A056391 -id:A056391 - cp's OEIS frontend

A049297 Number of nonisomorphic circulant digraphs (i.e., Cayley digraphs for the cyclic group) of order n.

1, 2, 3, 6, 6, 20, 14, 46, 51, 140, 108, 624, 352, 1400, 2172, 4262, 4116, 22040, 14602, 68016, 88376, 209936, 190746, 1062592, 839094, 2797000, 3728891, 11276704, 9587580, 67195520, 35792568

Offset: 1

Valery A. Liskovets

Terms may be computed by filtering potentially isomorphic graphs of A056391 through nauty. Terms computed in this way for a(25), a(27) agree with theoretical calculations by others. - Andrew Howroyd, Apr 23 2017

V. A. Liskovets, Some identities for enumerators of circulant graphs, arXiv:math/0104131 [math.CO], 2001.
V. A. Liskovets and R. Poeschel, On the enumeration of circulant graphs of prime-power and squarefree orders.
R. Poeschel, Publications.
V. Gatt, On the Enumeration of Circulant Graphs of Prime-Power Order: the case of p^3, arXiv:1703.06038 [math.CO], 2017.
Victoria Gatt, Mikhail Klin, Josef Lauri, Valery Liskovets, From Schur Rings to Constructive and Analytical Enumeration of Circulant Graphs with Prime-Cubed Number of Vertices, in Isomorphisms, Symmetry and Computations in Algebraic Graph Theory, (Pilsen, Czechia, WAGT 2016) Vol. 305, Springer, Cham, 37-65.
Brendan McKay, Nauty home page.

Cf. A049287-A049289, A056391, A038777.

LoadPackage("grape");
CirculantDigraphCount:= function(n) local g; # slow for n >= 10
g:=Graph( Group(()), [1..n], OnPoints, function(x,y) return (y-x) mod n = 1; end,false);
return Length(GraphIsomorphismClassRepresentatives(List(Combinations([1..n]), s->DistanceGraph(g,s))));
end; # Andrew Howroyd, Apr 23 2017

There is an easy formula for prime orders. Formulae are also known for squarefree and prime-squared orders.
From Andrew Howroyd, Apr 23 2017: (Start)
a(n) <= A056391(n).
a(n) = A056391(n) for squarefree n.
a(A000040(n)^2) = A038777(n).
(End)

Further values for (twice) squarefree and (twice) prime-squared orders can be found in the Liskovets reference.
a(14) corrected by Andrew Howroyd, Apr 23 2017
a(16)-a(31) from Andrew Howroyd, Apr 23 2017

A056487 a(n) = 5^(n/2) for n even, a(n) = 3*5^((n-1)/2) for n odd.

Original entry on oeis.org

1, 3, 5, 15, 25, 75, 125, 375, 625, 1875, 3125, 9375, 15625, 46875, 78125, 234375, 390625, 1171875, 1953125, 5859375, 9765625, 29296875, 48828125, 146484375, 244140625, 732421875, 1220703125, 3662109375, 6103515625, 18310546875, 30517578125, 91552734375

Offset: 0

Marks R. Nester

Apparently identical to A111386! Is this a theorem? - Klaus Brockhaus, Jul 21 2009

For n > 1, number of necklaces with n-1 beads and 5 colors that are the same when turned over and hence have reflection symmetry. - Herbert Kociemba, Nov 24 2016

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

Index entries for linear recurrences with constant coefficients, signature (0,5).

Column 5 of A284855.

Cf. A029744, A038754, A056391, A056451, A087205, A087206, A111386.

[n le 2 select 2*n-1 else 5*Self(n-2): n in [1..28]]; // Bruno Berselli, Mar 24 2011

A056487:=n->3^((1-(-1)^n)/2)*5^((2*n+(-1)^n-1)/4): seq(A056487(n), n=0..40); # Wesley Ivan Hurt, Nov 24 2016

