1...2
..3...2...1...2...1...2...1...3...3...2...3...1...3...3...3...1...2...1
..1...3...3...3...2...1...3...1...2...3...2...3...1...2...2...3...3...2
..3...2...2...2...3...3...2...3...3...2...3...2...3...3...3...2...2...3

J. Jeffries, Differentiating by prime numbers, Notices Amer. Math. Soc., 70:11 (2023), 1772-1779.

R. H. Hardin, Table of n, a(n) for n = 1..210
Wikipedia, p-derivation.
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A208535.

A208537[n_]:=((n-1)^7-(n-1))/7;Array[A208537,50] (* Paolo Xausa, Nov 14 2023 *)

Vec(6*x^3*(3 + 28*x + 58*x^2 + 28*x^3 + 3*x^4) / (1 - x)^8 + O(x^40)) \\ Colin Barker, Nov 11 2017

Empirical: a(n) = (1/7)*n^7 - 1*n^6 + 3*n^5 - 5*n^4 + 5*n^3 - 3*n^2 + (6/7)*n.
Empirical formula confirmed by Petros Hadjicostas, Nov 05 2017 (see A208535).
a(n+2) = delta(-n) = -delta(n) for n >= 0, where delta is the p-derivation over the integers with respect to prime p = 7. - Danny Rorabaugh, Nov 10 2017
From Colin Barker, Nov 11 2017: (Start)
G.f.: 6*x^3*(3 + 28*x + 58*x^2 + 28*x^3 + 3*x^4) / (1 - x)^8.
a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8) for n>8.
(End)
a(n) = ((n-1)^7 - (n-1))/7. (inspired by Hassler's formula in A208536) - Eric M. Schmidt, Dec 08 2017

A208544 T(n,k) = Number of n-bead necklaces of k colors allowing reversal, with no adjacent beads having the same color.

Original entry on oeis.org

1, 2, 0, 3, 1, 0, 4, 3, 0, 0, 5, 6, 1, 1, 0, 6, 10, 4, 6, 0, 0, 7, 15, 10, 21, 3, 1, 0, 8, 21, 20, 55, 24, 13, 0, 0, 9, 28, 35, 120, 102, 92, 9, 1, 0, 10, 36, 56, 231, 312, 430, 156, 30, 0, 0, 11, 45, 84, 406, 777, 1505, 1170, 498, 29, 1, 0, 12, 55, 120, 666, 1680, 4291, 5580, 4435

Offset: 1

R. H. Hardin, Feb 27 2012

Table starts
.1.2..3...4....5.....6......7......8.......9......10......11.......12.......13
.0.1..3...6...10....15.....21.....28......36......45......55.......66.......78
.0.0..1...4...10....20.....35.....56......84.....120.....165......220......286
.0.1..6..21...55...120....231....406.....666....1035....1540.....2211.....3081
.0.0..3..24..102...312....777...1680....3276....5904....9999....16104....24882
.0.1.13..92..430..1505...4291..10528...23052...46185...86185...151756...254618
.0.0..9.156.1170..5580..19995..58824..149796..341640..714285..1391940..2559414
.0.1.30.498.4435.25395.107331.365260.1058058.2707245.6278140.13442286.26942565

			All solutions for n=7, k=3:
..1....1....1....1....1....1....1....1....1
..2....2....2....2....2....2....2....2....2
..3....3....1....1....3....1....3....1....3
..1....1....2....2....1....2....2....3....2
..2....3....3....3....3....1....3....1....3
..3....1....1....2....2....2....2....2....1
..2....3....3....3....3....3....3....3....3

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (first 264 terms from R. H. Hardin)

Cf. A081720, A208535, A106512.
Main diagonal is A208538.
Columns 3..7 are A208539, A208540, A208541, A208542, A208543.
Row 2 is A000217(n-1).
Row 3 is A000292(n-2).
Row 4 is A002817(n-1).
Row 5 is A164938(n-1).
Row 6 is A027670(n-1).

