A037224 -id:A037224 - cp's OEIS frontend

A001813 Quadruple factorial numbers: a(n) = (2n)!/n!.

1, 2, 12, 120, 1680, 30240, 665280, 17297280, 518918400, 17643225600, 670442572800, 28158588057600, 1295295050649600, 64764752532480000, 3497296636753920000, 202843204931727360000, 12576278705767096320000, 830034394580628357120000, 58102407620643984998400000

Offset: 0

N. J. A. Sloane

Counts binary rooted trees (with out-degree <= 2), embedded in plane, with n labeled end nodes of degree 1. Unlabeled version gives Catalan numbers A000108.
Define a "downgrade" to be the permutation which places the items of a permutation in descending order. We are concerned with permutations that are identical to their downgrades. Only permutations of order 4n and 4n+1 can have this property; the number of permutations of length 4n having this property are equinumerous with those of length 4n+1. If a permutation p has this property then the reversal of this permutation also has it. a(n) = number of permutations of length 4n and 4n+1 that are identical to their downgrades. - Eugene McDonnell (eemcd(AT)mac.com), Oct 26 2003
Number of broadcast schemes in the complete graph on n+1 vertices, K_{n+1}. - Calin D. Morosan (cd_moros(AT)alumni.concordia.ca), Nov 28 2008
Hankel transform is A137565. - Paul Barry, Nov 25 2009
The e.g.f. of 1/a(n) = n!/(2*n)! is (exp(sqrt(x)) + exp(-sqrt(x)) )/2. - Wolfdieter Lang, Jan 09 2012
From Tom Copeland, Nov 15 2014: (Start)
Aerated with intervening zeros (1,0,2,0,12,0,120,...) = a(n) (cf. A123023 and A001147), the e.g.f. is e^(t^2), so this is the base for the Appell sequence with e.g.f. e^(t^2) e^(x*t) = exp(P(.,x),t) (reverse A059344, cf. A099174, A066325 also). P(n,x) = (a. + x)^n with (a.)^n = a_n and comprise the umbral compositional inverses for e^(-t^2)e^(x*t) = exp(UP(.,x),t), i.e., UP(n,P(.,t)) = x^n = P(n,UP(.,t)), e.g., (P(.,t))^n = P(n,t).
Equals A000407*2 with leading 1 added. (End)
a(n) is also the number of square roots of any permutation in S_{4*n} whose disjoint cycle decomposition consists of 2*n transpositions. - Luis Manuel Rivera Martínez, Mar 04 2015
Self-convolution gives A076729. - Vladimir Reshetnikov, Oct 11 2016
For n > 1, it follows from the formula dated Aug 07 2013 that a(n) is a Zumkeller number (A083207). - Ivan N. Ianakiev, Feb 28 2017
For n divisible by 4, a(n/4) is the number of ways to place n points on an n X n grid with pairwise distinct abscissae, pairwise distinct ordinates, and 90-degree rotational symmetry. For n == 1 (mod 4), the number of ways is a((n-1)/4) because the center point can be considered "fixed". For 180-degree rotational symmetry see A006882, for mirror symmetry see A000085, A135401, and A297708. - Manfred Scheucher, Dec 29 2017

			The following permutations of order 8 and their reversals have this property:
  1 7 3 5 2 4 0 6
  1 7 4 2 5 3 0 6
  2 3 7 6 1 0 4 5
  2 4 7 1 6 0 3 5
  3 2 6 7 0 1 5 4
  3 5 1 7 0 6 2 4

