A284855 -id:A284855 - cp's OEIS frontend

A285012 Irregular triangle read by rows: T(n,k) is the number of periodic palindromic structures of length n using exactly k different symbols, 1 <= k <= n/2 + 1.

1, 1, 1, 1, 1, 1, 3, 1, 1, 3, 1, 1, 6, 5, 1, 1, 7, 6, 1, 1, 13, 19, 7, 1, 1, 15, 25, 10, 1, 1, 25, 64, 43, 10, 1, 1, 31, 90, 65, 15, 1, 1, 50, 208, 220, 85, 13, 1, 1, 63, 301, 350, 140, 21, 1, 1, 99, 656, 1059, 618, 154, 17, 1, 1, 127, 966, 1701, 1050, 266, 28, 1

Offset: 1

Andrew Howroyd, Apr 07 2017

For example, aaabbb is not a (finite) palindrome but it is a periodic palindrome. Permuting the symbols will not change the structure.

Equivalently, the number of necklaces, up to permutation of the symbols, which when turned over are unchanged. When comparing with the turned over necklace a rotation is allowed but a permutation of the symbols is not.

			Triangle starts:
1
1   1
1   1
1   3    1
1   3    1
1   6    5     1
1   7    6     1
1  13   19     7     1
1  15   25    10     1
1  25   64    43    10     1
1  31   90    65    15     1
1  50  208   220    85    13    1
1  63  301   350   140    21    1
1  99  656  1059   618   154   17   1
1 127  966  1701  1050   266   28   1
1 197 2035  4803  4065  1488  258  21  1
1 255 3025  7770  6951  2646  462  36  1
1 391 6250 21105 24915 12857 3222 410 26 1
1 511 9330 34105 42525 22827 5880 750 45 1
...
Example for n=6, k=2:
Periodic symmetry means results are either in the form abccba or abcdcb.
There are 3 binary words in the form abccba that start with 0 and contain a 1 which are 001100, 010010, 011110. Of these, 011110 is equivalent to 001100 after rotation.
There are 7 binary words in the form abcdcb that start with 0 and contain a 1 which are 000100, 001010, 001110, 010001, 010101, 011011, 011111. Of these, 011111 is equivalent to 000100, 010001 is equivalent to 001010 and 011011 is equivalent to 010010 from the first set.
There are therefore a total of 7 + 3 - 4 = 6 equivalence classes so T(6,2) = 6.

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..2600

Columns 2..6 are A056508, A056509, A056510, A056511, A056512.
Partial row sums include A056503, A056504, A056505, A056506, A056507.
Row sums are A285013.
Cf. A284855, A284826, A284823, A284877, A285037, A304972.

\\ Ach is A304972, Prim is A285037.
Ach(n,k=n) = {my(M=matrix(n, k, n, k, n>=k)); for(n=3, n, for(k=2, k, M[n, k]=k*M[n-2, k] + M[n-2, k-1] + if(k>2, M[n-2, k-2]))); M}
Prim(n,k=n\2+1) = {my(A=Ach(n\2+1,k), S=matrix(n\2+1, k, n, k, stirling(n,k,2))); Mat(vectorv(n, n, sumdiv(n, d, moebius(d)*(S[(n/d+1)\2, ] + S[n/d\2+1, ] + if((n-d)%2, A[(n/d+1)\2, ] + A[n/d\2+1, ]))/if(d%2, 2, 1) )))}
T(n,k=n\2+1) = {my(A=Prim(n,k)); Mat(vectorv(n, n, sumdiv(n, d, A[d, ])))}
{ my(A=T(20)); for(n=1, matsize(A)[1], print(A[n,1..n\2+1])) } \\ Andrew Howroyd, Oct 02 2019

\\ column sequence using above code.
ColSeq(n, k=2) = { Vec(T(n,k)[,k]) } \\ Andrew Howroyd, Oct 02 2019

Column k is inverse Moebius transform of column k of A285037. - Andrew Howroyd, Oct 02 2019

A164090 a(n) = 2*a(n-2) for n > 2; a(1) = 2, a(2) = 3.

