A032121 -id:A032121 - cp's OEIS frontend

A277504 Array read by descending antidiagonals: T(n,k) is the number of unoriented strings with n beads of k or fewer colors.

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 4, 6, 6, 1, 0, 1, 5, 10, 18, 10, 1, 0, 1, 6, 15, 40, 45, 20, 1, 0, 1, 7, 21, 75, 136, 135, 36, 1, 0, 1, 8, 28, 126, 325, 544, 378, 72, 1, 0, 1, 9, 36, 196, 666, 1625, 2080, 1134, 136, 1, 0, 1, 10, 45, 288, 1225, 3996, 7875, 8320, 3321, 272, 1, 0

Offset: 0

Jean-François Alcover, Oct 18 2016

From Petros Hadjicostas, Jul 07 2018: (Start)
Column k of this array is the "BIK" (reversible, indistinct, unlabeled) transform of k,0,0,0,....
Consider the input sequence (c_k(n): n >= 1) with g.f. C_k(x) = Sum_{n>=1} c_k(n)*x^n. Let a_k(n) = BIK(c_k(n): n >= 1) be the output sequence under Bower's BIK transform. It can proved that the g.f. of BIK(c_k(n): n >= 1) is A_k(x) = (1/2)*(C_k(x)/(1-C_k(x)) + (C_k(x^2) + C_k(x))/(1-C_k(x^2))). (See the comments for sequence A001224.)
For column k of this two-dimensional array, the input sequence is defined by c_k(1) = k and c_k(n) = 0 for n >= 1. Thus, C_k(x) = k*x, and hence the g.f. of column k is (1/2)*(C_k(x)/(1-C_k(x)) + (C_k(x^2) + C_k(x))/(1-C_k(x^2))) = (1/2)*(k*x/(1-k*x) + (k*x^2 + k*x)/(1-k*x^2)) = (2 + (1-k)*x - 2*k*x^2)*k*x/(2*(1-k*x^2)*(1-k*x)).
Using the first form the g.f. above and the expansion 1/(1-y) = 1 + y + y^2 + ..., we can easily prove J.-F. Alcover's formula T(n,k) = (k^n + k^((n + mod(n,2))/2))/2.
(End)

			Array begins with T(0,0):
1 1   1     1      1       1        1         1         1          1 ...
0 1   2     3      4       5        6         7         8          9 ...
0 1   3     6     10      15       21        28        36         45 ...
0 1   6    18     40      75      126       196       288        405 ...
0 1  10    45    136     325      666      1225      2080       3321 ...
0 1  20   135    544    1625     3996      8575     16640      29889 ...
0 1  36   378   2080    7875    23436     58996    131328     266085 ...
0 1  72  1134   8320   39375   140616    412972   1050624    2394765 ...
0 1 136  3321  32896  195625   840456   2883601   8390656   21526641 ...
0 1 272  9963 131584  978125  5042736  20185207  67125248  193739769 ...
0 1 528 29646 524800 4884375 30236976 141246028 536887296 1743421725 ...
...

See A005418.

Robert A. Russell, Antidiagonals n=0..52, flattened (antidiagonals 1..50 from Andrew Howroyd)
C. G. Bower, Transforms (2)

Columns 0-6 are A000007, A000012, A005418(n+1), A032120, A032121, A032122, A056308.
Rows 0-20 are A000012, A001477, A000217 (triangular numbers), A002411 (pentagonal pyramidal numbers), A037270, A168178, A071232, A168194, A071231, A168372, A071236, A168627, A071235, A168663, A168664, A170779, A170780, A170790, A170791, A170801, A170802.
Main diagonal is A275549.
Transpose is A284979.
Cf. A284871, A284949.
Cf. A003992 (oriented), A293500 (chiral), A321391 (achiral).