T[n_, k_] := If[n == 1, k, (DivisorSum[n, EulerPhi[n/#]*(k-1)^#&]/n + If[ OddQ[n], 1-k, k*(k-1)^(n/2)/2])/2]; Table[T[n-k+1, k], {n, 1, 12}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Oct 30 2017, after Andrew Howroyd *)

T(n, k) = if(n==1, k, (sumdiv(n, d, eulerphi(n/d)*(k-1)^d)/n + if(n%2, 1-k, k*(k-1)^(n/2)/2))/2);
for(n=1, 10, for(k=1, 10, print1(T(n, k), ", ")); print) \\ Andrew Howroyd, Oct 14 2017

T(2n+1,k) = A208535(2n+1,k)/2 for n > 0, T(2n,k) = (A208535(2n,k) + (k*(k-1)^n)/2)/2. - Andrew Howroyd, Mar 12 2017
Empirical for row n:
n=1: a(k) = k
n=2: a(k) = (1/2)*k^2 - (1/2)*k
n=3: a(k) = (1/6)*k^3 - (1/2)*k^2 + (1/3)*k
n=4: a(k) = (1/8)*k^4 - (1/4)*k^3 + (3/8)*k^2 - (1/4)*k
n=5: a(k) = (1/10)*k^5 - (1/2)*k^4 + k^3 - k^2 + (2/5)*k
n=6: a(k) = (1/12)*k^6 - (1/2)*k^5 + (3/2)*k^4 - (7/3)*k^3 + (23/12)*k^2 - (2/3)*k
n=7: a(k) = (1/14)*k^7 - (1/2)*k^6 + (3/2)*k^5 - (5/2)*k^4 + (5/2)*k^3 - (3/2)*k^2 + (3/7)*k

A327396 Triangle read by rows: T(n,k) is the number of n-bead necklace structures with beads of exactly k colors and no adjacent beads having the same color.

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 3, 5, 2, 1, 0, 0, 3, 10, 8, 2, 1, 0, 1, 7, 33, 40, 18, 3, 1, 0, 0, 11, 83, 157, 104, 28, 3, 1, 0, 1, 19, 237, 650, 615, 246, 46, 4, 1, 0, 0, 31, 640, 2522, 3318, 1857, 495, 65, 4, 1, 0, 1, 63, 1817, 9888, 17594, 13311, 4911, 944, 97, 5, 1

Offset: 1

Andrew Howroyd, Oct 04 2019

Permuting the colors does not change the necklace structure.

Equivalently, the number of k-block partitions of an n-set up to rotations where no block contains cyclically adjacent elements of the n-set.

			Triangle begins:
  0;
  0, 1;
  0, 0,  1;
  0, 1,  1,    1;
  0, 0,  1,    1,    1;
  0, 1,  3,    5,    2,     1;
  0, 0,  3,   10,    8,     2,     1;
  0, 1,  7,   33,   40,    18,     3,    1;
  0, 0, 11,   83,  157,   104,    28,    3,   1;
  0, 1, 19,  237,  650,   615,   246,   46,   4,  1;
  0, 0, 31,  640, 2522,  3318,  1857,  495,  65,  4, 1;
  0, 1, 63, 1817, 9888, 17594, 13311, 4911, 944, 97, 5, 1;
  ...

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

Columns k=3..4 are A327397, A328130.
Partial row sums include A306888, A309673.
Row sums are A328150.
Cf. A152175, A261139, A208535.

R(n) = {Mat(Col([Vecrev(p/y, n) | p<-Vec(intformal(sum(m=1, n, eulerphi(m) * subst(serlaplace((y-1)*exp(-x + O(x*x^(n\m))) - y + exp(-x + sumdiv(m, d, y^d*(exp(d*x + O(x*x^(n\m)))-1)/d)) ), x, x^m))/x), -n)]))}
{ my(A=R(12)); for(n=1, #A, print(A[n, 1..n])) } \\ Andrew Howroyd, Oct 09 2019

A106512 Array read by antidiagonals: a(n,k) = number of k-colorings of a circle of n nodes (n >= 1, k >= 1).