Original entry on oeis.org

2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 128, 192, 256, 384, 512, 768, 1024, 1536, 2048, 3072, 4096, 6144, 8192, 12288, 16384, 24576, 32768, 49152, 65536, 98304, 131072, 196608, 262144, 393216, 524288, 786432, 1048576, 1572864, 2097152, 3145728

Offset: 1

Klaus Brockhaus, Aug 09 2009

Interleaving of A000079 without initial 1 and A007283.
Agrees from a(2) onward with A145751 for all terms listed there (up to 65536). Apparently equal to 2, 3 followed by A090989. Equals 2 followed by A163978.
Binomial transform is A000129 without first two terms, second binomial transform is A020727, third binomial transform is A164033, fourth binomial transform is A164034, fifth binomial transform is A164035.
Number of achiral necklaces or bracelets with n beads using up to 2 colors. For n=5, the eight achiral necklaces or bracelets are AAAAA, AAAAB, AAABB, AABAB, AABBB, ABABB, ABBBB, and BBBBB. - Robert A. Russell, Sep 22 2018

Vincenzo Librandi, Table of n, a(n) for n = 1..2000
Index entries for linear recurrences with constant coefficients, signature (0,2).

Cf. A000079 (powers of 2), A007283 (3*2^n), A029744, A145751, A090989, A163978, A000129, A020727, A164033, A164034, A164035, A052955, A027383, A060482, A070875.

Second column of A284855.

[ n le 2 select n+1 else 2*Self(n-2): n in [1..42] ];

a[n_] := If[EvenQ[n], 3*2^(n/2 - 1), 2^((n + 1)/2)]; Array[a, 42] (* Jean-François Alcover, Oct 12 2017 *)
RecurrenceTable[{a[1]==2,a[2]==3,a[n]==2a[n-2]},a,{n,50}] (* or *) LinearRecurrence[{0,2},{2,3},50] (* Harvey P. Dale, Mar 01 2018 *)

a(n) = if(n%2,2,3) * 2^((n-1)\2); \\ Andrew Howroyd, Oct 07 2017

a(n) = A029744(n+1).
a(n) = A052955(n-1) + 1.
a(n) = A027383(n-2) + 2 for n > 1.
a(n) = A060482(n-1) + 3 for n > 3.
a(n) = A070875(n) - A070875(n-1).
a(n) = (7 - (-1)^n)*2^((1/4)*(2*n - 1 + (-1)^n))/4.
G.f.: x*(2+3*x)/(1-2*x^2).
a(n) = A063759(n-1), n>1. - R. J. Mathar, Aug 17 2009
Sum_{n>=1} 1/a(n) = 5/3. - Amiram Eldar, Mar 28 2022

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

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

Andrew Howroyd, Oct 10 2017

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

			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

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

Columns 2..6 are A059076, A278639, A278640, A278641, A278642.

Cf. A075195, A081720, A284855.

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 *)

\\ 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;

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)

A284856 Array read by antidiagonals: T(n,k) = number of aperiodic necklaces (Lyndon words) with n beads and k colors that are the same when turned over.

Original entry on oeis.org

1, 2, 0, 3, 1, 0, 4, 3, 2, 0, 5, 6, 6, 3, 0, 6, 10, 12, 12, 6, 0, 7, 15, 20, 30, 24, 7, 0, 8, 21, 30, 60, 60, 42, 14, 0, 9, 28, 42, 105, 120, 138, 78, 18, 0, 10, 36, 56, 168, 210, 340, 252, 144, 28, 0, 11, 45, 72, 252, 336, 705, 620, 600, 234, 39, 0

Offset: 1

Andrew Howroyd, Apr 04 2017

Number of primitive (period n) periodic palindromes of length n using a maximum of k different symbols.

			Table starts:
1  2   3    4    5     6     7      8      9     10 ...
0  1   3    6   10    15    21     28     36     45 ...
0  2   6   12   20    30    42     56     72     90 ...
0  3  12   30   60   105   168    252    360    495 ...
0  6  24   60  120   210   336    504    720    990 ...
0  7  42  138  340   705  1302   2212   3528   5355 ...
0 14  78  252  620  1290  2394   4088   6552   9990 ...
0 18 144  600 1800  4410  9408  18144  32400  54450 ...
0 28 234 1008 3100  7740 16758  32704  58968  99900 ...
0 39 456 2490 9240 26985 66864 146916 294480 548955 ...
...

