A054252 -id:A054252 - cp's OEIS frontend

A008805 Triangular numbers repeated.

1, 1, 3, 3, 6, 6, 10, 10, 15, 15, 21, 21, 28, 28, 36, 36, 45, 45, 55, 55, 66, 66, 78, 78, 91, 91, 105, 105, 120, 120, 136, 136, 153, 153, 171, 171, 190, 190, 210, 210, 231, 231, 253, 253, 276, 276, 300, 300, 325, 325, 351, 351, 378, 378, 406, 406, 435, 435

Offset: 0

N. J. A. Sloane

Number of choices for nonnegative integers x,y,z such that x and y are even and x + y + z = n.
Diagonal sums of A002260, when arranged as a number triangle. - Paul Barry, Feb 28 2003
a(n) = number of partitions of n+4 such that the differences between greatest and smallest parts are 2: a(n-4) = A097364(n,2) for n>3. - Reinhard Zumkeller, Aug 09 2004
For n >= i, i=4,5, a(n-i) is the number of incongruent two-color bracelets of n beads, i from them are black (cf. A005232, A032279), having a diameter of symmetry. - Vladimir Shevelev, May 03 2011
Prefixing A008805 by 0,0,0,0 gives the sequence c(0), c(1), ... defined by c(n)=number of (w,x,y) such that w = 2x+2y, where w,x,y are all in {1,...,n}; see A211422. - Clark Kimberling, Apr 15 2012
Partial sums of positive terms of A142150. - Reinhard Zumkeller, Jul 07 2012
The sum of the first parts of the nondecreasing partitions of n+2 into exactly two parts, n >= 0. - Wesley Ivan Hurt, Jun 08 2013
Number of the distinct symmetric pentagons in a regular n-gon, see illustration for some small n in links. - Kival Ngaokrajang, Jun 25 2013
a(n) is the number of nonnegative integer solutions to the equation x + y + z = n such that x + y <= z. For example, a(4) = 6 because we have 0+0+4 = 0+1+3 = 0+2+2 = 1+0+3 = 1+1+2 = 2+0+2. - Geoffrey Critzer, Jul 09 2013
a(n) is the number of distinct opening moves in n X n tic-tac-toe. - I. J. Kennedy, Sep 04 2013
a(n) is the number of symmetry-allowed, linearly-independent terms at n-th order in the series expansion of the T2 X t2 vibronic perturbation matrix, H(Q) (cf. Opalka & Domcke). - Bradley Klee, Jul 20 2015
a(n-1) also gives the number of D_4 (dihedral group of order 4) orbits of an n X n square grid with squares coming in either of two colors and only one square has one of the colors. - Wolfdieter Lang, Oct 03 2016
Also, this sequence is the third column in the triangle of the coefficients of the sum of two consecutive Fibonacci polynomials F(n+1, x) and F(n, x) (n>=0) in ascending powers of x. - Mohammad K. Azarian, Jul 18 2018
In an n-person symmetric matching pennies game (a zero-sum normal-form game) with n > 2 symmetric and indistinguishable players, each with two strategies (viz. heads or tails), a(n-3) is the number of distinct subsets of players that must play the same strategy to avoid incurring losses (single pure Nash equilibrium in the reduced game). The total number of distinct partitions is A000217(n-1). - Ambrosio Valencia-Romero, Apr 17 2022
a(n) is the number of connected bipartite graphs with n+1 edges and a stable set of cardinality 2. - Christian Barrientos, Jun 15 2022
a(n) is the number of 132-avoiding odd Grassmannian permutations of size n+2. - Juan B. Gil, Mar 10 2023
Consider a regular n-gon with all diagonals drawn. Define a "layer" to be the set of all regions sharing an edge with the exterior. Removing a layer creates another layer. Count the layers, removing them until none remain. The number of layers is a(n-2). See illustration. - Christopher Scussel, Nov 07 2023

			a(5) = 6, since (5) + 2 = 7 has three nondecreasing partitions with exactly 2 parts: (1,6),(2,5),(3,4). The sum of the first parts of these partitions = 1 + 2 + 3 = 6. - _Wesley Ivan Hurt_, Jun 08 2013

