A011863 -id:A011863 - cp's OEIS frontend

A006003 a(n) = n*(n^2 + 1)/2.

0, 1, 5, 15, 34, 65, 111, 175, 260, 369, 505, 671, 870, 1105, 1379, 1695, 2056, 2465, 2925, 3439, 4010, 4641, 5335, 6095, 6924, 7825, 8801, 9855, 10990, 12209, 13515, 14911, 16400, 17985, 19669, 21455, 23346, 25345, 27455, 29679, 32020, 34481, 37065, 39775

Offset: 0

N. J. A. Sloane, Simon Plouffe

Write the natural numbers in groups: 1; 2,3; 4,5,6; 7,8,9,10; ... and add the groups. In other words, "sum of the next n natural numbers". - Felice Russo
Number of rhombi in an n X n rhombus, if 'crossformed' rhombi are allowed. - Matti De Craene (Matti.DeCraene(AT)rug.ac.be), May 14 2000
Also the sum of the integers between T(n-1)+1 and T(n), the n-th triangular number (A000217). Sum of n-th row of A000027 regarded as a triangular array.
Unlike the cubes which have a similar definition, it is possible for 2 terms of this sequence to sum to a third. E.g., a(36) + a(37) = 23346 + 25345 = 48691 = a(46). Might be called 2nd-order triangular numbers, thus defining 3rd-order triangular numbers (A027441) as n(n^3+1)/2, etc. - Jon Perry, Jan 14 2004
Also as a(n)=(1/6)*(3*n^3+3*n), n > 0: structured trigonal diamond numbers (vertex structure 4) (cf. A000330 = alternate vertex; A000447 = structured diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
The sequence M(n) of magic constants for n X n magic squares (numbered 1 through n^2) from n=3 begins M(n) = 15, 34, 65, 111, 175, 260, ... - Lekraj Beedassy, Apr 16 2005 [comment corrected by Colin Hall, Sep 11 2009]
The sequence Q(n) of magic constants for the n-queens problem in chess begins 0, 0, 0, 0, 34, 65, 111, 175, 260, ... - Paul Muljadi, Aug 23 2005
Alternate terms of A057587. - Jeremy Gardiner, Apr 10 2005
Also partial differences of A063488(n) = (2*n-1)*(n^2-n+2)/2. a(n) = A063488(n) - A063488(n-1) for n>1. - Alexander Adamchuk, Jun 03 2006
In an n X n grid of numbers from 1 to n^2, select -- in any manner -- one number from each row and column. Sum the selected numbers. The sum is independent of the choices and is equal to the n-th term of this sequence. - F.-J. Papp (fjpapp(AT)umich.edu), Jun 06 2006
Nonnegative X values of solutions to the equation (X-Y)^3 - (X+Y) = 0. To find Y values: b(n) = (n^3-n)/2. - Mohamed Bouhamida, May 16 2006
For the equation: m*(X-Y)^k - (X+Y) = 0 with X >= Y, k >= 2 and m is an odd number the X values are given by the sequence defined by a(n) = (m*n^k+n)/2. The Y values are given by the sequence defined by b(n) = (m*n^k-n)/2. - Mohamed Bouhamida, May 16 2006
If X is an n-set and Y a fixed 3-subset of X then a(n-3) is equal to the number of 4-subsets of X intersecting Y. - Milan Janjic, Jul 30 2007
(m*(2n)^k+n, m*(2n)^k-n) solves the Diophantine equation: 2m*(X-Y)^k - (X+Y) = 0 with X >= Y, k >= 2 where m is a positive integer. - Mohamed Bouhamida, Oct 02 2007
Also c^(1/2) in a^(1/2) + b^(1/2) = c^(1/2) such that a^2 + b = c. - Cino Hilliard, Feb 09 2008
a(n) = n*A000217(n) - Sum_{i=0..n-1} A001477(i). - Bruno Berselli, Apr 25 2010
a(n) is the number of triples (w,x,y) having all terms in {0,...,n} such that at least one of these inequalities fails: x+y < w, y+w < x, w+x < y. - Clark Kimberling, Jun 14 2012
Sum of n-th row of the triangle in A209297. - Reinhard Zumkeller, Jan 19 2013
The sequence starting with "1" is the third partial sum of (1, 2, 3, 3, 3, ...). - Gary W. Adamson, Sep 11 2015
a(n) is the largest eigenvalue of the matrix returned by the MATLAB command magic(n) for n > 0. - Altug Alkan, Nov 10 2015
a(n) is the number of triples (x,y,z) having all terms in {1,...,n} such that all these triangle inequalities are satisfied: x+y > z, y+z > x, z+x > y. - Heinz Dabrock, Jun 03 2016
Shares its digital root with the stella octangula numbers (A007588). See A267017. - Peter M. Chema, Aug 28 2016
Can be proved to be the number of nonnegative solutions of a system of three linear Diophantine equations for n >= 0 even: 2*a_{11} + a_{12} + a_{13} = n, 2*a_{22} + a_{12} + a_{23} = n and 2*a_{33} + a_{13} + a_{23} = n. The number of solutions is f(n) = (1/16)*(n+2)*(n^2 + 4n + 8) and a(n) = n*(n^2 + 1)/2 is obtained by remapping n -> 2*n-2. - Kamil Bradler, Oct 11 2016
For n > 0, a(n) coincides with the trace of the matrix formed by writing the numbers 1...n^2 back and forth along the antidiagonals (proved, see A078475 for the examples of matrix). - Stefano Spezia, Aug 07 2018
The trace of an n X n square matrix where the elements are entered on the ascending antidiagonals. The determinant is A069480. - Robert G. Wilson v, Aug 07 2018
Bisections are A317297 and A005917. - Omar E. Pol, Sep 01 2018
Number of achiral colorings of the vertices (or faces) of a regular tetrahedron with n available colors. An achiral coloring is identical to its reflection. - Robert A. Russell, Jan 22 2020
a(n) is the n-th centered triangular pyramidal number. - Lechoslaw Ratajczak, Nov 02 2021
a(n) is the number of words of length n defined on 4 letters {b,c,d,e} that contain one or no b's, one c or two d's, and any number of e's. For example, a(3) = 15 since the words are (number of permutations in parentheses): bce (6), bdd (3), cee (3), and dde (3). - Enrique Navarrete, Jun 21 2025