D. E. Knuth, The Art of Computer Programming, Vol. 4, Section 7.2.1.6, Eq. 32.
L. C. Larson, The number of essentially different nonattacking rook arrangements, J. Recreat. Math., 7 (No. 3, 1974), circa pages 180-181.
Eugene McDonnell, "Magic Squares and Permutations" APL Quote-Quad 7.3 (Fall, 1976)
R. W. Robinson, Counting arrangements of bishops, pp. 198-214 of Combinatorial Mathematics IV (Adelaide 1975), Lect. Notes Math., 560 (1976).
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, Table of n, a(n) for n = 0..100
Marcello Artioli, Giuseppe Dattoli, Silvia Licciardi, and Rosa Maria Pidatella, Hermite and Laguerre Functions: a Unifying Point of View, Università degli Studi di Catania, Sicily, Italy (2019).
Murray Bremner and Martin Markl, Distributive laws between the Three Graces, arXiv:1809.08191 [math.AT], 2018.
R. B. Brent, Generalizing Tuenter's Binomial Sums, J. Int. Seq. 18 (2015) # 15.3.2.
Peter J. Cameron, Some treelike objects, Quart. J. Math. Oxford Ser. (2) 38 (1987), no. 150, 155--183. MR0891613 (89a:05009). See p. 155. - _N. J. A. Sloane_, Apr 18 2014
Peter J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
Elliot J. Carr and Matthew J. Simpson, New homogenization approaches for stochastic transport through heterogeneous media, arXiv:1810.08890 [physics.bio-ph], 2018.
W. Y. C. Chen, L. W. Shapiro and L. L. M. Young, Parity reversing involutions on plane trees and 2-Motzkin paths, arXiv:math/0503300 [math.CO], 2005.
Ali Chouria, Vlad-Florin Drǎgoi, and Jean-Gabriel Luque, On recursively defined combinatorial classes and labelled trees, arXiv:2004.04203 [math.CO], 2020.
P. Cvitanovic, Group theory for Feynman diagrams in non-Abelian gauge theories, Phys. Rev. D14 (1976), 1536-1553.
Nick Early, Honeycomb tessellations and canonical bases for permutohedral blades, arXiv:1810.03246 [math.CO], 2018.
John Engbers, David Galvin, and Clifford Smyth, Restricted Stirling and Lah numbers and their inverses, arXiv:1610.05803 [math.CO], 2016. See p. 8.
P. Flajolet and R. Sedgewick, Analytic Combinatorics, 2009; see page 127
S. Goodenough and C. Lavault, On subsets of Riordan subgroups and Heisenberg--Weyl algebra, arXiv preprint arXiv:1404.1894 [cs.DM], 2014.
S. Goodenough and C. Lavault, Overview on Heisenberg—Weyl Algebra and Subsets of Riordan Subgroups, The Electronic Journal of Combinatorics, 22(4) (2015), #P4.16,
H. W. Gould, A class of binomial sums and a series transform, Utilitas Math., 45 (1994), 71-83. (Annotated scanned copy)
A. M. Ibrahim, Extension of factorial concept to negative numbers, Notes on Number Theory and Discrete Mathematics, Vol. 19, 2013, 2, 30-42.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 115
L. C. Larson, The number of essentially different nonattacking rook arrangements, J. Recreat. Math., 7 (No. 3, 1974), circa pages 180-181. [Annotated scan of pages 180 and 181 only]
Jesús Leaños, Rutilo Moreno, and Luis Manuel Rivera-Martínez, On the number of mth roots of permutations, Australas. J. Combin. 52 (2012), 41-54 (Theorem 1).
Jesús Leaños, Rutilo Moreno, and Luis Manuel Rivera-Martínez, On the number of mth roots of permutations, arXiv:1005.1531 [math.CO], 2010-2011.
Édouard Lucas, Théorie des Nombres, Gauthier-Villars, Paris, 1891, Vol. 1, p. 221.
R. J. Marsh and P. P. Martin, Pascal arrays: counting Catalan sets, arXiv:math/0612572 [math.CO], 2006.
Calin D. Morosan, On the number of broadcast schemes in networks, Information Processing Letters, Volume 100, Issue 5 (2006), 188-193.
R. A. Proctor, Let's Expand Rota's Twelvefold Way for Counting Partitions!, arXiv:math/0606404 [math.CO], 2006-2007.
C. Radoux, Déterminants de Hankel et théorème de Sylvester, Séminaire Lotharingien de Combinatoire, B28b (1992), 9 pp.
J. Riordan, Letter to N. J. A. Sloane, Feb 03 1975 (with notes by njas)
H. E. Salzer, Coefficients for expressing the first thirty powers in terms of the Hermite polynomials, Math. Comp., 3 (1948), 167-169.
Elena L. Wang and Guoce Xin, On Ward Numbers and Increasing Schröder Trees, arXiv:2507.15654 [math.CO], 2025. See pp. 12-13.
Index to divisibility sequences
Index entries for related partition-counting sequences

Cf. A037224, A048854, A001147, A007696, A008545, A122670 (essentially the same sequence), A000165, A047055, A047657, A084947, A084948, A084949, A010050, A000142, A008275, A000108, A000984, A008276, A000680, A094216.
Catalan(n-1)*M^(n-1)*n! for M=1,2,3,4,5,6: A001813, A052714 (or A144828), A221954, A052734, A221953, A221955.
Cf. A123023, A059344, A099174, A066325, A001700, A000407, A006882.
Cf. A000085, A135401, A297708. - Manfred Scheucher, Jan 07 2018

List([0..20],n->Factorial(2*n)/Factorial(n)); # Muniru A Asiru, Nov 01 2018

[Factorial(2*n)/Factorial(n): n in [0..20]]; // Vincenzo Librandi, Oct 09 2018

A001813 := n->(2*n)!/n!;
A001813 := n -> mul(k, k = select(k-> k mod 4 = 2,[$1 .. 4*n])):
seq(A001813(n), n=0..16);  # Peter Luschny, Jun 23 2011