H. D. Brunk, An Introduction to Mathematical Statistics, Ginn, Boston, 1960; p. 360.

Vincenzo Librandi, Table of n, a(n) for n = 0..3000
G. E. Andrews, M. Beck, and N. Robbins, Partitions with fixed differences between largest and smallest parts, arXiv preprint arXiv:1406.3374 [math.NT], 2014.
P. Flajolet and R. Sedgewick, Analytic Combinatorics, 2009; see page 46.
Juan B. Gil and Jessica A. Tomasko, Pattern-avoiding even and odd Grassmannian permutations, arXiv:2207.12617 [math.CO], 2022.
Jia Huang, Partially Palindromic Compositions, J. Int. Seq. (2023) Vol. 26, Art. 23.4.1. See pp. 4, 19.
Kival Ngaokrajang, The distinct symmetric 5-gons in a regular n-gon for n = 6..13
D. Opalka and W. Domcke, High-order expansion of T2xt2 Jahn-Teller potential energy surfaces in tetrahedral molecules, J. Chem. Phys., 132, 154108 (2010).
Christopher Scussel, Illustration of layers in regular n-gons with all diagonals drawn
Vladimir Shevelev, A problem of enumeration of two-color bracelets with several variations, arXiv:0710.1370 [math.CO], 2007-2011.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).
Index entries for Molien series

Cf. A000217, A002260, A002620, A006918 (partial sums), A054252, A135276, A142150, A158920 (binomial trans.).

List([0..60], n-> (2*n +3 +(-1)^n)*(2*n +7 +(-1)^n)/32); # G. C. Greubel, Sep 12 2019

import Data.List (transpose)
a008805 = a000217 . (`div` 2) . (+ 1)
a008805_list = drop 2 $ concat $ transpose [a000217_list, a000217_list]
-- Reinhard Zumkeller, Feb 01 2013

[(2*n+3+(-1)^n)*(2*n+7+(-1)^n)/32 : n in [0..50]]; // Wesley Ivan Hurt, Apr 22 2015

A008805:=n->(2*n+3+(-1)^n)*(2*n+7+(-1)^n)/32: seq(A008805(n), n=0..50); # Wesley Ivan Hurt, Apr 22 2015

CoefficientList[Series[1/(1-x^2)^2/(1-x), {x, 0, 50}], x]
Table[Binomial[Floor[n/2] + 2, 2], {n, 0, 57}] (* Michael De Vlieger, Oct 03 2016 *)

PARI
```
a(n)=(n\2+2)*(n\2+1)/2
```

def A008805(n): return (m:=(n>>1)+1)*(m+1)>>1 # Chai Wah Wu, Oct 20 2023

[(2*n +3 +(-1)^n)*(2*n +7 +(-1)^n)/32 for n in (0..60)] # G. C. Greubel, Sep 12 2019

G.f.: 1/((1-x)*(1-x^2)^2) = 1/((1+x)^2*(1-x)^3).
E.g.f.: (exp(x)*(2*x^2 +12*x+ 11) - exp(-x)*(2*x -5))/16.
a(-n) = a(-5+n).
a(n) = binomial(floor(n/2)+2, 2). - Vladimir Shevelev, May 03 2011
From Paul Barry, May 31 2003: (Start)
a(n) = ((2*n +5)*(-1)^n + (2*n^2 +10*n +11))/16.
a(n) = Sum_{k=0..n} ((k+2)*(1+(-1)^k))/4. (End)
From Paul Barry, Apr 16 2005: (Start)
a(n) = Sum_{k=0..n} floor((k+2)/2)*(1-(-1)^(n+k-1))/2.
a(n) = Sum_{k=0..floor(n/2)} floor((n-2k+2)/2). (End)
A signed version is given by Sum_{k=0..n} (-1)^k*floor(k^2/4). - Paul Barry, Aug 19 2003
a(n) = A108299(n-2,n)*(-1)^floor((n+1)/2) for n>1. - Reinhard Zumkeller, Jun 01 2005
a(n) = A004125(n+3) - A049798(n+2). - Carl Najafi, Jan 31 2013
a(n) = Sum_{i=1..floor((n+2)/2)} i. - Wesley Ivan Hurt, Jun 08 2013
a(n) = (1/2)*floor((n+2)/2)*(floor((n+2)/2)+1). - Wesley Ivan Hurt, Jun 08 2013
From Wesley Ivan Hurt, Apr 22 2015: (Start)
a(n) = a(n-1) +2*a(n-2) -2*a(n-3) -a(n-4) +a(n-5).
a(n) = (2*n +3 +(-1)^n)*(2*n +7 +(-1)^n)/32. (End)
a(n-1) = A054252(n,1) = A054252(n^2-1), n >= 1. See a Oct 03 2016 comment above. - Wolfdieter Lang, Oct 03 2016
a(n) = A000217(A008619(n)). - Guenther Schrack, Sep 12 2018
From Ambrosio Valencia-Romero, Apr 17 2022: (Start)
a(n) = a(n-1) if n odd, a(n) = a(n-1) + (n+2)/2 if n is even, for n > 0, a(0) = 1.
a(n) = (n+1)*(n+3)/8 if n odd, a(n) = (n+2)*(n+4)/8 if n is even, for n >= 0.
a(n) = A002620(n+2) - a(n-1), for n > 0, a(0) = 1.
a(n) = A142150(n+2) + a(n-1), for n > 0, a(0) = 1.
a(n) = A000217(n+3)/2 - A135276(n+3)/2. (End)

A014409 Number of inequivalent ways (mod D_4) a pair of checkers can be placed on an n X n board.

