A293500 -id:A293500 - cp's OEIS frontend

A309008 The same array as in A293500, but without the leading row and column of zeros. See that entry for more information.

1, 2, 3, 6, 9, 6, 12, 36, 24, 10, 28, 108, 120, 50, 15, 56, 351, 480, 300, 90, 21, 120, 1053, 2016, 1500, 630, 147, 28, 240, 3240, 8064, 7750, 3780, 1176, 224, 36, 496, 9720, 32640, 38750, 23220, 8232, 2016, 324, 45

Offset: 2

Bahman Ahmadi, Aug 05 2019

			The table begins:
   1,    3,    6,    10,     15,     21,      28,      36,      45, ...
   2,    9,   24,    50,     90,    147,     224,     324,     450, ...
   6,   36,  120,   300,    630,   1176,    2016,    3240,    4950, ...
  12,  108,  480,  1500,   3780,   8232,   16128,   29160,   49500, ...
  28,  351, 2016,  7750,  23220,  58653,  130816,  265356,  499500, ...
  56, 1053, 8064, 38750, 139320, 410571, 1046528, 2388204, 4995000, ...
  ...

Cf. A293500.

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

Original entry on oeis.org

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

A003992 Square array read by upwards antidiagonals: T(n,k) = n^k for n >= 0, k >= 0.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 4, 1, 0, 1, 4, 9, 8, 1, 0, 1, 5, 16, 27, 16, 1, 0, 1, 6, 25, 64, 81, 32, 1, 0, 1, 7, 36, 125, 256, 243, 64, 1, 0, 1, 8, 49, 216, 625, 1024, 729, 128, 1, 0, 1, 9, 64, 343, 1296, 3125, 4096, 2187, 256, 1, 0, 1, 10, 81, 512, 2401, 7776, 15625, 16384, 6561, 512, 1, 0

Offset: 0

Marc LeBrun

If the array is transposed, T(n,k) is the number of oriented rows of n colors using up to k different colors. The formula would be T(n,k) = [n==0] + [n>0]*k^n. The generating function for column k would be 1/(1-k*x). For T(3,2)=8, the rows are AAA, AAB, ABA, ABB, BAA, BAB, BBA, and BBB. - Robert A. Russell, Nov 08 2018

T(n,k) is the number of multichains of length n from {} to [k] in the Boolean lattice B_k. - Geoffrey Critzer, Apr 03 2020

			Rows begin:
[1, 0,  0,   0,    0,     0,      0,      0, ...],
[1, 1,  1,   1,    1,     1,      1,      1, ...],
[1, 2,  4,   8,   16,    32,     64,    128, ...],
[1, 3,  9,  27,   81,   243,    729,   2187, ...],
[1, 4, 16,  64,  256,  1024,   4096,  16384, ...],
[1, 5, 25, 125,  625,  3125,  15625,  78125, ...],
[1, 6, 36, 216, 1296,  7776,  46656, 279936, ...],
[1, 7, 49, 343, 2401, 16807, 117649, 823543, ...], ...

T. D. Noe, Rows n = 0..50 of triangle, flattened

Rows 0-49 are A000007, A000012, A000079, A000244, A000302, A000351, A000400, A000420, A001018, A001019, A011557, A001020, A001021, A001022, A001023, A001024, A001025, A001026, A001027, A001029, A009964-A009992, A087752.
Columns 0-26 are A000012, A001477, A000290, A000578, A000583, A000584, A001014, A001015, A001016, A001017, A008454, A008455, A008456, A010801-A010813, A089081.
Main diagonal is A000312. Other diagonals include A000169, A007778, A000272, A008788. Antidiagonal sums are in A026898.
Cf. A099555.
Transpose is A004248. See A051128, A095884, A009999 for other versions.
Cf. A277504 (unoriented), A293500 (chiral).

[[(n-k)^k: k in [0..n]]: n in [0..10]]; // G. C. Greubel, Nov 08 2018

Table[If[k == 0, 1, (n - k)^k], {n, 0, 11}, {k, 0, n}]//Flatten

T(n,k) = (n-k)^k \\ Charles R Greathouse IV, Feb 07 2017

E.g.f.: Sum T(n,k)*x^n*y^k/k! = 1/(1-x*exp(y)). - Paul D. Hanna, Oct 22 2004

E.g.f.: Sum T(n,k)*x^n/n!*y^k/k! = e^(x*e^y). - Franklin T. Adams-Watters, Jun 23 2006

More terms from David W. Wilson

Edited by Paul D. Hanna, Oct 22 2004

A305622 Triangle read by rows: T(n,k) is the number of chiral pairs of rows of n colors with exactly k different colors.