Table[(2n)!/n!, {n,0,20}] (* Harvey P. Dale, May 02 2011 *)

makelist(binomial(n+n, n)*n!,n,0,30); /* Martin Ettl, Nov 05 2012 */

a(n)=binomial(n+n,n)*n! \\ Charles R Greathouse IV, Jun 15 2011

first(n) = x='x+O('x^n); Vec(serlaplace((1 - 4*x)^(-1/2))) \\ Iain Fox, Jan 01 2018 (corrected by Iain Fox, Jan 11 2018)

from math import factorial
def A001813(n): return factorial(n<<1)//factorial(n) # Chai Wah Wu, Feb 14 2023

[binomial(2*n,n)*factorial(n) for n in range(0, 17)] # Zerinvary Lajos, Dec 03 2009

E.g.f.: (1-4*x)^(-1/2).
a(n) = (2*n)!/n! = Product_{k=0..n-1} (4*k + 2) = A081125(2*n).
Integral representation as n-th moment of a positive function on a positive half-axis: a(n) = Integral_{x=0..oo} x^n*exp(-x/4)/(sqrt(x)*2*sqrt(Pi)) dx, n >= 0. This representation is unique. - Karol A. Penson, Sep 18 2001
Define a'(1)=1, a'(n) = Sum_{k=1..n-1} a'(n-k)*a'(k)*C(n, k); then a(n)=a'(n+1). - Benoit Cloitre, Apr 27 2003
With interpolated zeros (1, 0, 2, 0, 12, ...) this has e.g.f. exp(x^2). - Paul Barry, May 09 2003
a(n) = A000680(n)/A000142(n)*A000079(n) = Product_{i=0..n-1} (4*i + 2) = 4^n*Pochhammer(1/2, n) = 4^n*GAMMA(n+1/2)/sqrt(Pi). - Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003
For asymptotics, see the Robinson paper.
a(k) = (2*k)!/k! = Sum_{i=1..k+1} |A008275(i,k+1)| * k^(i-1). - André F. Labossière, Jun 21 2007
a(n) = 12*A051618(a) n >= 2. - Zerinvary Lajos, Feb 15 2008
a(n) = A000984(n)*A000142(n). - Zerinvary Lajos, Mar 25 2008
a(n) = A016825(n-1)*a(n-1). - Roger L. Bagula, Sep 17 2008
a(n) = (-1)^n*A097388(n). - D. Morosan (cd_moros(AT)alumni.concordia.ca), Nov 28 2008
From Paul Barry, Jan 15 2009: (Start)
G.f.: 1/(1-2x/(1-4x/(1-6x/(1-8x/(1-10x/(1-... (continued fraction);
a(n) = (n+1)!*A000108(n). (End)
a(n) = Sum_{k=0..n} A132393(n,k)*2^(2n-k). - Philippe Deléham, Feb 10 2009
G.f.: 1/(1-2x-8x^2/(1-10x-48x^2/(1-18x-120x^2/(1-26x-224x^2/(1-34x-360x^2/(1-42x-528x^2/(1-... (continued fraction). - Paul Barry, Nov 25 2009
a(n) = A173333(2*n,n) for n>0; cf. A006963, A001761. - Reinhard Zumkeller, Feb 19 2010
From Gary W. Adamson, Jul 19 2011: (Start)
a(n) = upper left term of M^n, M = an infinite square production matrix as follows:
  2, 2, 0, 0, 0, 0, ...
  4, 4, 4, 0, 0, 0, ...
  6, 6, 6, 6, 0, 0, ...
  8, 8, 8, 8, 8, 0, ...
  ...
(End)
a(n) = (-2)^n*Sum_{k=0..n} 2^k*s(n+1,n+1-k), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
G.f.: 1/Q(0), where Q(k) = 1 + x*(4*k+2) - x*(4*k+4)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 18 2013
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - x*(8*k+4)/(x*(8*k+4) - 1 + 8*x*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 30 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 2*x/(2*x + 1/(2*k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
D-finite with recurrence: a(n) = (4*n-6)*a(n-2) + (4*n-3)*a(n-1), n>=2. - Ivan N. Ianakiev, Aug 07 2013
Sum_{n>=0} 1/a(n) = (exp(1/4)*sqrt(Pi)*erf(1/2) + 2)/2 = 1 + A214869, where erf(x) is the error function. - Ilya Gutkovskiy, Nov 10 2016
Sum_{n>=0} (-1)^n/a(n) = 1 - sqrt(Pi)*erfi(1/2)/(2*exp(1/4)), where erfi(x) is the imaginary error function. - Amiram Eldar, Feb 20 2021
a(n) = 1/([x^n] hypergeom([1], [1/2], x/4)). - Peter Luschny, Sep 13 2024
a(n) = 2^n*n!*JacobiP(n, -1/2, -n, 3). - Peter Luschny, Jan 22 2025
G.f.: 2F0(1,1/2;;4x). - R. J. Mathar, Jun 07 2025

More terms from James Sellers, May 01 2000

A037223 Number of solutions to non-attacking rooks problem on n X n board that are invariant under 180-degree rotation.