Original entry on oeis.org

0, 2, 8, 21, 49, 93, 171, 278, 446, 660, 970, 1347, 1863, 2471, 3269, 4188, 5356, 6678, 8316, 10145, 12365, 14817, 17743, 20946, 24714, 28808, 33566, 38703, 44611, 50955, 58185, 65912, 74648, 83946, 94384, 105453, 117801, 130853, 145331, 160590, 177430, 195132

Offset: 1

Borghard, William (bogey(AT)hostare.att.com)

Computed by Fred Hallden.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-6,0,6,-2,-2,1).

Cf. A054252, A019318, A082966, A242279, A242358, A054247.

[(2*n^4+14*n^2-12*n-1-(-1)^n*(2*n^2-4*n-1))/32 : n in [1..60]]; // Wesley Ivan Hurt, Dec 30 2023

LinearRecurrence[{2, 2, -6, 0, 6, -2, -2, 1}, {0, 2, 8, 21, 49, 93, 171, 278}, 40]
CoefficientList[Series[- x (x^5 + x^4 + 3 x^3 + x^2 + 4 x + 2)/((x - 1)^5 (x + 1)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Oct 15 2013 *)

a(n)=if(n%2, n^4 + 8*n^2 - 8*n - 1, n^4 + 6*n^2 - 4*n)/16  \\ Charles R Greathouse IV, Feb 09 2017

a(2*n) = n/2*(2*n^3 + 3*n - 1); a(2*n+1) = n/2*(2*n^3 + 4*n^2 + 7*n + 3).
a(0)=0, a(1)=2, a(2)=8, a(3)=21, a(4)=49, a(5)=93, a(6)=171, a(7)=278, a(n)=2*a(n-1)+2*a(n-2)-6*a(n-3)+0*a(n-4)+6*a(n-5)-2*a(n-6)- 2*a(n-7)+ a(n-8). - Harvey P. Dale, May 06 2012
G.f.: -x^2*(x^5+x^4+3*x^3+x^2+4*x+2) / ((x-1)^5*(x+1)^3). - Colin Barker, Jul 11 2013
From James Stein, May 22 2014: (Start)
For odd n: a(n) = (n^4 + 8*n^2 - 8*n - 1)/16;
For even n: a(n) = n*(n^3 + 6*n - 4)/16. (End)
a(n) = A054252(n, 2), n >= 0. - Wolfdieter Lang, Oct 03 2016
E.g.f.: (x*(1 + 13*x + 6*x^2 + x^3)*cosh(x) + (-1 + 3*x + 15*x^2 + 6*x^3 + x^4)*sinh(x))/16. - Stefano Spezia, Apr 14 2022
a(n) = (2*n^4+14*n^2-12*n-1-(-1)^n*(2*n^2-4*n-1))/32. - Wesley Ivan Hurt, Dec 30 2023

More terms and formula from Hugo van der Sanden

More terms from Colin Barker, Jul 11 2013

A236679 Number T(n,k) of equivalence classes of ways of placing k 2 X 2 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=2, 0<=k<=floor(n/2)^2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 4, 2, 1, 1, 3, 13, 20, 14, 1, 6, 37, 138, 277, 273, 143, 39, 7, 1, 1, 6, 75, 505, 2154, 5335, 7855, 6472, 2756, 459, 1, 10, 147, 1547, 10855, 50021, 153311, 311552, 416825, 361426, 200996, 71654, 16419, 2363, 211, 11, 1, 1, 10, 246, 3759, 39926, 291171

Offset: 2

Christopher Hunt Gribble, Jan 29 2014

Changing the offset from 2 to 1, this is also the sequence: "Triangle read by rows: T(n,k) is the number of nonequivalent ways to place k non-attacking kings on an n X n board." (For if each king is represented by a 2 X 2 tile with the king in the upper left corner, the kings do not attack each other.) For example, with offset 1, T(4,3) = 20 because there are 20 nonequivalent ways to place 3 kings on a 4 X 4 chessboard so that no king threatens any other. - Heinrich Ludwig and N. J. A. Sloane, Dec 21 2016