Table[3^((1 - (-1)^n)/2)*5^((2*n + (-1)^n - 1)/4), {n, 0, 30}] (* Wesley Ivan Hurt, Nov 24 2016 *)
CoefficientList[Series[(1 + 3 x)/(1 - 5 x^2), {x, 0, 31}], x] (* Michael De Vlieger, Nov 24 2016 *)
LinearRecurrence[{0, 5}, {1, 3}, 35] (* Vincenzo Librandi, Nov 25 2016 *)
k=5; Table[(k^Floor[(n+1)/2] + k^Ceiling[(n+1)/2]) / 2, {n, -1, 30}] (* Robert A. Russell, Sep 21 2018 *)

a(n)=if(n%2,3,1)*5^(n\2) \\ Charles R Greathouse IV, Oct 07 2015

def A056487(n): return 5**(n>>1)*(3 if n&1 else 1) # Chai Wah Wu, Oct 27 2022

a(n+2) = 5*a(n), a(0)=1, a(2)=3.
Binomial transform of A087205. Binomial transform is A087206. - Paul Barry, Aug 25 2003
G.f.: (1+3*x)/(1-5*x^2); a(n) = 5^(n/2)(1/2 + 3*sqrt(5)/10 + (1/2 - 3*sqrt(5)/10)(-1)^n). - Paul Barry, Mar 19 2004
2nd inverse binomial transform of Fibonacci(3n+2). - Paul Barry, Apr 16 2004
a(n+3) = a(n+2)*a(n+1)/a(n). - Reinhard Zumkeller, Mar 04 2011
a(n) = 3^((1 - (-1)^n)/2) * 5^((2*n + (-1)^n-1)/4). - Bruno Berselli, Mar 24 2011
a(n+1) = (k^floor((n+1)/2) + k^ceiling((n+1)/2)) / 2, where k=5 is the number of possible colors. - Robert A. Russell, Sep 22 2018
E.g.f.: cosh(sqrt(5)*x) + 3*sinh(sqrt(5)*x)/sqrt(5). - Stefano Spezia, Jun 06 2023

Changed one 'even' to 'odd' in the definition. - R. J. Mathar, Oct 06 2010

A056292 Number of n-bead necklace structures using a maximum of four different colored beads.

Original entry on oeis.org

1, 2, 3, 7, 11, 39, 103, 367, 1235, 4439, 15935, 58509, 215251, 799697, 2983217, 11187567, 42109451, 159082753, 602809327, 2290684251, 8726308317, 33318661277, 127479700199, 488672302909, 1876500180291, 7217308815887, 27799998949873, 107228568948547

Offset: 1

Marks R. Nester

nonn

Turning over the necklace is not allowed. Colors may be permuted without changing the necklace structure.

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

E. N. Gilbert and J. Riordan, Symmetry types of periodic sequences, Illinois J. Math., 5 (1961), 657-665.
Marko Riedel, Necklaces with swappable colors by Power Group Enumeration
Marko Riedel, Maple code for any necklace size, any number of swappable colors, by Power Group Enumeration.
N. J. A. Sloane, Maple code for this and related sequences

Cf. A000013, A001868.