Original entry on oeis.org

0, 0, 1, 0, 2, 3, 0, 6, 18, 12, 0, 12, 72, 120, 60, 0, 28, 267, 780, 900, 360, 0, 56, 885, 4188, 8400, 7560, 2520, 0, 120, 2880, 20400, 63000, 95760, 70560, 20160, 0, 240, 9000, 93120, 417000, 952560, 1164240, 725760, 181440, 0, 496, 27915, 409140, 2551440, 8217720, 14817600, 15120000, 8164800, 1814400, 0, 992, 85233, 1748220, 14802900, 64614960, 161247240, 239500800, 209563200, 99792000, 19958400

Offset: 1

Robert A. Russell, Jun 06 2018

If the row is achiral, i.e., the same as its reverse, we ignore it. If different from its reverse, we count it and its reverse as a chiral pair.

			The triangle begins:
  0;
  0,   1;
  0,   2,     3;
  0,   6,    18,     12;
  0,  12,    72,    120,      60;
  0,  28,   267,    780,     900,     360;
  0,  56,   885,   4188,    8400,    7560,     2520;
  0, 120,  2880,  20400,   63000,   95760,    70560,    20160;
  0, 240,  9000,  93120,  417000,  952560,  1164240,   725760,  181440;
  ...
For T(3,2)=2, the chiral pairs are AAB-BAA and ABB-BBA.  For T(3,3)=3, the chiral pairs are ABC-CBA, ACB-BCA, and BAC-CAB.

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

Columns 1-6 are A000004, A122746(n-2), A305623, A305624, A305625, and A305626.
Row sums are A327091.
Cf. A008277, A019538, A293500, A305621.

with(combinat):
a:=(n,k)->(factorial(k)/2)* (Stirling2(n,k)-Stirling2(ceil(n/2),k)): seq(seq(a(n,k),k=1..n),n=1..11); # Muniru A Asiru, Sep 27 2018

Table[(k!/2) (StirlingS2[n, k] - StirlingS2[Ceiling[n/2], k]), {n, 1, 15}, {k, 1, n}] // Flatten

T(n,k) = (k!/2) * (stirling(n,k,2) - stirling(ceil(n/2),k,2));
for (n=1, 10, for(k=1, n, print1(T(n,k), ", ")); print); \\ Michel Marcus, Sep 27 2018

T(n,k) = (k!/2) * (S2(n,k) - S2(ceiling(n/2),k)) where S2(n,k) is the Stirling subset number A008277.
T(n,k) = (A019538(n,k) - A019538(ceiling(n/2),k)) / 2.
T(n,k) = A019538(n,k) - A305621(n,k).
G.f. for column k: k! x^k / (2*Product_{i=1..k}(1-ix)) - k! (x^(2k-1)+x^(2k)) / (2*Product{i=1..k}(1-i x^2)). - Robert A. Russell, Sep 26 2018
T(n, k) = Sum_{i=0..k} (-1)^(k-i)*binomial(k,i)*A293500(n, i). - Andrew Howroyd, Sep 13 2019

A321391 Array read by antidiagonals: T(n,k) is the number of achiral rows of n colors using up to k colors.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 2, 1, 0, 1, 4, 3, 4, 1, 0, 1, 5, 4, 9, 4, 1, 0, 1, 6, 5, 16, 9, 8, 1, 0, 1, 7, 6, 25, 16, 27, 8, 1, 0, 1, 8, 7, 36, 25, 64, 27, 16, 1, 0, 1, 9, 8, 49, 36, 125, 64, 81, 16, 1, 0, 1, 10, 9, 64, 49, 216, 125, 256, 81, 32, 1, 0

Offset: 0

Robert A. Russell, Nov 08 2018

The antidiagonals go from top-right to bottom-left.

			The array begins with T(0,0):
1 1  1   1    1     1     1      1      1      1       1       1 ...
0 1  2   3    4     5     6      7      8      9      10      11 ...
0 1  2   3    4     5     6      7      8      9      10      11 ...
0 1  4   9   16    25    36     49     64     81     100     121 ...
0 1  4   9   16    25    36     49     64     81     100     121 ...
0 1  8  27   64   125   216    343    512    729    1000    1331 ...
0 1  8  27   64   125   216    343    512    729    1000    1331 ...
0 1 16  81  256   625  1296   2401   4096   6561   10000   14641 ...
0 1 16  81  256   625  1296   2401   4096   6561   10000   14641 ...
0 1 32 243 1024  3125  7776  16807  32768  59049  100000  161051 ...
0 1 32 243 1024  3125  7776  16807  32768  59049  100000  161051 ...
0 1 64 729 4096 15625 46656 117649 262144 531441 1000000 1771561 ...
For T(3,3)=9, the rows are AAA, ABA, ACA, BAB, BBB, BCB, CAC, CBC, and CCC.