It appears that rows 2n and 2n-1 both contain n^2 + 1 entries. Rotations and reflections of placements are not counted. If they are to be counted, see A193580. - Heinrich Ludwig, Dec 11 2016

			T(4,2) = 4 because the number of equivalence classes of ways of placing 2 2 X 2 square tiles in a 4 X 4 square under all symmetry operations of the square is 4. The portrayal of an example from each equivalence class is:
._______        _______        _______        _______
| . | . |      | . |___|      | . |   |      |_______|
|___|___|      |___| . |      |___|___|      | . | . |
|       |      |   |___|      |   | . |      |___|___|
|_______|      |_______|      |___|___|      |_______|
The first 6 rows of T(n,k) are:
.\ k  0    1    2    3    4    5    6    7    8    9
n
2     1    1
3     1    1
4     1    3    4    2    1
5     1    3   13   20   14
6     1    6   37  138  277  273  143   39    7    1
7     1    6   75  505 2154 5335 7855 6472 2756  459

Heinrich Ludwig, Table of n, a(n) for n = 2..107
Christopher Hunt Gribble, C++ program

Cf. A054252, A236560, A236757, A236800, A236829, A236865, A236915, A236936, A236939.
Row sums give A275869.
Columns 2..7: A279111, A279112, A279113, A279114, A279115, A279116.
Diagonal T(n,n) is A279117.
Cf. A193580.

It appears that:
T(n,0) = 1, n>= 2
T(n,1) = (floor((n-2)/2)+1)*(floor((n-2)/2+2))/2, n >= 2
T(c+2*2,2) = A131474(c+1)*(2-1) + A000217(c+1)*floor(2^2/4) + A014409(c+2), 0 <= c < 2, c even
T(c+2*2,2) = A131474(c+1)*(2-1) + A000217(c+1)*floor((2-1)(2-3)/4) + A014409(c+2), 0 <= c < 2, c odd
T(c+2*2,3) = (c+1)(c+2)/2(2*A002623(c-1)*floor((2-c-1)/2) + A131941(c+1)*floor((2-c)/2)) + S(c+1,3c+2,3), 0 <= c < 2 where
S(c+1,3c+2,3) =
A054252(2,3),  c = 0
A236679(5,3),  c = 1

More terms from Heinrich Ludwig, Dec 11 2016 (The former entry A279118 from Heinrich Ludwig was merged into this entry by N. J. A. Sloane, Dec 21 2016)

A082963 Number of n X n 0-1 matrices with half 1's and half 0's (rounded up/down if odd).

Original entry on oeis.org

1, 1, 2, 23, 1674, 652048, 1134460910, 7900674292378, 229078019084673798, 26549036304190836144544, 12611418068196090318131968752, 23955745839516317585042064530077352, 185026624806098273753009169783707528668060

Offset: 0

Vladeta Jovovic, May 27 2003

nonn

Andrew Howroyd, Table of n, a(n) for n = 0..50
James Grime, Maths Problem: Complete Noughts and Crosses (Burnside's Lemma)

Cf. A054247, A054252, A014409, A019318.

C(n,f)={(f(1)^(n^2) + 2*f(1)^((n%2)*n)*f(2)^((n\2)*n) + 2*f(1)^n*f(2)^binomial(n,2) + f(1)^(n%2)*f(2)^(n^2\2) + 2*f(1)^(n%2)*f(4)^(floor(n/2)*ceil(n/2)))/8}
a(n)={polcoef(C(n, k->1 + x^k), n^2\2)} \\ Andrew Howroyd, Feb 01 2020

a(n) = A054252(n, floor(n^2/2)).

Terms a(12) and beyond from Andrew Howroyd, Feb 01 2020

A054772 Triangle T(n,k) of n X n binary matrices with k=0..n^2 ones, up to rotational symmetry.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 10, 22, 34, 34, 22, 10, 3, 1, 1, 4, 32, 140, 464, 1092, 2016, 2860, 3238, 2860, 2016, 1092, 464, 140, 32, 4, 1, 1, 7, 78, 578, 3182, 13302, 44330, 120230, 270525, 510875, 817388, 1114548, 1300316, 1300316, 1114548, 817388