Adn[d_, n_] := Module[{ c, t1, t2}, t2 = 0; For[c = 1, c <= d, c++, If[Mod[d, c] == 0 , t2 = t2 + (x^c/c)*(E^(c*z) - 1)]]; t1 = E^t2; t1 = Series[t1, {z, 0, n+1}]; Coefficient[t1, z, n]*n!]; Pn[n_] := Module[{ d, e, t1}, t1 = 0; For[d = 1, d <= n, d++, If[Mod[n, d] == 0, t1 = t1 + EulerPhi[d]*Adn[d, n/d]/n]]; t1/(1 - x)]; Pnq[n_, q_] := Module[{t1}, t1 = Series[Pn[n], {x, 0, q+1}] ; Coefficient[t1, x, q]]; a[n_] := Pnq[n, 4]; Table[Print[an = a[n]]; an, {n, 1, 25}] (* Jean-François Alcover, Oct 04 2013, after N. J. A. Sloane's Maple code *)
(* This program uses Gilbert and Riordan's recurrence formula, which they recommend for calculations: *)
Adn[d_, n_] := Adn[d, n] = If[1==n, DivisorSum[d, x^# &],
  Expand[Adn[d, 1] Adn[d, n-1] + D[Adn[d, n-1], x] x]];
Table[SeriesCoefficient[DivisorSum[n, EulerPhi[#] Adn[#, n/#] &]
/(n (1 - x)), {x, 0, 4}], {n, 1, 40}] (* Robert A. Russell, Feb 24 2018 *)
From Robert A. Russell, May 29 2018: (Start)
Table[(1/n) DivisorSum[n, EulerPhi[#] Which[Divisible[#,12], 4 StirlingS2[n/#+3,4] - 24 StirlingS2[n/#+2,4] + 44 StirlingS2[n/#+1,4] - 24 StirlingS2[n/#,4], Divisible[#,6], 3 StirlingS2[n/#+3,4] - 18 StirlingS2[n/#+2,4] + 33 StirlingS2[n/#+1,4] - 18 StirlingS2[n/#,4], Divisible[#,4], 3 StirlingS2[n/#+3,4] - 19 StirlingS2[n/#+2,4] + 38 StirlingS2[n/#+1,4] - 24 StirlingS2[n/#,4], Divisible[#,3], 2 StirlingS2[n/#+3,4] - 13 StirlingS2[n/#+2,4] + 26 StirlingS2[n/#+1,4] - 15 StirlingS2[n/#,4], Divisible[#,2], 2 StirlingS2[n/#+3,4] - 13 StirlingS2[n/#+2,4] + 27 StirlingS2[n/#+1,4] - 18 StirlingS2[n/#,4], True, StirlingS2[n/#+3,4] - 8 StirlingS2[n/#+2,4] + 20 StirlingS2[n/#+1,4] - 15 StirlingS2[n/#,4]] &],{n, 1, 40}]
mx = 40; Drop[CoefficientList[Series[1 - Sum[(EulerPhi[d] / d) Which[
  Divisible[d, 12], Log[1 - 4x^d], Divisible[d, 6],
  3 Log[1 - 4x^d] / 4, Divisible[d, 4] ,
  (2 Log[1 - 4x^d] + Log[1 - x^d]) / 3, Divisible[d, 3],
  (3 Log[1 - 4x^d] + 2 Log[1 - 2x^d]) / 8,
  Divisible[d, 2], (5 Log[1 - 4x^d] + 4 Log[1 - x^d]) / 12,
  True, (Log[1 - 4x^d] + 6 Log[1 - 2x^d] + 8 Log[1 - x^d]) / 24], {d, 1, mx}], {x, 0, mx}], x], 1]
(End)

Use de Bruijn's generalization of Polya's enumeration theorem as discussed in reference.
From Robert A. Russell, May 29 2018: (Start)
a(n) = (1/n) * Sum_{d|n} phi(d) * ([d==0 mod 12] * (4*S2(n/d+3, 4) - 24*S2(n/d+2, 4) + 44*S2(n/d+1, 4) - 24*S2(n/d, 4)) + [d==6 mod 12] * (3*S2(n/d+3, 4) - 18*S2(n/d+2, 4) + 33*S2(n/d+1, 4) - 18*S2(n/d, 4)) + [d==4 mod 12 | d==8 mod 12] * (3*S2(n/d+3, 4) - 19*S2(n/d+2, 4) + 38*S2(n/d+1, 4) - 24*S2(n/d, 4)) + [d==3 mod 12 | d=9 mod 12] * (2*S2(n/d+3, 4) - 13*S2(n/d+2, 4) + 26*S2(n/d+1, 4) - 15*S2(n/d, 4)) + [d==2 mod 12 | d=10 mod 12] * (2*S2(n/d+3, 4) - 13*S2(n/d+2, 4) + 27*S2[n/d+1,4) - 18*S2(n/d, 4)) + [d mod 12 in {1,5,7,11}] * (S2(n/d+3, 4) - 8*S2(n/d+2, 4) + 20*S2(n/d+1, 4) - 15*S2(n/d, 4))), where S2(n, k) is the Stirling subset number, A008277.
G.f.: 1 - Sum_{d>0} (phi(d) / d) * ([d==0 mod 12] * log(1-4x^d) + [d==6 mod 12] * 3*log(1-4x^d) / 4 + [d==4 mod 12 | d==8 mod 12] * (2*log(1-4x^d) + log(1-x^d)) / 3 + [d==3 mod 12 | d=9 mod 12] * (3*log(1-4x^d) + 2*log(1-2x^d)) / 8 + [d==2 mod 12 | d=10 mod 12] * (5*log(1-4x^d) + 4*log(1-x^d)) / 12 + [d mod 12 in {1,5,7,11}] * (log(1-4x^d) + 6*log(1-2x^d) + 8*log(1-x^d)) / 24).
(End)

A056353 Number of bracelet structures using a maximum of three different colored beads.