Original entry on oeis.org

1, 1, 2, 2, 8, 8, 48, 48, 384, 384, 3840, 3840, 46080, 46080, 645120, 645120, 10321920, 10321920, 185794560, 185794560, 3715891200, 3715891200, 81749606400, 81749606400, 1961990553600, 1961990553600, 51011754393600, 51011754393600, 1428329123020800, 1428329123020800

Offset: 0

Miklos SZABO (mike(AT)ludens.elte.hu)

This is just A000165 doubled up. Normally such sequences do not get their own entry in the OEIS. This is an exception. - N. J. A. Sloane, Sep 23 2006
Also the number of permutations of (1,2,3,...,n) for which the reverse of the inverse is the same as the inverse of the reverse. - Ian Duff, Mar 09 2007
Conjecture: a(n) = Product_{1<=i<=n and phi(i)<=floor(i/2)}i. - Enrique Pérez Herrero, May 31 2012. This conjecture is WRONG, counterexample is n=105. [Vaclav Kotesovec, Sep 07 2012]

E. Lucas, Theorie des nombres, Gauthiers-Villars, Paris, 1891, Vol 1, p. 221.

Alois P. Heinz, Table of n, a(n) for n = 0..500
E. Lucas, Théorie des nombres, Gauthiers-Villars, Paris, 1891, Vol 1, p. 221.
R. W. Robinson, Counting arrangements of bishops, pp. 198-214 of Combinatorial Mathematics IV (Adelaide 1975), Lect. Notes Math., 560 (1976).
M. Szabo, Non-attacking Queens Problem Page

Cf. A000165, A033148, A037224, A032522, A037223.

[Factorial((n div 2) -1)*2^((n div 2)-1): n   in [2..35]]; // Vincenzo Librandi, Nov 17 2018

For Maple program see A000903.
# second Maple program:
a:= n-> (r-> r!*2^r)(iquo(n, 2)):
seq(a(n), n=0..30);  # Alois P. Heinz, Dec 23 2013

f[n_]:=Times@@Select[Range[n],EulerPhi[#]<=Floor[#/2]&]; Table[f[n],{n,1,30}] (* Conjectured: Enrique Pérez Herrero, May 31 2012 *)(* This conjecture and also program is WRONG for n=105, Vaclav Kotesovec, Sep 07 2012 *)
a[n_] := (2*Floor[n/2])!!; Table[a[n], {n, 0, 27}] (* Jean-François Alcover, Dec 23 2013, after N. J. A. Sloane's comment *)

a(2n) = a(2n+1) = n!*2^n.
E.g.f.: 1 + x + (1 + x + x^2)*exp(x^2/2)*sqrt(Pi/2)*erf(x/sqrt(2)), where erf denotes the error function. - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Nov 01 2002
For asymptotics see the Robinson paper.
E.g.f.: Q(0)  where Q(k)= 1 + x/(2*k + 1 - x*(2*k+1)/(x+1/Q(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Sep 21 2012
E.g.f.: 1/(W(0)-x)  where W(k)= x + 1/(1 + x/(2*k + 1 - x*(2*k+1)/W(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Sep 22 2012
a(n) = Product_{i=1..floor(n/2)} 2*i. - Wesley Ivan Hurt, Oct 19 2014
D-finite with recurrence: a(n) +a(n-1) -n*a(n-2) +(-n+2)*a(n-3)=0. - R. J. Mathar, Feb 20 2020

More terms from Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Nov 01 2002

Edited by N. J. A. Sloane, Sep 23 2006

A000903 Number of inequivalent ways of placing n nonattacking rooks on n X n board up to rotations and reflections of the board.

Original entry on oeis.org

1, 1, 2, 7, 23, 115, 694, 5282, 46066, 456454, 4999004, 59916028, 778525516, 10897964660, 163461964024, 2615361578344, 44460982752488, 800296985768776, 15205638776753680, 304112757426239984, 6386367801916347184

Offset: 1

N. J. A. Sloane

			For n=4 the 7 solutions may be taken to be 1234,1243,1324,1423,1432,2143,2413.