[[n le 0 select 1 else ((n-k)^k + (n-k)^Ceiling(k/2))/2: k in [0..n]]: n in [0..15]]; // G. C. Greubel, Nov 15 2018

Table[If[n>0, ((n-k)^k + (n-k)^Ceiling[k/2])/2, 1], {n, 0, 15}, {k, 0, n}] // Flatten (* updated Jul 10 2018 *) (* Adapted to T(0,k)=1 by Robert A. Russell, Nov 13 2018 *)

for(n=0,15, for(k=0,n, print1(if(n==0,1, ((n-k)^k + (n-k)^ceil(k/2))/2), ", "))) \\ G. C. Greubel, Nov 15 2018

T(n,k) = {(k^n + k^ceil(n/2)) / 2} \\ Andrew Howroyd, Sep 13 2019

T(n,k) = [n==0] + [n>0] * (k^n + k^ceiling(n/2)) / 2. [Adapted to T(0,k)=1 by Robert A. Russell, Nov 13 2018]
G.f. for column k: (1 - binomial(k+1,2)*x^2) / ((1-k*x)*(1-k*x^2)). - Petros Hadjicostas, Jul 07 2018 [Adapted to T(0,k)=1 by Robert A. Russell, Nov 13 2018]
From Robert A. Russell, Nov 13 2018: (Start)
T(n,k) = (A003992(k,n) + A321391(n,k)) / 2.
T(n,k) = A003992(k,n) - A293500(n,k) = A293500(n,k) + A321391(n,k).
G.f. for row n: (Sum_{j=0..n} S2(n,j)*j!*x^j/(1-x)^(j+1) + Sum_{j=0..ceiling(n/2)} S2(ceiling(n/2),j)*j!*x^j/(1-x)^(j+1)) / 2, where S2 is the Stirling subset number A008277.
G.f. for row n>0: x*Sum_{k=0..n-1} A145882(n,k) * x^k / (1-x)^(n+1).
E.g.f. for row n: (Sum_{k=0..n} S2(n,k)*x^k + Sum_{k=0..ceiling(n/2)} S2(ceiling(n/2),k)*x^k) * exp(x) / 2, where S2 is the Stirling subset number A008277.
T(0,k) = 1; T(1,k) = k; T(2,k) = binomial(k+1,2); for n>2, T(n,k) = k*(T(n-3,k)+T(n-2,k)-k*T(n-1,k)).
For k>n, T(n,k) = Sum_{j=1..n+1} -binomial(j-n-2,j) * T(n,k-j). (End)

Array transposed for greater consistency by Andrew Howroyd, Apr 04 2017

Origin changed to T(0,0) by Robert A. Russell, Nov 13 2018

A056450 a(n) = (3*2^n - (-2)^n)/2.

Original entry on oeis.org

1, 4, 4, 16, 16, 64, 64, 256, 256, 1024, 1024, 4096, 4096, 16384, 16384, 65536, 65536, 262144, 262144, 1048576, 1048576, 4194304, 4194304, 16777216, 16777216, 67108864, 67108864, 268435456, 268435456, 1073741824, 1073741824, 4294967296

Offset: 0

Marks R. Nester

Number of palindromes of length n using a maximum of four different symbols.
Number of achiral rows of n colors using up to four colors. - Robert A. Russell, Nov 09 2018
Interleaving of A000302 and 4*A000302.
Unsigned version of A141125.
Binomial transform is A164907. Second binomial transform is A164908. Third binomial transform is A057651. Fourth binomial transform is A016129.

			At length n=1 there are a(1)=4 palindromes, A, B, C, D.
At length n=2, there are a(2)=4 palindromes, AA, BB, CC, DD.
At length n=3, there are a(3)=16 palindromes, AAA, BBB, CCC, DDD, ABA, BAB, ... , CDC, DCD.

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
Sean A. Irvine, Walks on Graphs.
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (0,4).

Column k=4 of A321391.
Cf. A000302 (powers of 4), A056450, A141125, A164907, A164908, A057651, A016129.
Cf. A016116.
Essentially the same as A213173.
Cf. A000302 (oriented), A032121 (unoriented), A032087(n>1) (chiral).