Offset: 0

Vladeta Jovovic, May 18 2000

Row sums give A047937.
From Wolfdieter Lang, Oct 01 2016: (Start)
The formula is obtained from Pólya's counting theorem. See, e.g., the Harary-Palmer reference.
The cycle index for a square grid of n X n  squares G(n), n >= 1, under the cyclic group C_4 is
  (s[1]^(n^2)+s[2]^(n^2/2)+2*s[4]^(n^2/4))/4 if n is even,
  s[1]*(s[1]^(n^2-1) + s[2]^((n^2-1)/2) + 2*s[4]^((n^2-1)/4))/4 if n is odd. (Numerate the squares from 1 .. n^2 and compute for the C_4 rotations the cycle structure of the permutation from the symmetric group S(n^2)).
The figure counting series is c(x) = 1+x for coloring, say black and white (in the matrix case binary entries).
Therefore the counting series is C(n,x) = G(n) with substitution s[2^j] = c(x^(2*j)) = 1 + x^(2^j) for j=0,1,2. Row n gives the coefficients of C(n,x) in rising (or falling) order. (End)
A pedantic note: One should not use 0,1 matrices for this T(n,k) model because 1 (also |) is not C_4 invariant. Square grids with coloring of the squares, say black and white, or central entries o and + are better suited. - Wolfdieter Lang, Oct 02 2016

			[1],[1,1],[1,1,2,1,1],[1,3,10,22,34,34,22,10,3,1],...;
There are 10 inequivalent 3 X 3 binary matrices with 2 ones, up to rotational symmetry:
[0 0 0] [0 0 0] [0 0 0] [0 0 0] [0 0 0]
[0 0 0] [0 0 0] [0 0 0] [0 0 1] [0 0 1]
[0 1 1] [1 0 1] [1 1 0] [0 1 0] [1 0 0]
-------
[0 0 0] [0 0 0] [0 0 0] [0 0 0] [0 0 1]
[0 1 0] [0 1 0] [1 0 0] [1 0 1] [0 0 0]
[0 0 1] [0 1 0] [0 0 1] [0 0 0] [1 0 0].
- reformatted. _Wolfdieter Lang_, Oct 01 2016
See a remark above: use o for 0 and + for 1.
n=3: Cycle index G(3) = s[1]*(s[1]^8 + s[2]^4 + 2*s[4]^2)/4. C(3,x) = (1+x)*((1+x)^8 + (1+x^2)^4 + 2*(1+x^4)^2)/4 = 1 + 3*x + 10*x^2 + 22*x^3 + 34*x^4 + 34*x^5 + 22*x^6 + 10*x^7 + 3*x^8 + x^9. - _Wolfdieter Lang_, Oct 01 2016

F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, NY, 1973, p. 42, (2.4.6).

Cf. A054252, columns k=0..4: A000012, A004652, A212714, A011863, A275799.

See the comment above: T(n,k) = [x^k]C(n,x), with the counting series C(n,x) obtained from the cycle index for the n X n grid under C_4 rotations G(n;s[1],s[2],s[4]) with s[2^j] = 1 + x^(2^j) for j=0,1,2. - Wolfdieter Lang, Oct 01 2016

A236560 Number T(n,k) of equivalence classes of ways of placing k 3 X 3 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=3, 0<=k<=floor(n/3)^2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 3, 6, 2, 1, 1, 6, 21, 29, 14, 1, 6, 53, 161, 174, 1, 10, 111, 665, 1713, 1549, 608, 107, 11, 1, 1, 10, 201, 1961, 9973, 24267, 29437, 17438, 4756, 459

Offset: 3

Christopher Hunt Gribble, Jan 30 2014

The first 8 rows of T(n,k) are:
.\ k  0     1     2     3     4     5     6     7     8     9
n
   1     1
   1     1
   1     3
   1     3     6     2     1
   1     6    21    29    14
   1     6    53   161   174
   1    10   111   665  1713  1549   608   107    11     1
  1    10   201  1961  9973 24267 29437 17438  4756   459

			T(6,2) = 6 because the number of equivalence classes of ways of placing 2 3 X 3 square tiles in a 6 X 6 square under all symmetry operations of the square is 6. The portrayal of an example from each equivalence class is:
.___________      ___________      ___________
|     |     |    |     |_____|    |     |     |
|  .  |  .  |    |  .  |     |    |  .  |_____|
|_____|_____|    |_____|  .  |    |_____|     |
|           |    |     |_____|    |     |  .  |
|           |    |           |    |     |_____|
|___________|    |___________|    |_____|_____|
.
.___________      ___________      ___________
|     |     |    |_____ _____|    |_____      |
|  .  |     |    |     |     |    |     |_____|
|_____|_____|    |  .  |  .  |    |  .  |     |
|     |     |    |_____|_____|    |_____|  .  |
|     |  .  |    |           |    |     |_____|
|_____|_____|    |___________|    |_____|_____|

Christopher Hunt Gribble, C++ program

Cf. A054252, A236679, A236757, A236800, A236829, A236865, A236915, A236936, A236939.