L. C. Larson, The number of essentially different nonattacking rook arrangements, J. Recreat. Math., 7 (No. 3, 1974), circa pages 180-181.
R. C. Read, personal communication.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Z. Stankova and J. West, A new class of Wilf-equivalent permutations, J. Algeb. Combin., 15 (2002), 271-290.

N. J. A. Sloane, Table of n, a(n) for n = 1..100
P. J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
L. C. Larson, The number of essentially different nonattacking rook arrangements, J. Recreat. Math., 7 (No. 3, 1974), circa pages 180-181. [Annotated scan of pages 180 and 181 only]
E. Lucas, Théorie des Nombres, Gauthier-Villars, Paris, 1891, Vol. 1, p. 222.
E. Lucas, Théorie des nombres (annotated scans of a few selected pages)
R. C. Read, Letter to N. J. A. Sloane, Oct. 29, 1976
R. W. Robinson, Counting arrangements of bishops, pp. 198-214 of Combinatorial Mathematics IV (Adelaide 1975), Lect. Notes Math., 560 (1976).
R. W. Robinson, Counting arrangements of bishops, pp. 198-214 of Combinatorial Mathematics IV (Adelaide 1975), Lect. Notes Math., 560 (1976). (Annotated scanned copy)
Zvezdelina Stankova-Frenkel and Julian West, A new class of Wilf-equivalent permutations, arXiv:math/0103152 [math.CO], 2001. See Fig. 9.
Eric Weisstein's World of Mathematics, Rook Number.
Eric Weisstein's World of Mathematics, Rooks Problem.

Cf. A000142, A037223, A037224, A000085, A005635, A099952, A263685.

Maple programs for A000142, A037223, A122670, A001813, A000085, A000898, A000407, A000902, A000900, A000901, A000899, A000903
P:=n->n!; # Gives A000142
G:=proc(n) local k; k:=floor(n/2); k!*2^k; end; # Gives A037223, A000165
R:=proc(n) local m; if n mod 4 = 2 or n mod 4 = 3 then RETURN(0); fi; m:=floor(n/4); (2*m)!/m!; end; # Gives A122670, A001813
unprotect(D); D:=proc(n) option remember; if n <= 1 then 1 else D(n-1)+(n-1)*D(n-2); fi; end; # Gives A000085
B:=proc(n) option remember; if n <= 1 then RETURN(1); fi; if n mod 2 = 1 then RETURN(B(n-1)); fi; 2*B(n-2) + (n-2)*B(n-4); end; # Gives A000898 (doubled up)
rho:=n->R(n)/2; # Gives A000407, aerated
beta:=n->B(n)/2; # Gives A000902, doubled up
delta:=n->(D(n)-B(n))/2; # Gives A000900
unprotect(gamma); gamma:=n-> if n <= 1 then RETURN(0) else (G(n)-B(n)-R(n))/4; fi; # Gives A000901, doubled up
alpha:=n->P(n)/8-G(n)/8+B(n)/4-D(n)/4; # Gives A000899
unprotect(sigma); sigma:=n-> if n <= 1 then RETURN(1); else P(n)/8+G(n)/8+R(n)/4+D(n)/4; fi; #Gives A000903

c[n_] := Floor[n/2]! 2^Floor[n/2];
r[n_] := If[Mod[n, 4] > 1, 0, m = Floor[n/4]; If[m == 0, 1, (2 m)!/m!]];
d[0] = d[1] = 1; d[n_] := d[n] = (n - 1)d[n - 2] + d[n - 1];
a[1] = 1; a[n_] := (n! + c[n] + 2 r[n] + 2 d[n])/8;
Array[a, 21] (* Jean-François Alcover, Apr 06 2011, after Matthias Engelhardt, further improved by Robert G. Wilson v *)

If n>1 then a(n) = 1/8 * (F(n) + C(n) + 2 * R(n) + 2 * D(n)), where F(n) = A000142(n) [all solutions, i.e., factorials], C(n) = A037223(n) [central symmetric solutions], R(n) = A037224(n) [rotationally symmetric solutions] and D(n) = A000085(n) [symmetric solutions by reflection at a diagonal]. - Matthias Engelhardt, Apr 05 2000

For asymptotics see the Robinson paper.

More terms from David W. Wilson, Jul 13 2003

A033148 Number of rotationally symmetric solutions for queens on n X n board.