Original entry on oeis.org

0, 0, 0, 0, 2, 0, 0, 6, 0, 0, 0, 12, 6, 2, 0, 0, 20, 24, 18, 0, 0, 0, 30, 60, 84, 30, 2, 0, 0, 42, 120, 260, 240, 66, 0, 0, 0, 56, 210, 630, 1020, 732, 126, 2, 0, 0, 72, 336, 1302, 3120, 4100, 2184, 258, 0, 0, 0, 90, 504, 2408, 7770, 15630, 16380, 6564, 510, 2, 0, 0, 110

Offset: 1

Joshua Zucker, May 29 2005

Note that we keep one edge in the circular graph even when there's only one node (so there are 0 colorings of one node with k colors).

Number of closed walks of length n on the complete graph K_{k}. - Andrew Howroyd, Mar 12 2017

			From _Andrew Howroyd_, Mar 12 2017: (Start)
Table begins:
  0 0   0     0      0       0        0        0         0 ...
  0 2   6    12     20      30       42       56        72 ...
  0 0   6    24     60     120      210      336       504 ...
  0 2  18    84    260     630     1302     2408      4104 ...
  0 0  30   240   1020    3120     7770    16800     32760 ...
  0 2  66   732   4100   15630    46662   117656    262152 ...
  0 0 126  2184  16380   78120   279930   823536   2097144 ...
  0 2 258  6564  65540  390630  1679622  5764808  16777224 ...
  0 0 510 19680 262140 1953120 10077690 40353600 134217720 ...
(End)
a(4,3) = 18 because there are three choices for the first node's color (call it 1) and then two choices for the second node's color (call it 2) and then the remaining two nodes can be 12, 13, or 32. So in total there are 3*2*3 = 18 ways. a(3,4) = 4*3*2 = 24 because the three nodes must be three distinct colors.

Andrew Howroyd, Table of n, a(n) for n = 1..1274

Columns include A092297, A226493. Main diagonal is A118537.
Rows 2-7 are A002378, A007531, A091940, A061167, A131472, A133499.
Cf. A090860, A208535.

a(n, k) = (k-1)^n + (-1)^n * (k-1).

a(67) corrected by Andrew Howroyd, Mar 12 2017

A106368 Necklaces with n beads of 6 colors, no 2 adjacent beads the same color.

Original entry on oeis.org

6, 15, 40, 165, 624, 2635, 11160, 48915, 217040, 976887, 4438920, 20346485, 93900240, 435970995, 2034505656, 9536767665, 44878791360, 211927736135, 1003867701480, 4768372070757, 22706531350480, 108372083629275

Offset: 1

Christian G. Bower, Apr 29 2005

nonn

Andrew Howroyd, Table of n, a(n) for n = 1..200
M. Bernstein and N. J. A. Sloane, Some canonical sequences of integers, Vol. 226-228 (1995), 57-72.
Petros Hadjicostas, Proof of an explicit formula for Bower's CycleBG transform.
Index entries for sequences related to necklaces

Column 6 of A208535.

Cf. A000031.

a[1] = 6;
a[n_] := (1/n) Sum[EulerPhi[n/d]*(5*(-1)^d + 5^d), {d, Divisors[n]}];
Array[a, 30] (* Jean-François Alcover, Jul 02 2018, after Andrew Howroyd *)

a(n) = if(n==1, 6, sumdiv(n, d, eulerphi(n/d)*(5*(-1)^d + 5^d))/n); \\ Andrew Howroyd, Oct 14 2017