Original entry on oeis.org

1, 1, 2, 3, 6, 9, 22, 40, 100, 225, 582, 1464, 3960, 10585, 29252, 80819, 226530, 636321, 1800562, 5107480, 14548946, 41538916, 118929384, 341187048, 980842804, 2824561089, 8147557742, 23536592235, 68087343148, 197216119545, 571924754778, 1660419530056, 4825588205920

Offset: 0

Marks R. Nester

nonn

Turning over will not create a new bracelet. Permuting the colors of the beads will not change the structure.

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

Andrew Howroyd, Table of n, a(n) for n = 0..200
R. M. Thompson and R. T. Downs, Systematic generation of all nonequivalent closest packed stacking sequences of length N using group theory, Acta Cryst. B57 (2001), 766-771; B58 (2002), 153.

Cf. A002076, A000011, A027671, A114438, A152176.

Use de Bruijn's generalization of Polya's enumeration theorem as discussed in reference.

a(n) = Sum_{k=1..3} A152176(n, k) for n > 0. - Andrew Howroyd, Oct 25 2019

a(0)=1 prepended and terms a(28) and beyond from Andrew Howroyd, Oct 25 2019

A056451 Number of palindromes using a maximum of five different symbols.

Original entry on oeis.org

1, 5, 5, 25, 25, 125, 125, 625, 625, 3125, 3125, 15625, 15625, 78125, 78125, 390625, 390625, 1953125, 1953125, 9765625, 9765625, 48828125, 48828125, 244140625, 244140625, 1220703125, 1220703125, 6103515625, 6103515625, 30517578125, 30517578125, 152587890625, 152587890625

Offset: 0

Marks R. Nester

Number of achiral rows of n colors using up to five colors. For a(3) = 25, the rows are AAA, ABA, ACA, ADA, AEA, BAB, BBB, BCB, BDB, BEB, CAC, CBC, CCC, CDC, CEC, DAD, DBD, DCD, DDD, DED, EAE, EBE, ECE, EDE, and EEE. - Robert A. Russell, Nov 09 2018

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

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (0,5).

Column k=5 of A321391.
Cf. A000007, A016116, A056391, A056453, A056454, A056455, A056456, A057427,
Cf. A000351 (oriented), A032122 (unoriented), A032088(n>1) (chiral).

[5^Floor((n+1)/2): n in [0..40]]; // Vincenzo Librandi, Aug 16 2011

LinearRecurrence[{0,5},{1,5},30] (* or *) Riffle[5^Range[0, 20], 5^Range[20]] (* Harvey P. Dale, Jul 28 2018 *)
Table[5^Ceiling[n/2], {n,0,40}] (* Robert A. Russell, Nov 07 2018 *)

vector(40, n, n--; 5^floor((n+1)/2)) \\ G. C. Greubel, Nov 07 2018

a(n) = 5^floor((n+1)/2).
a(n) = 5*a(n-2). - Colin Barker, May 06 2012
G.f.: (1+5*x) / (1-5*x^2). - Colin Barker, May 06 2012 [Adapted to offset 0 by Robert A. Russell, Nov 07 2018]
a(n) = C(5,0)*A000007(n) + C(5,1)*A057427(n) + C(5,2)*A056453(n) + C(5,3)*A056454(n) + C(5,4)*A056455(n) + C(5,5)*A056456(n). - Robert A. Russell, Nov 08 2018
E.g.f.: cosh(sqrt(5)*x) + sqrt(5)*sinh(sqrt(5)*x). - Stefano Spezia, Jun 06 2023

a(0)=1 prepended by Robert A. Russell, Nov 07 2018

A056454 Number of palindromes of length n using exactly three different symbols.

Original entry on oeis.org

0, 0, 0, 0, 6, 6, 36, 36, 150, 150, 540, 540, 1806, 1806, 5796, 5796, 18150, 18150, 55980, 55980, 171006, 171006, 519156, 519156, 1569750, 1569750, 4733820, 4733820, 14250606, 14250606, 42850116, 42850116, 128746950, 128746950, 386634060, 386634060, 1160688606

Offset: 1

Marks R. Nester

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

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,5,-5,-6,6).

Cf. A056449, A001117.