Original entry on oeis.org

1, 0, 0, 2, 2, 0, 0, 0, 0, 0, 0, 8, 8, 0, 0, 64, 128, 0, 0, 480, 704, 0, 0, 3328, 3264, 0, 0, 32896, 43776, 0, 0, 406784, 667904, 0, 0, 5845504, 8650752, 0, 0, 77184000, 101492736, 0, 0, 1261588480, 1795233792, 0, 0, 21517426688, 35028172800, 0, 0, 406875119616, 652044443648, 0, 0, 8613581094912, 12530550128640, 0, 0, 194409626533888, 291826098503680, 0, 0

Offset: 1

Miklos SZABO (mike(AT)ludens.elte.hu)

From Don Knuth, Jul 17 2015: (Start)
Ahrens proved that a(n)=0 unless n=4k or 4k+1. He also proved that in the latter case, a(n) is a multiple of 2^k. He found all solutions when n was less than 20.
Kraitchik carried the calculations further (for n less than 28). In his book he tabulated only the values a(n)/2^k. He had correct entries for n=21 and n=25, but his values for n=20 and n=24 were 1 too small -- of course he had calculated everything by hand! (End)

W. Ahrens, Mathematische Unterhaltungen und Spiele, 2nd edition, volume 1, Teubner, 1910, pages 249-258.
Maurice Kraitchik, Le problème des reines, Bruxelles: L'Échiquier, 1926, page 18.

Tricia M. Brown, Kaleidoscopes, Chessboards, and Symmetry, Journal of Humanistic Mathematics, Volume 6 Issue 1 ( January 2016), pages 110-126.
P. Capstick and K. McCann, The problem of the n queens, apparently unpublished, no date (circa 1990?) [Scanned copy]
Gheorghe Coserea, Solutions for n=20.
Gheorghe Coserea, Solutions for n=24.
Gheorghe Coserea, MiniZinc model for generating solutions.
YuhPyng Shieh, Cyclic Complete Mappings Counting Problems
M. Szabo, Non-attacking Queens Problem Page

Cf. A002562, A032522, A037223, A037224, A260189.

More terms from Jieh Hsiang and YuhPyng Shieh (arping(AT)turing.csie.ntu.edu.tw), May 20 2002

A032522 Number of point symmetric solutions to non-attacking queens problem on n X n board.

Original entry on oeis.org

1, 0, 0, 2, 2, 4, 8, 4, 16, 12, 48, 80, 136, 420, 1240, 3000, 8152, 18104, 44184, 144620, 375664, 1250692, 3581240, 11675080, 34132592, 115718268, 320403024, 1250901440, 3600075088, 14589438024, 43266334696, 181254386312

Offset: 1

Miklos SZABO (mike(AT)ludens.elte.hu)

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
R. J. Walker, An enumerative technique for a class of combinatorial problems, pp. 91-94 of Proc. Sympos. Applied Math., vol. 10, Amer. Math. Soc., 1960.

W. Schubert, Table of n, a(n) for n = 1..40
Tricia M. Brown, Kaleidoscopes, Chessboards, and Symmetry, Journal of Humanistic Mathematics, Volume 6 Issue 1 ( January 2016), pages 110-126.
Gheorghe Coserea, Solutions for n=10.
Gheorghe Coserea, Solutions for n=11.
Gheorghe Coserea, MiniZinc model for generating solutions.
W. Schubert, N-Queens page
M. Szabo, Non-attacking Queens Problem Page

Cf. A002562, A033148, A037224, A037223.

More terms for n = 33..36 from W. Schubert, Jul 31 2009

A064280 Number of nonequivalent solutions to the order n checkerboard problem up to reflection and rotation: place n pieces on an n X n board so there is exactly one piece in each row, column and main diagonal.

Original entry on oeis.org

1, 0, 0, 1, 4, 12, 86, 696, 6150, 61760, 673256, 8137200, 105074420, 1479237312, 22077680616, 354753059584, 6007578698408, 108500041654272, 2055204828592832, 41215470268919040, 863378484993573840, 19036646809582054400, 436944006380312366240

Offset: 1

Jud McCranie, Sep 24 2001

nonn

For even n>=4: A007016(n) = 8*A064280(n).

For even n, the diagonals do not intersect and there can be no symmetrical solutions. For odd n, a symmetrical solution will have a rook on the central square and the remaining n-1 rooks must be placed so as to avoid the main diagonals. See A292080 for information on counting non-attacking rook configurations with no rook on either main diagonal. - Andrew Howroyd, Sep 12 2017

			The 4 X 4 solution is unique, up to equivalence, with pieces at (1,1), (2,3), (3,4) and (4,2).