			G.f. = x + 5*x^2 + 15*x^3 + 34*x^4 + 65*x^5 + 111*x^6 + 175*x^7 + 260*x^8 + ...
For a(2)=5, the five tetrahedra have faces AAAA, AAAB, AABB, ABBB, and BBBB with colors A and B. - _Robert A. Russell_, Jan 31 2020

J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 15, p. 5, Ellipses, Paris 2008.
F.-J. Papp, Colloquium Talk, Department of Mathematics, University of Michigan-Dearborn, March 6, 2005.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..1000
J. D. Bell, A translation of Leonhard Euler's "De Quadratis Magicis", E795, arXiv:math/0408230 [math.CO], 2004-2005.
James Grime and Brady Haran, Magic Hexagon, Numberphile video (2014).
Milan Janjic, Two Enumerative Functions.
Milan Janjic and Boris Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013. - _N. J. A. Sloane_, Feb 13 2013
Milan Janjic and Boris Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, Journal of Integer Sequences, 17 (2014), Article 14.3.5. - _Felix Fröhlich_, Oct 11 2016
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber. 30 (1897), 1917-1926.
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber. 30 (1897), 1917-1926. (Annotated scanned copy)
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (11).
Ashish Kumar Pandey and Brajesh Kumar Sharma, A Note On Magic Squares And Magic Constants, Appl. Math. E-Notes (2023) Vol. 23, Art. No. 53, 577-582. See p. 577.
A. J. Turner and J. F. Miller, Recurrent Cartesian Genetic Programming Applied to Famous Mathematical Sequences, preprint, Proceedings of the Companion Publication of the 2015 Annual Conference on Genetic and Evolutionary Computation.
Eric Weisstein's World of Mathematics, Magic Constant.
Wikipedia, Floyd's triangle. - _Paul Muljadi_, Jan 25 2010
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Index entries for sequences related to magic squares.
Index to sequences related to polygonal numbers.