Magma
```
[ (3*2^n-(-2)^n)/2: n in [0..31] ];
```

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

Table[4^Ceiling[n/2], {n,0,40}] (* or *)
CoefficientList[Series[(1 + 4 x)/((1 + 2 x) (1 - 2 x)), {x, 0, 31}], x] (* or *)
LinearRecurrence[{0, 4}, {1, 4}, 40] (* Robert A. Russell, Nov 07 2018 *)

a(n)=4^((n+1)\2) \\ Charles R Greathouse IV, Apr 08 2012

a(n)=(3*2^n-(-2)^n)/2 \\ Charles R Greathouse IV, Oct 03 2016

a(n) = 4^floor((n+1)/2).
a(n) = 4*a(n-2) for n > 1; a(0) = 1, a(1) = 4.
G.f.: (1+4*x) / (1-4*x^2). - R. J. Mathar, Jan 19 2011 [Adapted to offset 0 by Robert A. Russell, Nov 07 2018]
a(n+3) = a(n+2)*a(n+1)/a(n). - Reinhard Zumkeller, Mar 04 2011
a(n) = 4*abs(A164111(n-1)). - R. J. Mathar, Jan 19 2011
a(n) = C(4,0)*A000007(n) + C(4,1)*A057427(n) + C(4,2)*A056453(n) + C(4,3)*A056454(n) + C(4,4)*A056455(n). - Robert A. Russell, Nov 08 2018

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

Edited by N. J. A. Sloane, Sep 29 2019

A056323 Number of reversible string structures with n beads using a maximum of four different colors.

Original entry on oeis.org

1, 1, 2, 4, 11, 31, 107, 379, 1451, 5611, 22187, 87979, 350891, 1400491, 5597867, 22379179, 89500331, 357952171, 1431743147, 5726775979, 22906841771, 91626580651, 366505274027, 1466017950379, 5864067607211

Offset: 0

Marks R. Nester

A string and its reverse are considered to be equivalent. Permuting the colors will not change the structure. Thus aabc, cbaa and bbac are all considered to be identical.
Number of set partitions of an unoriented row of n elements with four or fewer nonempty subsets. - Robert A. Russell, Oct 28 2018
There are nonrecursive formulas, generating functions, and computer programs for A124303 and A305750, which can be used in conjunction with the formula. - Robert A. Russell, Oct 28 2018
From Allan Bickle, Jun 02 2022: (Start)
a(n) is the number of (unlabeled) 4-paths with n+6 vertices. (A 4-path with order n at least 6 can be constructed from a 5-clique by iteratively adding a new 4-leaf (vertex of degree 4) adjacent to an existing 4-clique containing an existing 4-leaf.)
Recurrences appear in the papers by Bickle, Eckhoff, and Markenzon et al. (End)

			For a(4)=11, the 7 achiral patterns are AAAA, AABB, ABAB, ABBA, ABCA, ABBC, and ABCD. The 4 chiral pairs are AAAB-ABBB, AABA-ABAA, AABC-ABCC, and ABAC-ABCB.

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]

Allan Bickle, How to Count k-Paths, J. Integer Sequences, 25 (2022) Article 22.5.6.
Allan Bickle, A Survey of Maximal k-degenerate Graphs and k-Trees, Theory and Applications of Graphs 0 1 (2024) Article 5.
J. Eckhoff, Extremal interval graphs, J. Graph Theory 17 1 (1993), 117-127.
L. Markenzon, O. Vernet, and P. R. da Costa Pereira, A clique-difference encoding scheme for labelled k-path graphs, Discrete Appl. Math. 156 (2008), 3216-3222.
Index entries for linear recurrences with constant coefficients, signature (5,0,-20,16).

Cf. A032121.
Column 4 of A320750.
Cf. A124303 (oriented), A320934 (chiral), A305750 (achiral).
The numbers of unlabeled k-paths for k = 2..7 are given in A005418, A001998, A056323, A056324, A056325, and A345207, respectively.
The sequences above converge to A103293(n+1).

Ach[n_, k_] := Ach[n, k] = If[n<2, Boole[n==k && n>=0], k Ach[n-2,k] + Ach[n-2,k-1] + Ach[n-2,k-2]] (* A304972 *)
k=4; Table[Sum[StirlingS2[n,j]+Ach[n,j],{j,0,k}]/2,{n,0,40}] (* Robert A. Russell, Oct 28 2018 *)
LinearRecurrence[{5, 0, -20, 16}, {1, 1, 2, 4, 11}, 40] (* Robert A. Russell, Oct 28 2018 *)