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: A056493, A056494, A056495, A056496, A056497.

b[d_, k_] := If[EvenQ[d], (k^(d/2) + k^(d/2 + 1))/2, k^((d + 1)/2)];
a[n_, k_] := DivisorSum[n, MoebiusMu[n/#] b[#, k] &];
Table[a[n - k + 1, k], {n, 1, 11}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Jun 06 2017, translated from PARI *)

b(d,k) = if(d % 2 == 0, (k^(d/2) + k^(d/2+1))/2, k^((d+1)/2));
a(n,k) = sumdiv(n,d, moebius(n/d) * b(d,k));
for(n=1, 10, for(k=1, 10, print1( a(n,k),", ");); print(););

T(n, k) = Sum_{d | n} mu(n/d) * A284855(d, k).

A056486 a(n) = (9*2^n + (-2)^n)/4 for n>0.

Original entry on oeis.org

4, 10, 16, 40, 64, 160, 256, 640, 1024, 2560, 4096, 10240, 16384, 40960, 65536, 163840, 262144, 655360, 1048576, 2621440, 4194304, 10485760, 16777216, 41943040, 67108864, 167772160, 268435456, 671088640, 1073741824, 2684354560, 4294967296, 10737418240

Offset: 1

Marks R. Nester

Old name was: "Number of periodic palindromes using a maximum of four different symbols".

Number of necklaces with n beads and 4 colors that are the same when turned over and hence have reflection symmetry. - Herbert Kociemba, Nov 24 2016

			G.f. = 4*x + 10*x^2 + 16*x^3 + 40*x^4 + 64*x^5 + 160*x^6 + 256*x^7 + 640*x^8 + ...
For example, aaabbb is not a (finite) palindrome but it is a periodic palindrome.

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 (0,4).

Cf. A029744, A038754, A056450.

Column 4 of A284855.

[(9*2^n + (-2)^n)/4 : n in [1..50]]; // Wesley Ivan Hurt, Nov 24 2016

A056486:=n->(9*2^n + (-2)^n)/4: seq(A056486(n), n=1..50); # Wesley Ivan Hurt, Nov 24 2016

CoefficientList[Series[-1+(1+4*x+6*x^2)/(1-4*x^2),{x,0,30}],x] (* Herbert Kociemba, Nov 24 2016 *)
k=4; Table[(k^Floor[(n+1)/2] + k^Ceiling[(n+1)/2]) / 2, {n, 1, 30}] (* Robert A. Russell, Sep 21 2018 *)

a(n) = (9*2^n+(-2)^n)/4; \\ Altug Alkan, Sep 21 2018

[2^(n-2)*(9+(-1)^n) for n in range(1,51)] # G. C. Greubel, Mar 23 2024

a(n) = 4^((n+1)/2) for n odd, a(n) = 4^(n/2)*5/2 for n even.
From Colin Barker, Jul 08 2012: (Start)
a(n) = 4*a(n-2).
G.f.: 2*x*(2+5*x)/((1-2*x)*(1+2*x)). (End)
G.f.: -1 + (1+4*x+6*x^2)/(1-4*x^2). - Herbert Kociemba, Nov 24 2016
E.g.f.: 5*sinh(x)^2 + 2*sinh(2*x). - Ilya Gutkovskiy, Nov 24 2016
a(n) = ( 4^floor((n+1)/2) + 4^ceiling((n+1)/2) )/2. - Robert A. Russell, Sep 21 2018

Better name from Ralf Stephan, Jul 18 2013

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

Alford Arnold

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

			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

C. G. Bower, Transforms (2)

Columns 2-6 are A000029, A027671, A032275, A032276, and A056341.
Rows 2-7 are A000217, A000292, A002817, A060446, A027670, and A060532.
Cf. A000031.
Cf. A081720, A081721.
T(n,k) = (A075195(n,k) + A284855(n,k)) / 2.

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 *)

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

A056488 Number of periodic palindromes using a maximum of six different symbols.

Original entry on oeis.org

6, 21, 36, 126, 216, 756, 1296, 4536, 7776, 27216, 46656, 163296, 279936, 979776, 1679616, 5878656, 10077696, 35271936, 60466176, 211631616, 362797056, 1269789696, 2176782336, 7618738176, 13060694016, 45712429056, 78364164096, 274274574336, 470184984576

Offset: 1

Marks R. Nester

Also number of necklaces with n beads and 6 colors that are the same when turned over and hence have reflection symmetry. - Herbert Kociemba, Nov 24 2016

			G.f. = 6*x + 21*x^2 + 36*x^3 + 126*x^4 + 216*x^5 + 756*x^6 + 1296*x^7 + ...
For example, aaabbb is not a (finite) palindrome but it is a periodic palindrome.

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 (0,6).

Column 6 of A284855.