Cf. A000330, A000537, A066886, A057587, A027480, A002817 (partial sums).
Cf. A000578 (cubes).
Cf. A007742, A005449, A005448, A118465, A226449, A034262, A080992, A267017.
(1/12)*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, this sequence, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Antidiagonal sums of array in A000027. Row sums of the triangular view of A000027.
Cf. A063488 (sum of two consecutive terms), A005917 (bisection), A317297 (bisection).
Cf. A105374 / 8.
Cf. A000292, A011863, A001620, A248177.
Tetrahedron colorings: A006008 (oriented), A000332(n+3) (unoriented), A000332 (chiral), A037270 (edges).
Other polyhedron colorings: A337898 (cube faces, octahedron vertices), A337897 (octahedron faces, cube vertices), A337962 (dodecahedron faces, icosahedron vertices), A337960 (icosahedron faces, dodecahedron vertices).
Row 3 of A325001 (simplex vertices and facets) and A337886 (simplex faces and peaks).

a_n:=List([0..nmax], n->n*(n^2 + 1)/2); # Stefano Spezia, Aug 12 2018

a006003 n = n * (n ^ 2 + 1) `div` 2
a006003_list = scanl (+) 0 a005448_list
-- Reinhard Zumkeller, Jun 20 2013

% Also works with FreeMat.
for(n=0:nmax); tm=n*(n^2 + 1)/2; fprintf('%d\t%0.f\n', n, tm); end
% Stefano Spezia, Aug 12 2018

[n*(n^2 + 1)/2 : n in [0..50]]; // Wesley Ivan Hurt, Sep 11 2015

[Binomial(n,3)+Binomial(n-1,3)+Binomial(n-2,3): n in [2..60]]; // Vincenzo Librandi, Sep 12 2015

Table[ n(n^2 + 1)/2, {n, 0, 45}]
LinearRecurrence[{4,-6,4,-1}, {0,1,5,15},50] (* Harvey P. Dale, May 16 2012 *)
CoefficientList[Series[x (1 + x + x^2)/(x - 1)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Sep 12 2015 *)
With[{n=50},Total/@TakeList[Range[(n(n^2+1))/2],Range[0,n]]] (* Requires Mathematica version 11 or later *) (* Harvey P. Dale, Nov 28 2017 *)

a(n):=n*(n^2 + 1)/2$ makelist(a(n), n, 0, nmax); /* Stefano Spezia, Aug 12 2018 */