Use de Bruijn's generalization of Polya's enumeration theorem as discussed in reference.
For n > 0, a(n) = (16 + (-2)^n + 15*2^n + 4^n)/48. - Colin Barker, Nov 24 2012
G.f.: (1 - 4x - 3x^2 + 14x^3 - 5x^4) / ((1-x)*(1-4x)*(1-4x^2)). - Colin Barker, Nov 24 2012 [Adapted to offset 0 by Robert A. Russell, Nov 09 2018]
From Robert A. Russell, Oct 28 2018: (Start)
a(n) = (A124303(n) + A305750(n)) / 2.
a(n) = A124303(n) - A320934(n) = A320934(n) + A305750(n).
a(n) = Sum_{j=0..k} (S2(n,j) + Ach(n,j)) / 2, where k=4 is the maximum number of colors, S2 is the Stirling subset number A008277, and Ach(n,k) = [n>=0 & n<2 & n==k] + [n>1]*(k*Ach(n-2,k) + Ach(n-2,k-1) + Ach(n-2,k-2)).
a(n) = A000007(n) + A057427(n) + A056326(n) + A056327(n) + A056328(n). (End)

a(0)=1 prepended by Robert A. Russell, Nov 09 2018

A032087 Number of reversible strings with n beads of 4 colors. If more than 1 bead, not palindromic.

Original entry on oeis.org

4, 6, 24, 120, 480, 2016, 8064, 32640, 130560, 523776, 2095104, 8386560, 33546240, 134209536, 536838144, 2147450880, 8589803520, 34359607296, 137438429184, 549755289600, 2199021158400, 8796090925056, 35184363700224, 140737479966720, 562949919866880

Offset: 1

Christian G. Bower

From Petros Hadjicostas, Jun 30 2018: (Start)
Using the formulae in C. G. Bower's web link below about transforms, it can be proved that, for k >= 2, the BHK[k] transform of sequence (c(n): n >= 1), which has g.f. C(x) = Sum_{n >= 1} c(n)*x^n, has generating function B_k(x) = (1/2)*(C(x)^k - C(x^2)^{k/2}) if k is even, and B_k(x) = C(x)*B_{k-1}(x) = (C(x)/2)*(C(x)^{k-1} - C(x^2)^{(k-1)/2}) if k is odd. For k=1, Bower assumes that the BHK[k=1] transform of (c(n): n >= 1) is itself, which means that the g.f. of the output sequence is C(x). (This assumption is not accepted by all mathematicians because a sequence of length 1 is not only reversible but palindromic as well.)
Since a(m) = BHK(c(n): n >= 1)(m) = Sum_{k=1..m} BHK[k](c(n): n >= 1)(m) for m = 1,2,3,..., it can be easily proved (using sums of infinite geometric series) that the g.f. of BHK(c(n): n >= 1) is A(x) = (C(x)^2 - C(x^2))/(2*(1-C(x))*(1-C(x^2))) + C(x). (The extra C(x) is due of course to the special assumption made for the BHK[k=1] transform.)
Here, BHK(c(n): n >= 1)(m) indicates the m-th element of the output sequence when the transform is BHK and the input sequence is (c(n): n >= 1). Similarly, BHK[k](c(n): n >= 1)(m) indicates the m-th element of the output sequence when the transform is BHK[k] (i.e., with k boxes) and the input sequence is (c(n): n >= 1).
For the current sequence, c(1) = 4, and c(n) = 0 for all n >= 2, and thus, C(x) = 4*x. Substituting into the above formula for A(x), and doing the algebra, we get A(x) = 2*x*(2-5*x-8*x^2+32*x^3) / ((1+2*x)*(1-2*x)*(1-4*x)), which is R. J. Mathar's formula below.
(End)
The formula for a(n) for this sequence was Ralf Stephan's conjecture 72. It was solved by Elizabeth Wilmer (see Proposition 1 in one of the links below). She does not accept Bower's assertion that a string of length 1 is not palindromic. - Petros Hadjicostas, Jul 05 2018

Colin Barker, Table of n, a(n) for n = 1..1000
C. G. Bower, Transforms (2)
Ralf Stephan, Prove or disprove: 100 conjectures from the OEIS, arXiv:math/0409509 [math.CO], 2004.
Elizabeth Wilmer, Notes on Stephan's conjectures 72, 73 and 74
Elizabeth Wilmer, Notes on Stephan's conjectures 72, 73 and 74 [cached copy]
Index entries for linear recurrences with constant coefficients, signature (4,4,-16).