Columns 0-6 are A000007, A000012, A060546, A056449, A056450, A056451, A056452.

Cf. A003992 (oriented), A277504 (unoriented), A293500 (chiral).

Table[If[n>0, (n-k)^Ceiling[k/2], 1], {n, 0, 12}, {k, 0, n}] // Flatten

T(n,k) = [n==0] + [n>0] * k^ceiling(n/2).

The generating function for column k is (1+k*x) / (1-k*x^2).

A032086 Number of reversible strings with n beads of 3 colors. If more than 1 bead, not palindromic.

Original entry on oeis.org

3, 3, 9, 36, 108, 351, 1053, 3240, 9720, 29403, 88209, 265356, 796068, 2390391, 7171173, 21520080, 64560240, 193700403, 581101209, 1743362676, 5230088028, 15690441231, 47071323693, 141214502520, 423643507560

Offset: 1

Christian G. Bower

Andrew Howroyd, Table of n, a(n) for n = 1..200
C. G. Bower, Transforms (2)
Index entries for linear recurrences with constant coefficients, signature (3, 3, -9).

Column 3 of A293500 for n>1.

Cf. A032120.

Join[{3}, LinearRecurrence[{3, 3, -9}, {3, 9, 36}, 24]] (* Jean-François Alcover, Oct 11 2017 *)

a(n) = if(n<2, [3][n], (3^n - 3^(ceil(n/2)))/2); \\ Andrew Howroyd, Oct 10 2017

"BHK" (reversible, identity, unlabeled) transform of 3, 0, 0, 0, ...
Conjectures from Colin Barker, Apr 02 2012: (Start)
a(n) = 3*a(n-1) + 3*a(n-2) - 9*a(n-3) for n > 4.
G.f.: 3*x*(1 - 2*x - 3*x^2 + 9*x^3)/((1 - 3*x)*(1 - 3*x^2)).
(End)
Conjectures from Colin Barker, Mar 09 2017: (Start)
a(n) = (2*3^n - 2*3^(n/2)) / 4 for n > 2 and even.
a(n) = (2*3^n - 2*3^((n+1)/2)) / 4 for n > 2 and odd.
(End)
The above conjectures are true: The second set follows from the definition and the first set can be derived from that. - Andrew Howroyd, Oct 10 2017
a(n) = (3^n - 3^(ceiling(n/2))) / 2 = (A000244(n) - A056449(n)) / 2 for n>1. - Robert A. Russell and Danny Rorabaugh, Jun 22 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

A032088 Number of reversible strings with n beads of 5 colors. If more than 1 bead, not palindromic.

Original entry on oeis.org

5, 10, 50, 300, 1500, 7750, 38750, 195000, 975000, 4881250, 24406250, 122062500, 610312500, 3051718750, 15258593750, 76293750000, 381468750000, 1907347656250, 9536738281250, 47683710937500

Offset: 1

Christian G. Bower

Andrew Howroyd, Table of n, a(n) for n = 1..200
C. G. Bower, Transforms (2)
Index entries for linear recurrences with constant coefficients, signature (5, 5, -25).

Column 5 of A293500 for n>1.
Cf. A032122.
Equals (A000351 - A056451) / 2 for n>1.

Join[{5}, LinearRecurrence[{5, 5, -25}, {10, 50, 300}, 19]] (* Jean-François Alcover, Oct 11 2017 *)

a(n) = if(n<2, [5][n], (5^n - 5^(ceil(n/2)))/2); \\ Andrew Howroyd, Oct 10 2017

"BHK" (reversible, identity, unlabeled) transform of 5, 0, 0, 0, ...
Conjectures from Colin Barker, Jul 07 2012: (Start)
a(n) = 5*a(n-1) + 5*a(n-2) - 25*a(n-3) for n > 4.
G.f.: 5*x*(1 - 3*x - 5*x^2 + 25*x^3)/((1 - 5*x)*(1 - 5*x^2)).
(End)
Conjectures from Colin Barker, Mar 09 2017: (Start)
a(n) = 5^(n/2)*(5^(n/2) - 1) / 2 for n > 1 and even.
a(n) = -5*(5^(n/2-1/2) - 5^(n-1)) / 2 for n > 1 and odd.
(End)
The above conjectures are true: The second set follows from the definition and the first set can be derived from that. - Andrew Howroyd, Oct 10 2017
a(n) = (5^n - 5^(ceiling(n/2))) / 2 = (A000351(n) - A056451(n)) / 2 for n>1. - Robert A. Russell and Danny Rorabaugh, Jun 22 2018

A320524 Number of chiral pairs of a row of n colors using 6 or fewer colors.