[StirlingSecond((n+1) div 2, 3)*Factorial(3): n in [1..40]]; // Vincenzo Librandi, Sep 26 2018

A056454:= n-> 3!*Stirling2(floor((n+1)/2),3); # (C. Ronaldo)

LinearRecurrence[{1,5,-5,-6,6},{0,0,0,0,6},40] (* Harvey P. Dale, Sep 02 2016 *)
k=3; Table[k! StirlingS2[Ceiling[n/2],k],{n,1,30}] (* Robert A. Russell, Sep 25 2018 *)

a(n) = 3!*stirling((n+1)\2, 3, 2); \\ Altug Alkan, Sep 25 2018

a(n) = 3! * Stirling2( [(n+1)/2], 3).
G.f.: 6*x^5/((1-x)*(1-2*x^2)*(1-3*x^2)). - Colin Barker, May 05 2012
G.f.: k!(x^(2k-1)+x^(2k))/Product_{i=1..k}(1-i*x^2), where k=3 is the number of symbols. - Robert A. Russell, Sep 25 2018

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

Andrew Howroyd, Apr 04 2017

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)

			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

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]

Andrew Howroyd, Table of n, a(n) for n = 1..1275
Christian G. Bower, Transforms (2).
Petros Hadjicostas, Generalized colored circular palindromic compositions, Moscow Journal of Combinatorics and Number Theory, 9(2) (2020), 173-186.
D. M. Y. Sommerville, On certain periodic properties of cyclic compositions of numbers, Proc. London Math. Soc. S2-7(1) (1909), 263-313.

Columns 2-6 are A164090, A182751(n-1), A056486, A056487(n-1), A056488.

Cf. A037306, A081720, A075195, A119963, A284856, A277504.

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 *)

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(););

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)

A288620 Triangle read by rows: T(n,k) = number of step shifted (decimated) sequence structures of length n using exactly k different symbols.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 5, 4, 1, 1, 5, 8, 3, 1, 1, 19, 50, 37, 9, 1, 1, 13, 54, 63, 26, 4, 1, 1, 47, 284, 479, 299, 83, 11, 1, 1, 51, 525, 1316, 1183, 454, 82, 8, 1, 1, 139, 2370, 8597, 10701, 5761, 1492, 196, 13, 1, 1, 107, 2872, 14619, 24736, 17998, 6429, 1198, 119, 6, 1

Offset: 1

Andrew Howroyd, Jun 11 2017

See A056371 for an explanation of step shifts. Permuting the symbols will not change the structure.

			Triangle begins
1;
1,  1;
1,  2,   1;
1,  5,   4,    1;
1,  5,   8,    3,    1;
1, 19,  50,   37,    9,   1;
1, 13,  54,   63,   26,   4,  1;
1, 47, 284,  479,  299,  83, 11, 1;
1, 51, 525, 1316, 1183, 454, 82, 8, 1;
...

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

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

Columns 2-6 are A056396, A056397, A056398, A056399, A056400.
Row sums are A288621.
Partial row sums include A056391, A056392, A056393, A056394, A056395.
Cf. A056371, A288627, A285522, A285548, A132191.

\\ see A056391 for Polya enumeration functions
T(n,k) = NonequivalentStructsExactly(StepShiftPerms(n), k); \\ Andrew Howroyd, Oct 14 2017

A288627 Triangle read by rows: T(n,k) = number of step cyclic shifted sequence structures of length n using exactly k different symbols.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 3, 2, 1, 1, 2, 3, 1, 1, 1, 7, 14, 11, 3, 1, 1, 4, 11, 13, 6, 1, 1, 1, 13, 52, 83, 52, 18, 3, 1, 1, 10, 72, 162, 148, 59, 13, 2, 1, 1, 25, 274, 930, 1140, 630, 171, 28, 3, 1, 1, 14, 281, 1369, 2306, 1681, 612, 118, 14, 1, 1

Offset: 1

Andrew Howroyd, Jun 11 2017

See A056371 for an explanation of step shifts. Under step cyclic shifts, abcde, bdace, bcdea, cdeab and daceb etc. are equivalent. Permuting the symbols will not change the structure.