Column 4 of A293500 for n>1.

Cf. A000302, A026337 (bisection), A032121, A056450, A088037.

A032087:= func< n | n eq 1 select 4 else 2^(2*n-1) -(3-(-1)^n)*2^(n-2) >;
[A032087(n): n in [1..30]]; // G. C. Greubel, Oct 02 2024

Join[{4}, LinearRecurrence[{4, 4, -16}, {6, 24, 120}, 24]] (* Jean-François Alcover, Oct 11 2017 *)

Vec(2*x*(2 - 5*x - 8*x^2 + 32*x^3) / ((1 - 2*x)*(1 + 2*x)*(1 - 4*x)) + O(x^30)) \\ Colin Barker, Mar 08 2017

def A032087(n): return 2^(2*n-1) -3*2^(n-2) +(-2)^(n-2) +4*int(n==1)
[A032087(n) for n in range(1,31)] # G. C. Greubel, Oct 02 2024

"BHK" (reversible, identity, unlabeled) transform of 4, 0, 0, 0, ...
a(2*n+1) = 2^(4*n+1) - 2^(2*n+1), a(2*n) = 2^(4*n-1) - 2^(2*n) + 2^(2*n-1), a(1)=4.
a(n) = (A000302(n) - A056450(n))/2 for n > 1.
From R. J. Mathar, Mar 20 2009: (Start)
a(n) = 4*a(n-1) + 4*a(n-2) - 16*a(n-3) for n > 4.
G.f.: 2*x*(2-5*x-8*x^2+32*x^3)/((1-2*x)*(1+2*x)*(1-4*x)). (End)
From Colin Barker, Mar 08 2017: (Start)
a(n) = 2^(n-1) * (2^n-1) for n > 1 and even.
a(n) = 2^(2*n-1) - 2^n for n > 1 and odd. (End)
E.g.f.: (1/4)*( exp(-2*x) - 3*exp(2*x) + 2*exp(4*x) ) + 4*x. - G. C. Greubel, Oct 02 2024

A056311 Number of reversible strings with n beads using exactly four different colors.

Original entry on oeis.org

0, 0, 0, 12, 120, 780, 4212, 20424, 93360, 409380, 1749780, 7338792, 30394560, 124705140, 508291812, 2061607224, 8332140720, 33585777060, 135116412660, 542785800072, 2178110589600, 8733345234900

Offset: 1

Marks R. Nester

nonn

A string and its reverse are considered to be equivalent.

			For n=4, the 12 rows are 12 permutations of ABCD that do not include any mutual reversals.  Each of the 12 chiral pairs, such as ABCD-DCBA, is then counted just once. - _Robert A. Russell_, Sep 25 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 = 1..1000

Cf. A032121.
Column 4 of A305621.
Equals (A000919 + A056455) / 2 = A000919 - A305624 = A305624 + A056455.

k=4; Table[(StirlingS2[i,k]+StirlingS2[Ceiling[i/2],k])k!/2,{i,k,30}] (* Robert A. Russell, Nov 25 2017 *)
CoefficientList[Series[12 x^3 (3 x + 1) (8 x^4 - 3 x^3 - 2 x^2 - x + 1) / ((x - 1) (4 x - 1) (3 x - 1) (2 x + 1) (2 x - 1) (3 x^2 - 1) (2 x^2 - 1)), {x, 0, 33}], x] (* Vincenzo Librandi, Sep 26 2018 *)

Equals A032121(n) - 4*A032120(n) + 6*A005418(n+1) - 4.
G.f.: 12*x^4*(3*x+1)*(8*x^4-3*x^3-2*x^2-x+1)/ ((x-1) * (4*x-1) * (3*x-1) * (2*x+1) * (2*x -1) * (3*x^2-1) * (2*x^2-1)). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 27 2009 [Corrected by R. J. Mathar, Sep 16 2009]
a(n) = k! (S2(n,k) + S2(ceiling(n/2),k)) / 2, where k=4 is the number of colors and S2 is the Stirling subset number. - Robert A. Russell, Sep 25 2018

A056312 Number of reversible strings with n beads using exactly five different colors.