{a(n) = n * (n^2 + 1) / 2}; /* Michael Somos, Dec 24 2011 */

concat(0, Vec(x*(1+x+x^2)/(x-1)^4 + O(x^20))) \\ Felix Fröhlich, Oct 11 2016

def A006003(n): return n*(n**2+1)>>1 # Chai Wah Wu, Mar 25 2024

a(n) = binomial(n+2, 3) + binomial(n+1, 3) + binomial(n, 3). [corrected by Michel Marcus, Jan 22 2020]
G.f.: x*(1+x+x^2)/(x-1)^4. - Floor van Lamoen, Feb 11 2002
Partial sums of A005448. - Jonathan Vos Post, Mar 16 2006
Binomial transform of [1, 4, 6, 3, 0, 0, 0, ...] = (1, 5, 15, 34, 65, ...). - Gary W. Adamson, Aug 10 2007
a(n) = -a(-n) for all n in Z. - Michael Somos, Dec 24 2011
a(n) = Sum_{k = 1..n} A(k-1, k-1-n) where A(i, j) = i^2 + i*j + j^2 + i + j + 1. - Michael Somos, Jan 02 2012
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), with a(0)=0, a(1)=1, a(2)=5, a(3)=15. - Harvey P. Dale, May 16 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 3. - Ant King, Jun 13 2012
a(n) = A000217(n) + n*A000217(n-1). - Bruno Berselli, Jun 07 2013
a(n) = A057145(n+3,n). - Luciano Ancora, Apr 10 2015
E.g.f.: (1/2)*(2*x + 3*x^2 + x^3)*exp(x). - G. C. Greubel, Dec 18 2015; corrected by Ilya Gutkovskiy, Oct 12 2016
a(n) = T(n) + T(n-1) + T(n-2), where T means the tetrahedral numbers, A000292. - Heinz Dabrock, Jun 03 2016
From Ilya Gutkovskiy, Oct 11 2016: (Start)
Convolution of A001477 and A008486.
Convolution of A000217 and A158799.
Sum_{n>=1} 1/a(n) = H(-i) + H(i) = 1.343731971048019675756781..., where H(k) is the harmonic number, i is the imaginary unit. (End)
a(n) = A000578(n) - A135503(n). - Miquel Cerda, Dec 25 2016
Euler transform of length 3 sequence [5, 0, -1]. - Michael Somos, Dec 25 2016
a(n) = A037270(n)/n for n > 0. - Kritsada Moomuang, Dec 15 2018
a(n) = 3*A000292(n-1) + n. - Bruce J. Nicholson, Nov 23 2019
a(n) = A011863(n) - A011863(n-2). - Bruce J. Nicholson, Dec 22 2019
From Robert A. Russell, Jan 22 2020: (Start)
a(n) = C(n,1) + 3*C(n,2) + 3*C(n,3), where the coefficient of C(n,k) is the number of tetrahedron colorings using exactly k colors.
a(n) = C(n+3,4) - C(n,4).
a(n) = 2*A000332(n+3) - A006008(n) = A006008(n) - 2*A000332(n) = A000332(n+3) - A000332(n).
a(n) = A325001(3,n). (End)
From Amiram Eldar, Aug 21 2023: (Start)
Sum_{n>=1} 1/a(n) = 2 * (A248177 + A001620).
Product_{n>=2} (1 - 1/a(n)) = cosh(sqrt(7)*Pi/2)*cosech(Pi)/4.
Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(7)*Pi/2)*cosech(Pi). (End)

Better description from Albert Rich (Albert_Rich(AT)msn.com), Mar 1997

A188392 T(n,k) = number of (n*k) X k binary arrays with rows in nonincreasing order and n ones in every column.

Original entry on oeis.org

1, 2, 1, 5, 3, 1, 15, 16, 4, 1, 52, 139, 39, 5, 1, 203, 1750, 862, 81, 6, 1, 877, 29388, 35775, 4079, 150, 7, 1, 4140, 624889, 2406208, 507549, 15791, 256, 8, 1, 21147, 16255738, 238773109, 127126912, 5442547, 52450, 410, 9, 1, 115975, 504717929, 32867762616

Offset: 1

R. H. Hardin, Mar 30 2011

			Array begins:
========================================================================
n\k| 1  2   3      4         5            6            7               8
---+--------------------------------------------------------------------
1  | 1  2   5     15        52          203           877           4140
2  | 1  3  16    139      1750        29388        624889       16255738
3  | 1  4  39    862     35775      2406208     238773109    32867762616
4  | 1  5  81   4079    507549    127126912   55643064708 38715666455777
5  | 1  6 150  15791   5442547   4762077620 8738543204786
6  | 1  7 256  52450  46757209 135029200594
7  | 1  8 410 154279 335279744
8  | 1  9 625 411180
9  | 1 10 915
     ...
All solutions for 6 X 2
..1..1....1..1....1..0....1..1
..1..1....1..1....1..0....1..0
..1..0....1..1....1..0....1..0
..0..1....0..0....0..1....0..1
..0..0....0..0....0..1....0..1
..0..0....0..0....0..1....0..0

Andrew Howroyd, Table of n, a(n) for n = 1..181 (terms 1..69 from R. H. Hardin)

Rows 1..8 are A000110, A020554, A165434, A165435, A165436, A165437, A188393, A188394.
Columns 3..7 are A011863(n+1), A175707, A188389, A188390, A188391.
Main diagonal gives A188388.
Cf. A188445, A219727, A330942.