Original entry on oeis.org

0, 15, 90, 630, 3780, 23220, 139320, 839160, 5034960, 30229200, 181375200, 1088367840, 6530207040, 39181942080, 235091652480, 1410554113920, 8463324683520, 50779973295360, 304679839772160, 1828079189798400, 10968475138790400, 65810851739735040, 394865110438410240, 2369190668072417280, 14215144008434503680, 85290864083258757120

Offset: 1

Robert A. Russell, Oct 14 2018

A chiral row is different from its reverse.

			For a(2)=15, the chiral pairs are the fifteen combinations of six colors taken two at a time, e.g., AB-BA.

Index entries for linear recurrences with constant coefficients, signature (6,6,-36).

Column 6 of A293500.

Cf. A000400 (oriented), A056308 (unoriented), A056452 (achiral).

[(6^n - 6^Ceiling(n / 2)) / 2: n in [1..25]]; // Vincenzo Librandi, Oct 15 2018

k = 6; Table[(k^n - k^Ceiling[n/2])/2, {n, 1, 30}]
LinearRecurrence[{6, 6, -36}, {0, 15, 90}, 30]

m=40; v=concat([0,15,90], vector(m-3)); for(n=4, m, v[n] = 6*v[n-1] +6*v[n-2] -36*v[n-3]); v \\ G. C. Greubel, Oct 17 2018

a(n) = (k^n - k^ceiling(n/2)) / 2, where k=6 is maximum number of colors.
G.f.: k*x^2*(k-1) / (2*(1-k*x)*(1-k*x^2)), where k=6.
a(n) = (A000400(n) - A056452(n)) / 2 = A000400(n) - A056308(n) = A056308(n) - A056452(n).

A321672 Number of chiral pairs of rows of length 5 using up to n colors.

Original entry on oeis.org

0, 0, 12, 108, 480, 1500, 3780, 8232, 16128, 29160, 49500, 79860, 123552, 184548, 267540, 378000, 522240, 707472, 941868, 1234620, 1596000, 2037420, 2571492, 3212088, 3974400, 4875000, 5931900, 7164612, 8594208, 10243380, 12136500

Offset: 0

Robert A. Russell, Nov 16 2018

			For a(0)=0 and  a(1)=0, there are no chiral rows using fewer than two colors. For a(2)=12, the chiral pairs are AAAAB-BAAAA, AAABA-ABAAA, AAABB-BBAAA, AABAB-BABAA, AABBA-ABBAA, AABBB-BBBAAA, ABAAB-BAABA, ABABB-BBABA, ABBAB-BABBA, ABBBB-BBBBA, BAABB-BBAAB, and BABBB-BBBAB.

Index entries for linear recurrences with constant coefficients, signature (6, -15, 20, -15, 6, -1).

Row 5 of A293500.

Cf. A000584 (oriented), A168178 (unoriented), A000578 (achiral).

Table[(n^5-n^3)/2,{n,0,40}]
LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 0, 12, 108, 480, 1500}, 40]

a(n)=(n^5-n^3)/2 \\ Charles R Greathouse IV, Oct 21 2022

a(n) = (n^5 - n^3) / 2.
a(n) = (A000584(n) - A000578(n)) / 2.
a(n) = A000584(n) - A168178(n) = A168178(n) - A000578(n).
G.f.: (Sum_{j=1..5} S2(5,j)*j!*x^j/(1-x)^(j+1) - Sum_{j=1..3} S2(3,j)*j!*x^j/(1-x)^(j+1)) / 2, where S2 is the Stirling subset number A008277.
G.f.: x * Sum_{k=1..4} A145883(5,k) * x^k / (1-x)^6.
E.g.f.: (Sum_{k=1..5} S2(5,k)*x^k - Sum_{k=1..3} S2(3,k)*x^k) * exp(x) / 2, where S2 is the Stirling subset number A008277.
For n>5, a(n) = Sum_{j=1..6} -binomial(j-7,j) * a(n-j).

cp's OEIS Frontend

A309008 The same array as in A293500, but without the leading row and column of zeros. See that entry for more information.

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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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A003992 Square array read by upwards antidiagonals: T(n,k) = n^k for n >= 0, k >= 0.

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A305622 Triangle read by rows: T(n,k) is the number of chiral pairs of rows of n colors with exactly k different colors.

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A321391 Array read by antidiagonals: T(n,k) is the number of achiral rows of n colors using up to k colors.

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A032086 Number of reversible strings with n beads of 3 colors. If more than 1 bead, not palindromic.

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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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