Original entry on oeis.org

0, 0, 0, 0, 60, 900, 8400, 63000, 417120, 2551560, 14804700, 82764900, 450518460, 2404510500, 12646078200, 65771496000, 339165516120, 1737486149760, 8855359634100, 44952367981500, 227475768907860, 1148269329527100, 5785013373810000, 29100047092479000

Offset: 1

Marks R. Nester

A string and its reverse are considered to be equivalent.

			For n=5, the 60 rows are 60 permutations of ABCDE that do not include any mutual reversals.  Each of the 60 chiral pairs, such as ABCDE-EDCBA, is then counted just once.

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 (13,-45,-75,695,-575,-3195,5595,4706,-14918,2160,12840,-7200).

Cf. A005418, A032120, A032121, A032122.
Column 5 of A305621.
Equals (A001118 + A056456) / 2 = A001118 - A305625 = A305625 + A056456.

[60*(StirlingSecond(n, 5)+StirlingSecond(Ceiling(n/2), 5)): n in [1..30]]; // Vincenzo Librandi, Sep 30 2018

k=5; Table[(StirlingS2[i,k]+StirlingS2[Ceiling[i/2],k])k!/2,{i,30}] (* Robert A. Russell, Nov 25 2017 *) adapted
CoefficientList[Series[-60*x^4*(120*x^7 - 17*x^6 - 50*x^5 - 32*x^4 + 20*x^3 + 10*x^2 - 2*x - 1)/((x - 1)*(2*x - 1)*(2*x + 1)*(3*x - 1)*(4*x - 1)*(5*x - 1)*(2*x^2 - 1)*(3*x^2 - 1)*(5*x^2 - 1)), {x, 0, 30}], x] (* Stefano Spezia, Sep 29 2018 *)

a(n) = 60*(stirling(n, 5, 2) + stirling(ceil(n/2), 5, 2)); \\ Altug Alkan, Sep 27 2018

a(n) = A032122(n) - 5*A032121(n) + 10*A032120(n) - 10*A005418(n+1) + 5.
G.f.: -60*x^5*(120*x^7 - 17*x^6 - 50*x^5 - 32*x^4 + 20*x^3 + 10*x^2 - 2*x - 1)/((x - 1)*(2*x - 1)*(2*x + 1)*(3*x - 1)*(4*x - 1)*(5*x - 1)*(2*x^2 - 1)*(3*x^2 - 1)*(5*x^2 - 1)). [Colin Barker, Sep 03 2012]
a(n) = k! (S2(n,k) + S2(ceiling(n/2),k)) / 2, where k=5 is the number of colors and S2 is the Stirling subset number. - Robert A. Russell, Sep 25 2018

A056313 Number of reversible strings with n beads using exactly six different colors.

Original entry on oeis.org

0, 0, 0, 0, 0, 360, 7560, 95760, 952560, 8217720, 64615680, 476515080, 3355679880, 22837101840, 151449674040, 984573656640, 6302070915840, 39847411326600, 249509384858160, 1550188410555960, 9570844671224760

Offset: 1

Marks R. Nester

A string and its reverse are considered to be equivalent.

			For n=6, the 360 rows are 360 permutations of ABCDEF that do not include any mutual reversals.  Each of the 360 chiral pairs, such as ABCDEF-FEDCBA, is then counted just once. - _Robert A. Russell_, Sep 25 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]

Index entries for linear recurrences with constant coefficients, signature (19, -117, 81, 1883, -5915, -6615, 53235, -30394, -191744, 264852, 258804, -634248, 43920, 505440, -259200).

Cf. A056308, A056322.
Column 6 of A305621.
Equals (A000920 + A056457) / 2 = A000920 - A305626 = A305626 + A056457.