It appears that:
T(n,0) = 1, n>= 3
T(n,1) = (floor((n-3)/2)+1)*(floor((n-3)/2+2))/2, n >= 3
T(c+2*3,2) = A131474(c+1)*(3-1) + A000217(c+1)*floor(3^2/4) + A014409(c+2), 0 <= c < 3, c even
T(c+2*3,2) = A131474(c+1)*(3-1) + A000217(c+1)*floor((3-1)(3-3)/4) + A014409(c+2), 0 <= c < 3, c odd
T(c+2*3,3) = (c+1)(c+2)/2(2*A002623(c-1)*floor((3-c-1)/2) + A131941(c+1)*floor((3-c)/2)) + S(c+1,3c+2,3), 0 <= c < 3 where
S(c+1,3c+2,3) =
A054252(2,3),  c = 0
A236679(5,3),  c = 1
A236560(8,3),  c = 2

A236757 Number T(n,k) of equivalence classes of ways of placing k 4 X 4 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=4, 0<=k<=floor(n/4)^2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 3, 1, 6, 9, 3, 1, 1, 6, 29, 35, 14, 1, 10, 75, 209, 174, 1, 10, 147, 765, 1234, 1, 15, 270, 2340, 7639, 6169, 1893, 242, 17, 1, 1, 15, 438, 5806, 34342, 79821, 80722, 36569, 7106, 459

Offset: 4

Christopher Hunt Gribble, Jan 30 2014

The first 10 rows of T(n,k) are:
.\ k  0     1     2     3     4     5     6     7     8     9
n
   1     1
   1     1
   1     3
   1     3
   1     6     9     3     1
   1     6    29    35    14
  1    10    75   209   174
  1    10   147   765  1234
  1    15   270  2340  7639  6169  1893   242    17     1
  1    15   438  5806 34342 79821 80722 36569  7106   459

			T(8,3) = 3 because the number of equivalence classes of ways of placing 3 4 X 4 square tiles in an 8 X 8 square under all symmetry operations of the square is 3. The portrayal of an example from each equivalence class is:
._____________          _____________          _____________
|      |      |        |      |______|        |      |      |
|   .  |   .  |        |   .  |      |        |   .  |______|
|      |      |        |      |   .  |        |      |      |
|______|______|        |______|      |        |______|   .  |
|      |      |        |      |______|        |      |      |
|   .  |      |        |   .  |      |        |   .  |______|
|      |      |        |      |      |        |      |      |
|______|______|        |______|______|        |______|______|

Christopher Hunt Gribble, C++ program

Cf. A054252, A236679, A236560, A236800, A236829, A236865, A236915, A236936, A236939.

It appears that:
T(n,0) = 1, n>= 4
T(n,1) = (floor((n-4)/2)+1)*(floor((n-4)/2+2))/2, n >= 4
T(c+2*4,2) = A131474(c+1)*(4-1) + A000217(c+1)*floor(4^2/4) + A014409(c+2), 0 <= c < 4, c even
T(c+2*4,2) = A131474(c+1)*(4-1) + A000217(c+1)*floor((4-1)(4-3)/4) + A014409(c+2), 0 <= c < 4, c odd
T(c+2*4,3) = (c+1)(c+2)/2(2*A002623(c-1)*floor((4-c-1)/2) + A131941(c+1)*floor((4-c)/2)) + S(c+1,3c+2,3), 0 <= c < 4 where
S(c+1,3c+2,3) =
A054252(2,3),  c = 0
A236679(5,3),  c = 1
A236560(8,3),  c = 2
A236757(11,3), c = 3

A236800 Number T(n,k) of equivalence classes of ways of placing k 5 X 5 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=5, 0<=k<=floor(n/5)^2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 3, 1, 6, 1, 6, 12, 3, 1, 1, 10, 40, 44, 14, 1, 10, 97, 245, 174, 1, 15, 193, 925, 1234, 1, 15, 339, 2640, 6124, 1, 21, 555, 6617, 27074, 19336, 4785, 461, 23, 1

Offset: 5

Christopher Hunt Gribble, Jan 31 2014

The first 11 rows of T(n,k) are:
.\ k  0     1     2     3     4     5     6     7     8     9
n
   1     1
   1     1
   1     3
   1     3
   1     6
  1     6    12     3     1
  1    10    40    44    14
  1    10    97   245   174
  1    15   193   925  1234
  1    15   339  2640  6124
  1    21   555  6617 27074 19336  4785   461    23     1