WeighT(v)={Vec(exp(x*Ser(dirmul(v, vector(#v,n,(-1)^(n-1)/n))))-1,-#v)}
D(p, n, k)={my(v=vector(n)); for(i=1, #p, v[p[i]]++); WeighT(v)[n]^k/prod(i=1, #v, i^v[i]*v[i]!)}
T(n, k)={my(m=n*k, q=Vec(exp(O(x*x^m) + intformal((x^n-1)/(1-x)))/(1-x))); if(n==0, 1, sum(j=0, m, my(s=0); forpart(p=j, s+=D(p,n,k), [1,n]); s*q[#q-j]))} \\ Andrew Howroyd, Dec 12 2018

A188416 T(n,k)=Number of (n*k)Xk binary arrays with rows in nonincreasing order, n ones in every column and no more than 3 ones in any row.

Original entry on oeis.org

1, 2, 1, 5, 3, 1, 14, 16, 4, 1, 46, 124, 39, 5, 1, 166, 1308, 723, 81, 6, 1, 652, 17443, 21965, 3217, 150, 7, 1, 2780, 282349, 968230, 252493, 11712, 256, 8, 1, 12644, 5396178, 57967660, 33313408, 2181522, 36659, 410, 9, 1, 61136, 119324602, 4505774905

Offset: 1

R. H. Hardin Mar 30 2011

Table starts
.1..2....5.....14........46..........166............652............2780
.1..3...16....124......1308........17443.........282349.........5396178
.1..4...39....723.....21965.......968230.......57967660......4505774905
.1..5...81...3217....252493.....33313408.....6720357785...1943505834628
.1..6..150..11712...2181522....793369090...498407294632.497114063711190
.1..7..256..36659..15083334..14100262412.25784105550034
.1..8..410.101829..87171345.197263686098
.1..9..625.256901.434626896
.1.10..915.598561
.1.11.1296

			All solutions for 6X2
..1..1....1..1....1..0....1..1
..1..0....1..1....1..0....1..1
..1..0....1..1....1..0....1..0
..0..1....0..0....0..1....0..1
..0..1....0..0....0..1....0..0
..0..0....0..0....0..1....0..0

R. H. Hardin, Table of n, a(n) for n = 1..80

Column 3 is A011863(n+1)

Row 1 is A001680

A219086 a(n) = floor((n + 1/2)^4).

Original entry on oeis.org

0, 5, 39, 150, 410, 915, 1785, 3164, 5220, 8145, 12155, 17490, 24414, 33215, 44205, 57720, 74120, 93789, 117135, 144590, 176610, 213675, 256289, 304980, 360300, 422825, 493155, 571914, 659750, 757335, 865365, 984560, 1115664

Offset: 0

Clark Kimberling, Jan 01 2013

a(n) is the number k such that {k^p} < 1/2 < {(k+1)^p}, where p = 1/4 and { } = fractional part.  Equivalently, the jump sequence of f(x) = x^(1/4), in the sense that these are the nonnegative integers k for which round(k^p) < round((k+1)^p).  For details and a guide to related sequences, see A219085.
-4*a(n) gives the real part of (n+n*i)*((n+1)+n*i)*(n+(n+1)*i)*((n+1)+(n+1)*i). The imaginary part is always zero. - Jon Perry, Feb 05 2014
Numbers k such that 16*k+1 is a fourth power. - Bruno Berselli, May 29 2018
The row sums of "Floyd's Triangle", which is a triangular array of natural numbers beginning with the number 1, produce the sequence A006003. A006003 can be bisected to get the Rhombic Dodecahedron Sequence A005917, whose n-th partial sum is n^4, and A317297, whose n-th partial sum is a(n). Interleave n^4 or A000583 back with {a(n)} to get A011863, whose first differences are A019298. Finally, A011863(n)-A011863(n-2) = A006003(n-1). - Bruce J. Nicholson, Dec 22 2019

			0^(1/4) = 0.000...; 1^(1/4) = 1.000...
5^(1/4) = 1.495...; 6^(1/4) = 1.565...
39^(1/4) = 2.499...; 40^(1/4) = 2.514...