Cf. A029744, A038754, A056452.

[IsEven(n) select 6^(n div 2)*7/2 else 6^((n+1) div 2): n in [1..30]]; // Vincenzo Librandi, Sep 22 2018

LinearRecurrence[{0,6},{6,21},30] (* Harvey P. Dale, Feb 02 2015 *)
k = 6; Table[(k^Floor[(n + 1)/2] + k^Ceiling[(n + 1)/2]) / 2, {n, 30}] (* Robert A. Russell, Sep 21 2018 *)
If[EvenQ[#], 6^(# / 2) 7/2, 6^((# + 1) / 2)]&/@Range[30] (* Vincenzo Librandi, Sep 22 2018 *)

a(n) = if(n%2, 6^((n+1)/2), 7*6^(n/2)/2); \\ Altug Alkan, Sep 21 2018

a(n) = 6^((n+1)/2) for n odd, a(n) = 6^(n/2)*7/2 for n even.
From Colin Barker, Jul 08 2012: (Start)
a(n) = 6*a(n-2).
G.f.: 3*x*(2+7*x)/(1-6*x^2). (End)
a(n) = (k^floor((n+1)/2) + k^ceiling((n+1)/2)) / 2, where k = 6 is the number of possible colors. - Robert A. Russell, Sep 22 2018

More terms from Vincenzo Librandi, Sep 22 2018

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

Original entry on oeis.org

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

Offset: 0

Robert A. Russell, Dec 18 2018

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

Index entries for sequences related to bracelets

Cf. A075195 (oriented), A293496(chiral), A284855 (achiral).
Cf. A051137 (ascending antidiagonals).
Columns 0-6 are A000007, A000012, A000029, A027671, A032275, A032276, and A056341.
Rows 0-7 are A000012, A001477, A000217, A000292, A002817, A060446, A027670, and A060532.
Main diagonal gives A081721.

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

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

A056508 Number of periodic palindromic structures of length n using exactly two different symbols.

Original entry on oeis.org

0, 1, 1, 3, 3, 6, 7, 13, 15, 25, 31, 50, 63, 99, 127, 197, 255, 391, 511, 777, 1023, 1551, 2047, 3090, 4095, 6175, 8191, 12323, 16383, 24639, 32767, 49221, 65535, 98431, 131071, 196743, 262143, 393471, 524287, 786697, 1048575, 1573375, 2097151, 3146255, 4194303

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.

For odd n, a palindrome cannot be the complement of itself, so a(n) is given by A284855(n,2)/2 - 1. - Andrew Howroyd, Apr 08 2017

			From _Andrew Howroyd_, Apr 07 2017: (Start)
Example for n=6:
Periodic symmetry means results are either in the form abccba or abcdcb.
There are 3 binary words in the form abccba that start with 0 and contain a 1 which are 001100, 010010, 011110. Of these, 011110 is equivalent to 001100 after rotation.
There are 7 binary words in the form abcdcb that start with 0 and contain a 1 which are 000100, 001010, 001110, 010001, 010101, 011011, 011111. Of these, 011111 is equivalent to 000100, 010001 is equivalent to 001010 and 011011 is equivalent to 010010 from the first set.
There are therefore a total of 7 + 3 - 4 = 6 equivalence classes so a(6) = 6.
(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]

Giovanni Resta, Table of n, a(n) for n = 1..1000

Column 2 of A285012.

Cf. A052551.

(* 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_] := If[OddQ[n], b[n]/2, (1/2)*(b[n] + c[n/2])] - 1;
Array[a, 45] (* Jean-François Alcover, Jun 29 2018, after Andrew Howroyd *)

a(n) = A056503(n) - 1.

a(2n + 1) = 2^n - 1. - Andrew Howroyd, Apr 07 2017

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

cp's OEIS Frontend

A285012 Irregular triangle read by rows: T(n,k) is the number of periodic palindromic structures of length n using exactly k different symbols, 1 <= k <= n/2 + 1.

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A164090 a(n) = 2*a(n-2) for n > 2; a(1) = 2, a(2) = 3.

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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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A293496 Array read by antidiagonals: T(n,k) = number of chiral pairs of necklaces with n beads using a maximum of k colors.

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A284856 Array read by antidiagonals: T(n,k) = number of aperiodic necklaces (Lyndon words) with n beads and k colors that are the same when turned over.

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A056486 a(n) = (9*2^n + (-2)^n)/4 for n>0.

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