k=6; Table[(StirlingS2[i,k]+StirlingS2[Ceiling[i/2],k])k!/2,{i,k,30}] (* Robert A. Russell, Nov 25 2017 *)

a(n) = my(k=6); k!/2*(stirling(n, k, 2) + stirling(ceil(n/2), k, 2)); \\ Altug Alkan, Sep 27 2018

a(n) = A056308(n) - 6*A032122(n) + 15*A032121(n) - 20*A032120(n) + 15*A005418(n+1) - 6.
G.f.: 360*x^6*(8*x^2 - x - 1)*(90*x^7 - 9*x^6 - 29*x^5 - 34*x^4 + 15*x^3 + 9*x^2 - x - 1)/((x - 1)*(2*x - 1)*(2*x + 1)*(3*x - 1)*(4*x - 1)*(5*x - 1)*(6*x - 1)*(2*x^2 - 1)*(3*x^2 - 1)*(5*x^2 - 1)*(6*x^2 - 1)). - Colin Barker, Sep 03 2012
a(n) = k! (S2(n,k) + S2(ceiling(n/2),k)) / 2, where k=6 is the number of colors and S2 is the Stirling subset number. - Robert A. Russell, Sep 25 2018

A056315 Number of primitive (aperiodic) reversible strings with n beads using a maximum of four different colors.

Original entry on oeis.org

4, 6, 36, 126, 540, 2034, 8316, 32760, 131544, 524250, 2099196, 8388450, 33562620, 134217594, 536903100, 2147483520, 8590065660, 34359735816, 137439477756, 549755813250, 2199025344348, 8796093020154

Offset: 1

Marks R. Nester

nonn

A string and its reverse are considered to be equivalent.

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]

Column 4 of A284871.

Cf. A045625, A056314.

Sum mu(d)*A032121(n/d) where d|n.

A321579 Number of n-tuples of 4 elements excluding reverse duplicates and those consisting of repetitions of the same element only.

Original entry on oeis.org

0, 0, 6, 36, 132, 540, 2076, 8316, 32892, 131580, 524796, 2099196, 8390652, 33562620, 134225916, 536903676, 2147516412, 8590065660, 34359869436, 137439477756, 549756338172, 2199025352700, 8796095119356, 35184380477436, 140737496743932

Offset: 0

Mikk Heidemaa, Nov 13 2018

Also the number of distinct DNA or RNA sequences of length n if the reverse copies and homopolymeric oligonucleotides (i.e., repetitions of the same nucleobases: aaa..., ccc..., ggg..., and ttt... (or uuu...)) are excluded.

			a(2) = 6 because {a,c,g,t} give six 2-tuples (duples): {a,c}, {a,g}, {a,t}, {c,g}, {c,t}, {g,t} as 4: {a,a}, {c,c}, {g,g}, {t,t} (consisting of the same element only) and 6 reverse duplicates: {c,a}, {g,a}, {t,a}, {g,c}, {t,c}, {t,g} are excluded ({c,a} is the duplicate of {a,c}, etc.), leaving 6 from 16 possible 2-tuples.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,0,-20,16).

Cf. A032121.

a[n_]:=(2^(# - 2)*((-1)^(# + 1) + 3) + 2^(2*# - 1) - 4)&/@ Range@n; a[25] (* or *)
CoefficientList[Series[6*(8*x^3-x^2-x)/(16*x^4-20*x^3+5*x-1), {x, 0, 20}], x]
LinearRecurrence[{5,0,-20,16},{0,0,6,36,132},30] (* Harvey P. Dale, Mar 20 2023 *)

concat([0,0], Vec(6*x^2*(1 + x - 8*x^2) / ((1 - x)*(1 - 2*x)*(1 + 2*x)*(1 - 4*x)) + O(x^40))) \\ Colin Barker, Nov 14 2018

a(n) = (2^(n-2)*((-1)^(n+1) + 3) + 2^(2*n-1) - 4) for n > 0.
a(n) = A032121(n) - 4 for n > 2.
G.f.: 6*x^2*(8*x^2 - x - 1)/((x-1)*(2*x+1)*(2*x-1)*(4*x-1)).
a(n) = 5*a(n-1) - 20*a(n-3) + 16*a(n-4). - Colin Barker, Nov 14 2018

cp's OEIS Frontend

A277504 Array read by descending antidiagonals: T(n,k) is the number of unoriented strings with n beads of k or fewer colors.

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A056450 a(n) = (3*2^n - (-2)^n)/2.

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A056323 Number of reversible string structures with n beads using a maximum of four different colors.

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A032087 Number of reversible strings with n beads of 4 colors. If more than 1 bead, not palindromic.

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A056311 Number of reversible strings with n beads using exactly four different colors.

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A056312 Number of reversible strings with n beads using exactly five different colors.

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