Andrew Howroyd, Table of n, a(n) for n = 1..100
Geoffrey Chase, Checkerboard Problem Solved, Creative Computing 6(1), Jan 1980, 122.
Bahairiv Joshi, Unique Solutions to the Checkerboard Problem, Creative Computing 6(10), Oct 1980, 124-125.
Abijah Reed, Comments on Checkerboard Problem Solved, Creative Computing 6(5), May 1980, 94.

A007016 gives the number of solutions including symmetrical ones.

Cf. A000166, A007016, A037224, A053871, A292080.

sf = Subfactorial;
x[n_] := x[n] = Integrate[If[EvenQ[n], (x^2 - 4*x + 2)^(n/2), (x - 1)*(x^2 - 4*x + 2)^((n - 1)/2)]/E^x, {x, 0, Infinity}];
F[n_ /; EvenQ[n]] := With[{m = n/2}, m*(x[2*m] - (2*m - 3)*x[2*m - 1])];
F[n_ /; OddQ[n]] := With[{m = (n - 1)/2}, (2*m + 1)*x[2*m] + 3*m*x[2*m - 1] - 2*m*(m - 1)*x[2*m - 2]];
d[n_] := (-1)^n HypergeometricPFQ[{1/2, -n}, {}, 2];
R[n_] := If[OddQ[n], 0, If[n == 0, 1, (n - 1)!*2/(n/2 - 1)!]];
a[1] = 1; a[n_] := With[{m = Quotient[n, 2]}, (F[n] + If[EvenQ[n], 0, 2^m * sf[m] + 2*R[m] + 2*d[m] + 2*Boole[m == 0]])/8];
Array[a, 30] (* Jean-François Alcover, Sep 15 2019 *)

\\ here sf is A000166, F is A007016, D is A053871, R(n) is A037224(2n).
sf(n) = {n! * polcoeff( exp(-x + x * O(x^n)) / (1 - x), n)}
F(n) = {my(v = vector(n)); for(n=4, length(v), v[n] = (n-1)*v[n-1] + 2*if(n%2==1, (n-1)*v[n-2], (n-2)*if(n==4,1,v[n-4]))); if(n<4, [1,0,0][n], if(n%2==0, n*(v[n] - (n-3)*v[n-1]), 2*n*v[n-1] + 3*(n-1)*v[n-2] - (n-1)*(n-3)*v[n-3])/2)}
D(n) = {sum(k=0, n, (-1)^(n-k) * binomial(n,k) * (2*k)!/(2^k*k!))}
R(n) = {if(n%2==1, 0, if(n==0, 1, (n-1)!*2/(n/2-1)!))}
a(n) = {(F(n) + if(n%2==0, 0, my(m=n\2); 2^m * sf(m) + 2*R(m) + 2*D(m) + 2*(m==0)))/8} \\ Andrew Howroyd, Sep 12 2017

a(2n) = A007016(2n) / 8, a(2n+1) = (A007016(2n+1) + 2^n * A000166(n) + 2*A037224(2*n) + 2*A053871(n)) / 8 for n > 0. - Andrew Howroyd, Sep 12 2017

a(11)-a(12) from Lars Blomberg, Jan 13 2013

Name clarified and terms a(13) and beyond from Andrew Howroyd, Sep 12 2017

A122670 If n mod 4 = 2 or n mod 4 = 3 then a(n) = 0 else let m=floor(n/4), then a(n) = (2*m)!/m!.

Original entry on oeis.org

1, 1, 0, 0, 2, 2, 0, 0, 12, 12, 0, 0, 120, 120, 0, 0, 1680, 1680, 0, 0, 30240, 30240, 0, 0, 665280, 665280, 0, 0, 17297280, 17297280, 0, 0, 518918400, 518918400, 0, 0, 17643225600, 17643225600, 0, 0, 670442572800, 670442572800, 0, 0, 28158588057600, 28158588057600, 0, 0, 1295295050649600

Offset: 0

N. J. A. Sloane, Sep 23 2006

nonn

Number of solutions to the rook problem on an n X n board having a certain symmetry group (see Robinson for details).

A037224 is an essentially identical sequence.

R. W. Robinson, Counting arrangements of bishops, pp. 198-214 of Combinatorial Mathematics IV (Adelaide 1975), Lect. Notes Math., 560 (1976).

If the duplicates and zeros are omitted we get A001813.

Cf. A000898, A000899, A000900, A000901, A000902, A000903, A037224.

R:=proc(n) local m; if n mod 4 = 2 or n mod 4 = 3 then RETURN(0); fi; m:=floor(n/4); (2*m)!/m!; end;
For Maple program see A000903.