CycleBG transform of (6, 0, 0, 0, ...)
CycleBG transform T(A) = invMOEBIUS(invEULER(Carlitz(A)) + A(x^2) - A) + A.
Carlitz transform T(A(x)) has g.f. 1/(1-Sum_{k>0}(-1)^(k+1)*A(x^k)).
General formula for the CycleBG transform: T(A)(x) = A(x) - Sum_{k>=0} A(x^{2k+1}) + Sum_{k>=1} (phi(k)/k)*log(Carlitz(A)(x^k)). For a proof, see the links. (For this sequence, A(x) = 6*x.) - Petros Hadjicostas, Sep 16 2017
a(n) = (1/n) * Sum_{d | n} totient(n/d) * (5*(-1)^d + 5^d) for n > 1. - Andrew Howroyd, Mar 12 2017
G.f.: m*x - Sum_{k>=1} (phi(k)/k)*((m-1)*B(x^k)+C(x^k)) = m*x - (m-1)*x/(1-x^2) - Sum_{k>=1} (phi(k)/k)*C(x^k), where B(x) = log(1+x) and C(x) = log(1-(m-1)*x). Here m=6 is the number of colors. - Petros Hadjicostas, Sep 16 2017

A330618 Triangle read by rows: T(n,k) is the number of n-bead necklaces using exactly k colors with no adjacent beads having the same color.

Original entry on oeis.org

0, 0, 1, 0, 0, 2, 0, 1, 3, 6, 0, 0, 6, 24, 24, 0, 1, 11, 80, 180, 120, 0, 0, 18, 240, 960, 1440, 720, 0, 1, 33, 696, 4410, 11340, 12600, 5040, 0, 0, 58, 1960, 18760, 73920, 137760, 120960, 40320, 0, 1, 105, 5508, 76368, 433944, 1209600, 1753920, 1270080, 362880

Offset: 1

Andrew Howroyd, Dec 20 2019

In the case of n = 1, the single bead is considered to be cyclically adjacent to itself giving T(1,1) = 0. If compatibility with A208535 is wanted then T(1,1) should be 1.

			Triangle begins:
  0;
  0, 1;
  0, 0,  2;
  0, 1,  3,    6;
  0, 0,  6,   24,    24;
  0, 1, 11,   80,   180,   120;
  0, 0, 18,  240,   960,  1440,    720;
  0, 1, 33,  696,  4410, 11340,  12600,   5040;
  0, 0, 58, 1960, 18760, 73920, 137760, 120960, 40320;
  ...

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (first 50 rows)

Column 3 is A093367.
Row sums are A330620.
Cf. A087854, A208535, A327396, A330341.

\\ here U(n,k) is A208535(n,k) for n > 1.
U(n, k)={sumdiv(n, d, eulerphi(n/d)*(k-1)^d)/n - if(n%2, k-1)}
T(n,k)={sum(j=1, k, (-1)^(k-j)*binomial(k,j)*U(n,j))}

T(n,k) = Sum_{j=1..k} (-1)^(k-j)*binomial(k,j)*A208535(n,j) for n > 1.

T(n,n) = (n-1)! for n > 1.

A106365 Number of necklaces with n beads of 3 colors, no 2 adjacent beads the same color.

Original entry on oeis.org

3, 3, 2, 6, 6, 14, 18, 36, 58, 108, 186, 352, 630, 1182, 2190, 4116, 7710, 14602, 27594, 52488, 99878, 190746, 364722, 699252, 1342182, 2581428, 4971066, 9587580, 18512790, 35792568, 69273666, 134219796, 260301174, 505294128, 981706830

Offset: 1

Christian G. Bower, Apr 29 2005

nonn

Andrew Howroyd, Table of n, a(n) for n = 1..500
Petros Hadjicostas, Proof of an explicit formula for Bower's CycleBG transform
Index entries for sequences related to necklaces

Column 3 of A208535.