			T(10,3) = 3 because the number of equivalence classes of ways of placing 3 5 X 5 square tiles in an 10 X 10 square under all symmetry operations of the square is 3. The portrayal of an example from each equivalence class is:
._______________      _______________      _______________
|       |       |    |       |_______|    |       |       |
|       |       |    |       |       |    |       |_______|
|   .   |   .   |    |   .   |       |    |   .   |       |
|       |       |    |       |   .   |    |       |       |
|_______|_______|    |_______|       |    |_______|   .   |
|       |       |    |       |_______|    |       |       |
|       |       |    |       |       |    |       |_______|
|   .   |       |    |   .   |       |    |   .   |       |
|       |       |    |       |       |    |       |       |
|_______|_______|    |_______|_______|    |_______|_______|

Christopher Hunt Gribble, C++ program

Cf. A054252, A236679, A236560, A236757, A236829, A236865, A236915, A236936, A236939.

It appears that:
T(n,0) = 1, n>= 5
T(n,1) = (floor((n-5)/2)+1)*(floor((n-5)/2+2))/2, n >= 5
T(c+2*5,2) = A131474(c+1)*(5-1) + A000217(c+1)*floor(5^2/4) + A014409(c+2), 0 <= c < 5, c even
T(c+2*5,2) = A131474(c+1)*(5-1) + A000217(c+1)*floor((5-1)(5-3)/4) + A014409(c+2), 0 <= c < 5, c odd
T(c+2*5,3) = (c+1)(c+2)/2(2*A002623(c-1)*floor((5-c-1)/2) + A131941(c+1)*floor((5-c)/2)) + S(c+1,3c+2,3), 0 <= c < 5 where
S(c+1,3c+2,3) =
A054252(2,3),  c = 0
A236679(5,3),  c = 1
A236560(8,3),  c = 2
A236757(11,3), c = 3
A236800(14,3), c = 4

A236829 Number T(n,k) of equivalence classes of ways of placing k 6 X 6 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=6, 0<=k<=floor(n/6)^2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 3, 1, 6, 1, 6, 1, 10, 16, 4, 1, 1, 10, 51, 50, 14, 1, 15, 125, 293, 174, 1, 15, 239, 1065, 1234, 1, 21, 423, 3075, 6124, 1, 21, 672, 7371, 23259, 1, 28, 1030, 16093, 81480, 51615, 10596, 808, 31, 1

Offset: 6

Christopher Hunt Gribble, Jan 31 2014

The first 13 rows of T(n,k) are:
.\ k  0     1     2     3     4     5     6     7     8     9
n
   1     1
   1     1
   1     3
   1     3
  1     6
  1     6
  1    10    16     4     1
  1    10    51    50    14
  1    15   125   293   174
  1    15   239  1065  1234
  1    21   423  3075  6124
  1    21   672  7371 23259
  1    28  1030 16093 81480 51615 10596   808    31     1

			T(12,3) = 4 because the number of equivalence classes of ways of placing 3 6 X 6 square tiles in a 12 X 12 square under all symmetry operations of the square is 4. The portrayal of an example from each equivalence class is:
._________________          _________________
|        |        |        |        |________|
|        |        |        |        |        |
|    .   |    .   |        |    .   |        |
|        |        |        |        |    .   |
|        |        |        |        |        |
|________|________|        |________|        |
|        |        |        |        |________|
|        |        |        |        |        |
|    .   |        |        |    .   |        |
|        |        |        |        |        |
|        |        |        |        |        |
|________|________|        |________|________|
.
._________________          _________________
|        |        |        |        |        |
|        |________|        |        |        |
|    .   |        |        |    .   |________|
|        |        |        |        |        |
|        |    .   |        |        |        |
|________|        |        |________|    .   |
|        |        |        |        |        |
|        |________|        |        |        |
|    .   |        |        |    .   |________|
|        |        |        |        |        |
|        |        |        |        |        |
|________|________|        |________|________|

Christopher Hunt Gribble, C++ program

Cf. A054252, A236679, A236560, A236757, A236800, A236865, A236915, A236936, A236939.