Table[If[MemberQ[{2,3},Mod[n,4]],0,((2Floor[n/4])!/Floor[n/4]!)],{n,0,50}] (* Harvey P. Dale, Dec 30 2023 *)

For asymptotics see the Robinson paper.

a(n) = (1/2 + (-1)^(n/2 - 1/4 + (-1)^n/4)/2) * ((n/2 - 3/4 + (-1)^n/4 + (-1)^(n/2 - 1/4 + (-1)^n/4)/2)! / ((n/4 - 3/8 + (-1)^n/8 + (-1)^(n/2 - 1/4 + (-1)^n/4)/4)!)). - Wesley Ivan Hurt, Mar 30 2015

A292080 Number of nonequivalent ways to place n non-attacking rooks on an n X n board with no rook on 2 main diagonals up to rotations and reflections of the board.

Original entry on oeis.org

1, 0, 0, 0, 2, 2, 14, 84, 630, 6096, 55336, 672160, 7409300, 104999520, 1366363752, 22068387264, 331233939624, 6005919062528, 102144359744192, 2054811316442112, 39053339674065360, 863259240785840640, 18132529836143846560, 436899062862222484480

Offset: 0

Andrew Howroyd, Sep 12 2017

nonn

For odd n, there are no symmetrical configurations of non-attacking rooks without a rook in the main diagonal, so a(2n+1) = A003471(2n+1) / 8. For even n, configurations with rotational and diagonal symmetry are possible.

			Case n=4: The 2 nonequivalent solutions are:
   _ x _ _     _ x _ _
   x _ _ _     _ _ _ x
   _ _ _ x     x _ _ _
   _ _ x _     _ _ x _
Case n=5: The 2 nonequivalent solutions are:
   _ x _ _ _   _ x _ _ _
   x _ _ _ _   _ _ _ _ x
   _ _ _ x _   x _ _ _ _
   _ _ _ _ x   _ _ x _ _
   _ _ x _ _   _ _ _ x _

Andrew Howroyd, Table of n, a(n) for n = 0..100
Andrew Howroyd, Nonequivalent and Symmetric Solutions

Cf. A000166, A000903, A003471, A037224, A053871, A064280.

sf[n_] := n! * SeriesCoefficient[Exp[-x ] / (1 - x), {x, 0, n}];
F[n_] := (Clear[v]; v[_] = 0; For[m = 4, m <= n, m++, v[m] = (m - 1)*v[m - 1] + 2*If[OddQ[m], (m - 1)*v[m - 2], (m - 2)*If[m == 4, 1, v[m - 4]]]]; v[n]);
d[n_] := Sum[(-1)^(n-k)*Binomial[n, k]*(2k)!/(2^k*k!), {k, 0, n}];
R[n_] := If[OddQ[n], 0, (n - 1)!*2/(n/2 - 1)!];
a[0] = 1; a[n_] := (F[n] + If[OddQ[n], 0, m = n/2; 2^m * sf[m] + 2*R[m] + 2*d[m]])/8;
Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Dec 28 2017, after Andrew Howroyd *)

\\ here sf is A000166, F is A003471, D is A053871, R(n) is A037224(2n).
sf(n) = {n! * polcoeff( exp(-x + x * O(x^n)) / (1 - x), n)}
F(n) = {my(v = vector(n)); for(n=4,length(v),v[n]=(n-1)*v[n-1]+2*if(n%2==1,(n-1)*v[n-2],(n-2)*if(n==4,1,v[n-4]))); v[n]}
D(n) = {sum(k=0, n, (-1)^(n-k) * binomial(n,k) * (2*k)!/(2^k*k!))}
R(n) = {if(n%2==1, 0, (n-1)!*2/(n/2-1)!)}
a(n) = {(F(n) + if(n%2==1, 0, my(m=n/2); 2^m * sf(m) + 2*R(m) + 2*D(m)))/8}

a(2n+1) = A003471(2n+1) / 8, a(2n) = (A003471(2n) + 2^n * A000166(n) + 2*A037224(2*n) + 2*A053871(n)) / 8.

cp's OEIS Frontend

A001813 Quadruple factorial numbers: a(n) = (2n)!/n!.

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A037223 Number of solutions to non-attacking rooks problem on n X n board that are invariant under 180-degree rotation.

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A000903 Number of inequivalent ways of placing n nonattacking rooks on n X n board up to rotations and reflections of the board.

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A033148 Number of rotationally symmetric solutions for queens on n X n board.

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A032522 Number of point symmetric solutions to non-attacking queens problem on n X n board.

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A064280 Number of nonequivalent solutions to the order n checkerboard problem up to reflection and rotation: place n pieces on an n X n board so there is exactly one piece in each row, column and main diagonal.

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A122670 If n mod 4 = 2 or n mod 4 = 3 then a(n) = 0 else let m=floor(n/4), then a(n) = (2*m)!/m!.