			Triangle begins
1;
1,  1;
1,  1,   1;
1,  3,   2,   1;
1,  2,   3,   1,    1;
1,  7,  14,  11,    3,   1;
1,  4,  11,  13,    6,   1,   1;
1, 13,  52,  83,   52,  18,   3,  1;
1, 10,  72, 162,  148,  59,  13,  2, 1;
1, 25, 274, 930, 1140, 630, 171, 28, 3, 1;
...

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

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

Columns 2-6 are A056434, A056435, A056436, A056437, A056438.
Row sums are A288628.
Partial row sums include A056429, A056430, A056431, A056432, A056433.
Cf. A056391, A056371, A288620, A285522, A285548, A132191.

\\ see A056391 for Polya enumeration functions
T(n,k) = NonequivalentStructsExactly(CyclicStepShiftPerms(n), k); \\ Andrew Howroyd, Oct 14 2017

A056503 Number of periodic palindromic structures of length n using a maximum of two different symbols.

Original entry on oeis.org

1, 2, 2, 4, 4, 7, 8, 14, 16, 26, 32, 51, 64, 100, 128, 198, 256, 392, 512, 778, 1024, 1552, 2048, 3091, 4096, 6176, 8192, 12324, 16384, 24640, 32768, 49222, 65536, 98432, 131072, 196744, 262144, 393472, 524288, 786698, 1048576, 1573376, 2097152, 3146256, 4194304

Offset: 1

Marks R. Nester

nonn

For example, aaabbb is not a (finite) palindrome but it is a periodic palindrome. Permuting the symbols will not change the structure.
A periodic palindrome is just a necklace that is equivalent to its reverse. The number of binary periodic palindromes of length n is given by A164090(n). A binary periodic palindrome can only be equivalent to its complement when there are an equal number of 0's and 1's. - Andrew Howroyd, Sep 29 2017
Number of cyclic compositions (necklaces of positive integers) summing to n that can be rotated to form a palindrome. - Gus Wiseman, Sep 16 2018

			From _Gus Wiseman_, Sep 16 2018: (Start)
The sequence of palindromic cyclic compositions begins:
  (1)  (2)   (3)    (4)     (5)      (6)       (7)
       (11)  (111)  (22)    (113)    (33)      (115)
                    (112)   (122)    (114)     (133)
                    (1111)  (11111)  (222)     (223)
                                     (1122)    (11113)
                                     (11112)   (11212)
                                     (111111)  (11122)
                                               (1111111)
(End)

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

Andrew Howroyd, Table of n, a(n) for n = 1..200

Row sums of A179181.
Cf. A016116, A045674, A056508, A164090, A285012.
Cf. A000740, A000837, A008965, A025065, A059966, A242414, A296302, A317085, A317086, A317087, A318731.

(* b = A164090, c = A045674 *)
b[n_] := (1/4)*(7 - (-1)^n)*2^((1/4)*(2*n + (-1)^n - 1));
c[0] = 1; c[n_] := c[n] = If[EvenQ[n], 2^(n/2-1) + c[n/2], 2^((n-1)/2)];
a[n_?OddQ] := b[n]/2; a[n_?EvenQ] := (1/2)*(b[n] + c[n/2]);
Array[a, 45] (* Jean-François Alcover, Oct 08 2017, after Andrew Howroyd *)
Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],Function[q,And[Array[OrderedQ[{q,RotateRight[q,#]}]&,Length[q]-1,1,And],Array[SameQ[RotateRight[q,#],Reverse[RotateRight[q,#]]]&,Length[q],1,Or]]]]],{n,15}] (* Gus Wiseman, Sep 16 2018 *)

a(2n+1) = A164090(2n+1)/2 = 2^n, a(2n) = (A164090(2n) + A045674(n))/2. - Andrew Howroyd, Sep 29 2017

a(17)-a(45) from Andrew Howroyd, Apr 07 2017

cp's OEIS Frontend

A049297 Number of nonisomorphic circulant digraphs (i.e., Cayley digraphs for the cyclic group) of order n.

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A056487 a(n) = 5^(n/2) for n even, a(n) = 3*5^((n-1)/2) for n odd.

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A056292 Number of n-bead necklace structures using a maximum of four different colored beads.

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A056353 Number of bracelet structures using a maximum of three different colored beads.

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A056451 Number of palindromes using a maximum of five different symbols.

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A056454 Number of palindromes of length n using exactly three different symbols.

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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.

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