Cf. A000031, A001867.

a[n_] := If[n==1, 3, Sum[EulerPhi[n/d]*(2*(-1)^d+2^d), {d, Divisors[n]}]/n ];
Array[a, 35] (* Jean-François Alcover, Jul 06 2018, after Andrew Howroyd *)

a(n) = if(n==1, 3, sumdiv(n, d, eulerphi(n/d)*(2*(-1)^d + 2^d))/n); \\ Andrew Howroyd, Oct 14 2017

CycleBG transform of (3, 0, 0, 0, ...)
CycleBG transform T(A) = invMOEBIUS(invEULER(Carlitz(A)) + A(x^2) - A) + A.
Carlitz transform T(A(x)) has g.f. 1/(1-sum(k>0, (-1)^(k+1)*A(x^k))).
a(n) = (1/n)*sum_{d divides n} phi(n/d)*A092297(d) (n>1). - Seiichi Azuma, Oct 25 2014
a(n) = -1+(-1)^n+A000031(n) (n>1). - Seiichi Azuma, Oct 25 2014 [Corrected by Petros Hadjicostas, Feb 16 2018.]
From Petros Hadjicostas, Feb 16 2018: (Start)
General formula for the CycleBG transform: T(A)(x) = A(x) - Sum_{k>=0} A(x^(2k+1)) + Sum_{k>=1} (phi(k)/k)*log(Carlitz(A)(x^k)). For a proof, see the links above. (For this sequence, A(x) = 3*x.)
G.f.: Sum_{n>=1} a(n)*x^n = 3*x - 2*x/(1-x^2) - Sum_{n>=1} (phi(n)/n)*log(1-2*x^n) = 3*x - Sum_{n>=1} (phi(n)/n)*(2*log(1+x^n) + log(1-2*x^n)).
(End)

A106366 Number of necklaces with n beads of 4 colors, no 2 adjacent beads the same color.

Original entry on oeis.org

4, 6, 8, 24, 48, 130, 312, 834, 2192, 5934, 16104, 44368, 122640, 341802, 956632, 2690844, 7596480, 21524542, 61171656, 174342216, 498112272, 1426419858, 4093181688, 11767920118, 33891544416, 97764131646, 282429537944

Offset: 1

Christian G. Bower, Apr 29 2005

nonn

Andrew Howroyd, Table of n, a(n) for n = 1..200
Index entries for sequences related to necklaces

Column 4 of A208535.

Cf. A000031, A001868, A226493.

a[n_] := If[n==1, 4, Sum[EulerPhi[n/d]*(3*(-1)^d+3^d), {d, Divisors[n]}]/n ];
Array[a, 35] (* Jean-François Alcover, Jul 06 2018, after Andrew Howroyd *)

a(n) = if(n==1, 4, sumdiv(n, d, eulerphi(n/d)*(3*(-1)^d + 3^d))/n); \\ Andrew Howroyd, Oct 14 2017

CycleBG transform of (4, 0, 0, 0, ...)
CycleBG transform T(A) = invMOEBIUS(invEULER(Carlitz(A)) + A(x^2) - A) + A.
Carlitz transform T(A(x)) has g.f. 1/(1-Sum_{k>0} (-1)^(k+1)*A(x^k)).
a(n) = (1/n) * Sum_{d | n} totient(n/d) * (3*(-1)^d + 3^d) for n > 1. - Andrew Howroyd, Mar 12 2017

cp's OEIS Frontend

A111492 Triangle read by rows: a(n,k) = (k-1)! * C(n,k).

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A208536 Number of 5-bead necklaces of n colors not allowing reversal, with no adjacent beads having the same color.

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A208537 Number of 7-bead necklaces of n colors not allowing reversal, with no adjacent beads having the same color.

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A208544 T(n,k) = Number of n-bead necklaces of k colors allowing reversal, with no adjacent beads having the same color.

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A327396 Triangle read by rows: T(n,k) is the number of n-bead necklace structures with beads of exactly k colors and no adjacent beads having the same color.

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A106512 Array read by antidiagonals: a(n,k) = number of k-colorings of a circle of n nodes (n >= 1, k >= 1).

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A106368 Necklaces with n beads of 6 colors, no 2 adjacent beads the same color.

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A330618 Triangle read by rows: T(n,k) is the number of n-bead necklaces using exactly k colors with no adjacent beads having the same color.