It appears that:
T(n,0) = 1, n>= 6
T(n,1) = (floor((n-6)/2)+1)*(floor((n-6)/2+2))/2, n >= 6
T(c+2*6,2) = A131474(c+1)*(6-1) + A000217(c+1)*floor(6^2/4) + A014409(c+2), 0 <= c < 6, c even
T(c+2*6,2) = A131474(c+1)*(6-1) + A000217(c+1)*floor((6-1)(6-3)/4) + A014409(c+2), 0 <= c < 6, c odd
T(c+2*6,3) = (c+1)(c+2)/2(2*A002623(c-1)*floor((6-c-1)/2) + A131941(c+1)*floor((6-c)/2)) + S(c+1,3c+2,3), 0 <= c < 6 where
S(c+1,3c+2,3) =
A054252(2,3),  c = 0
A236679(5,3),  c = 1
A236560(8,3),  c = 2
A236757(11,3), c = 3
A236800(14,3), c = 4
A236829(17,3), c = 5

A236865 Number T(n,k) of equivalence classes of ways of placing k 7 X 7 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=7, 0<=k<=floor(n/7)^2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 3, 1, 6, 1, 6, 1, 10, 1, 10, 20, 4, 1, 1, 15, 65, 59, 14, 1, 15, 153, 329, 174, 1, 21, 295, 1225, 1234, 1, 21, 507, 3465, 6124, 1, 28, 810, 8358, 23259, 1, 28, 1214, 17710, 73204

Offset: 7

Christopher Hunt Gribble, Jan 31 2014

The first 13 rows of T(n,k) are:
.\ k  0     1     2     3     4     5     6     7     8     9
n
   1     1
   1     1
   1     3
  1     3
  1     6
  1     6
  1    10
  1    10    20     4     1
  1    15    65    59    14
  1    15   153   329   174
  1    21   295  1225  1234
  1    21   507  3465  6124
  1    28   810  8358 23259
  1    28  1214 17710 73204

			T(14,3) = 4 because the number of equivalent classes of ways of placing 3 7 X 7 square tiles in an 14 X 14 square under all symmetry operations of the square is 4. The portrayal of an example from each equivalence class is:
.___________________          ___________________
|         |         |        |         |_________|
|         |         |        |         |         |
|         |         |        |         |         |
|    .    |    .    |        |    .    |         |
|         |         |        |         |    .    |
|         |         |        |         |         |
|_________|_________|        |_________|         |
|         |         |        |         |_________|
|         |         |        |         |         |
|         |         |        |         |         |
|    .    |         |        |    .    |         |
|         |         |        |         |         |
|         |         |        |         |         |
|_________|_________|        |_________|_________|
.
.___________________          ___________________
|         |         |        |         |         |
|         |_________|        |         |         |
|         |         |        |         |_________|
|    .    |         |        |    .    |         |
|         |         |        |         |         |
|         |    .    |        |         |         |
|_________|         |        |_________|    .    |
|         |         |        |         |         |
|         |_________|        |         |         |
|         |         |        |         |_________|
|    .    |         |        |    .    |         |
|         |         |        |         |         |
|         |         |        |         |         |
|_________|_________|        |_________|_________|

Christopher Hunt Gribble, C++ program

Cf. A054252, A236679, A236560, A236757, A236800, A236829, A236915, A236936, A236939.

It appears that:
T(n,0) = 1, n>= 7
T(n,1) = (floor((n-7)/2)+1)*(floor((n-7)/2+2))/2, n >= 7
T(c+2*7,2) = A131474(c+1)*(7-1) + A000217(c+1)*floor(7^2/4) + A014409(c+2), 0 <= c < 7, c even
T(c+2*7,2) = A131474(c+1)*(7-1) + A000217(c+1)*floor((7-1)(7-3)/4) + A014409(c+2), 0 <= c < 7, c odd
T(c+2*7,3) = (c+1)(c+2)/2(2*A002623(c-1)*floor((7-c-1)/2) + A131941(c+1)*floor((7-c)/2)) + S(c+1,3c+2,3), 0 <= c < 7 where
S(c+1,3c+2,3) =
A054252(2,3),  c = 0
A236679(5,3),  c = 1
A236560(8,3),  c = 2
A236757(11,3), c = 3
A236800(14,3), c = 4
A236829(17,3), c = 5
A236865(20,3), c = 6

cp's OEIS Frontend

A008805 Triangular numbers repeated.

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A014409 Number of inequivalent ways (mod D_4) a pair of checkers can be placed on an n X n board.

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A236679 Number T(n,k) of equivalence classes of ways of placing k 2 X 2 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=2, 0<=k<=floor(n/2)^2, read by rows.

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A082963 Number of n X n 0-1 matrices with half 1's and half 0's (rounded up/down if odd).

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A054772 Triangle T(n,k) of n X n binary matrices with k=0..n^2 ones, up to rotational symmetry.

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A236560 Number T(n,k) of equivalence classes of ways of placing k 3 X 3 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=3, 0<=k<=floor(n/3)^2, read by rows.

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A236757 Number T(n,k) of equivalence classes of ways of placing k 4 X 4 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=4, 0<=k<=floor(n/4)^2, read by rows.

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