Clark Kimberling, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A123865, A219085, A234459, A000217, A001844.

Cf. A317297, A006003, A005917, A000583, A011863, A019298.

A219086:=n->floor((n + (1/2))^4); seq(A219086(n), n=0..50); # Wesley Ivan Hurt, Apr 05 2014

Table[Floor[(n + 1/2)^4], {n, 0, 100}]
LinearRecurrence[{5,-10,10,-5,1},{0,5,39,150,410},40] (* Harvey P. Dale, Jan 15 2023 *)

a(n)=floor((n + 1/2)^4) \\ Charles R Greathouse IV, Apr 15 2014

G.f.: (5*x^3 + 14*x^2 + 5*x)/(1 - x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = (2*n^4 + 4*n^3 + 3*n^2 + n)/2. - J. M. Bergot, Apr 05 2014
a(n) = Sum_{i=0..n} i*(4*i^2 + 1) = n*(n + 1)*(2*n^2 + 2*n + 1)/2. - Bruno Berselli, Feb 09 2017
a(n) = lcm((2*n + 1)^2 - 1, (2*n + 1)^2 + 1)/8 for n>=1. - Lechoslaw Ratajczak, Mar 26 2017
a(n) = A000217(n) * A001844(n). - Bruce J. Nicholson, May 14 2017
E.g.f.: (1/2)*exp(x)*x*(10 + 29*x + 16*x^2 + 2*x^3). - Stefano Spezia, Dec 27 2019
a(n) = ((2*n+1)^4 - 1)/16. - Jianing Song, Jan 03 2023
Sum_{n>=1} 1/a(n) = 6 - 2*Pi*tanh(Pi/2). - Amiram Eldar, Jan 08 2023

A055898 Triangle: Number of directed site animals on a square lattice with n+1 total sites and k sites supported in one particular way.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 5, 1, 1, 10, 16, 7, 1, 1, 15, 39, 31, 9, 1, 1, 21, 81, 101, 51, 11, 1, 1, 28, 150, 272, 209, 76, 13, 1, 1, 36, 256, 636, 696, 376, 106, 15, 1, 1, 45, 410, 1340, 1980, 1496, 615, 141, 17, 1, 1, 55, 625, 2600, 5000, 5032, 2850, 939, 181, 19, 1, 1

Offset: 0

Christian G. Bower, Jun 13 2000

			Triangle begins:
  1;
  1, 1;
  1, 3, 1;
  1, 6, 5, 1;
  1,10,16, 7, 1;
  ...

A. J. Guttmann, Indicators of solvability for lattice models, Discrete Math., 217 (2000), 167-189 (A_sq of Section 6).
M. Bousquet-Mélou, New enumerative results on two-dimensional directed animals
M. Bousquet-Mélou, New enumerative results on two-dimensional directed animals, Discr. Math., 180 (1998), 73-106.

Row sums give A005773. Columns 0-8: A000012, A000217, A011863(n-1), A055899-A055904. Cf. A055905, A055907.

nmax = 10;
A[x_, y_] = (1/2) x ((1 - (4 x/((1 + x) (1 + x - x y))))^(-1/2) - 1);
g = A[x, y] + O[x]^(nmax+3);
row[n_] := CoefficientList[Coefficient[g, x, n+2], y];
Table[row[n], {n, 0, nmax}] // Flatten (* Jean-François Alcover, Jul 24 2018 *)

T(n,m):=sum(binomial(n-k,m)*sum(binomial(n-m-k,i)*(-1)^(n+m-i)*binomial(k+i,k)*binomial(2*i+1,i+1),i,0,n-m-k),k,0,n); /* Vladimir Kruchinin, Jan 26 2022 */

G.f.: A(x, y)=(1/2x)((1-(4x/((1+x)(1+x-xy))))^(-1/2) - 1).

T(n,m) = Sum_{k=0..n} C(n-k,m)*Sum_{i=0..n-m-k} (-1)^(n+m-i) *C(n-m-k,i) *C(k+i,k) *C(2*i+1,i+1). - Vladimir Kruchinin, Jan 26 2022

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

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