A259212 -id:A259212 - cp's OEIS frontend

A000179 Ménage numbers: a(0) = 1, a(1) = -1, and for n >= 2, a(n) = number of permutations s of [0, ..., n-1] such that s(i) != i and s(i) != i+1 (mod n) for all i.

1, -1, 0, 1, 2, 13, 80, 579, 4738, 43387, 439792, 4890741, 59216642, 775596313, 10927434464, 164806435783, 2649391469058, 45226435601207, 817056406224416, 15574618910994665, 312400218671253762, 6577618644576902053, 145051250421230224304, 3343382818203784146955, 80399425364623070680706, 2013619745874493923699123

Offset: 0

N. J. A. Sloane

According to rook theory, John Riordan considered a(1) to be -1. - Vladimir Shevelev, Apr 02 2010
This is also the value that the formulas of Touchard and of Wyman and Moser give and is compatible with many recurrences. - William P. Orrick, Aug 31 2020
Or, for n >= 3, the number of 3 X n Latin rectangles the second row of which is full cycle with a fixed order of its elements, e.g., the cycle (x_2,x_3,...,x_n,x_1) with x_1 < x_2 < ... < x_n. - Vladimir Shevelev, Mar 22 2010
Muir (p. 112) gives essentially this recurrence (although without specifying any initial conditions). Compare A186638. - N. J. A. Sloane, Feb 24 2011
Sequence discovered by Touchard in 1934. - L. Edson Jeffery, Nov 13 2013
Although these are also known as Touchard numbers, the problem was formulated by Lucas in 1891, who gave the recurrence formula shown below. See Cerasoli et al., 1988. - Stanislav Sykora, Mar 14 2014
An equivalent problem was formulated by Tait; solutions to Tait's problem were given by Muir (1878) and Cayley (1878). - William P. Orrick, Aug 31 2020
From Vladimir Shevelev, Jun 25 2015: (Start)
According to the ménage problem, 2*n!*a(n) is the number of ways of seating n married couples at 2*n chairs around a circular table, men and women in alternate positions, so that no husband is next to his wife.
It is known [Riordan, ch. 7] that a(n) is the number of arrangements of n non-attacking rooks on the positions of the 1's in an n X n (0,1)-matrix A_n with 0's in positions (i,i), i = 1,...,n, (i,i+1), i = 1,...,n-1, and (n,1). This statement could be written as a(n) = per(A_n). For example, A_5 has the form
    001*11
    1*0011
    11001*                        (1)
    11*100
    0111*0,
where 5 non-attacking rooks are denoted by {1*}.
We can indicate a one-to-one correspondence between arrangements of n non-attacking rooks on the 1's of a matrix A_n and arrangements of n married couples around a circular table by the rules of the ménage problem, after the ladies w_1, w_2, ..., w_n have taken the chairs numbered
    2*n, 2, 4, ..., 2*n-2         (2)
respectively. Suppose we consider an arrangement of rooks: (1,j_1), (2,j_2), ..., (n,j_n). Then the men m_1, m_2, ..., m_n took chairs with numbers
    2*j_i - 3 (mod 2*n),          (3)
where the residues are chosen from the interval[1,2*n]. Indeed {j_i} is a permutation of 1,...,n. So {2*j_i-3}(mod 2*n) is a permutation of odd positive integers <= 2*n-1. Besides, the distance between m_i and w_i cannot be 1. Indeed, the equality |2*(j_i-i)-1| = 1 (mod 2*n) is possible if and only if either j_i=i or j_i=i+1 (mod n) that correspond to positions of 0's in matrix A_n.
For example, in the case of positions of {1*} in(1) we have j_1=3, j_2=1, j_3=5, j_4=2, j_5=4. So, by(2) and (3) the chairs 1,2,...,10 are taken by m_4, w_2, m_1, w_3, m_5, w_4, m_3, w_5, m_2, w_1, respectively. (End)
The first 20 terms of this sequence were calculated in 1891 by E. Lucas (see [Lucas, p. 495]). - Peter J. C. Moses, Jun 26 2015
From Ira M. Gessel, Nov 27 2018: (Start)
If we invert the formula
  Sum_{ n>=0 } u_n z^n = ((1-z)/(1+z)) F(z/(1+z)^2)
that Don Knuth mentions (see link) (i.e., set x=z/(1+z)^2 and solve for z in terms of x), we get a formula for F(z) = Sum_{n >= 0} n! z^n as a sum with all positive coefficients of (almost) powers of the Catalan number generating function.
The exact formula is (5) of the Yiting Li article.
This article also gives a combinatorial proof of this formula (though it is not as simple as one might want). (End)

			a(2) = 0; nothing works. a(3) = 1; (201) works. a(4) = 2; (2301), (3012) work. a(5) = 13; (20413), (23401), (24013), (24103), (30412), (30421), (34012), (34021), (34102), (40123), (43012), (43021), (43102) work.

W. W. R. Ball and H. S. M. Coxeter, Mathematical Recreations and Essays, 13th Ed. Dover, p. 50.
M. Cerasoli, F. Eugeni and M. Protasi, Elementi di Matematica Discreta, Nicola Zanichelli Editore, Bologna 1988, Chapter 3, p. 78.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 185, mu(n).
Kaplansky, Irving and Riordan, John, The probleme des menages, Scripta Math. 12, (1946). 113-124. See u_n.
E. Lucas, Théorie des nombres, Paris, 1891, pp. 491-495.
P. A. MacMahon, Combinatory Analysis. Cambridge Univ. Press, London and New York, Vol. 1, 1915 and Vol. 2, 1916; see vol. 1, p 256.
T. Muir, A Treatise on the Theory of Determinants. Dover, NY, 1960, Sect. 132, p. 112. - N. J. A. Sloane, Feb 24 2011
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 197.
V. S. Shevelev, Reduced Latin rectangles and square matrices with equal row and column sums, Diskr. Mat. (J. of the Akademy of Sciences of Russia) 4(1992), 91-110. - Vladimir Shevelev, Mar 22 2010
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).
H. M. Taylor, A problem on arrangements, Mess. Math., 32 (1902), 60ff.
J. Touchard, Permutations discordant with two given permutations, Scripta Math., 19 (1953), 108-119.
J. H. van Lint, Combinatorial Theory Seminar, Eindhoven University of Technology, Springer Lecture Notes in Mathematics, Vol. 382, 1974. See page 10.

Seiichi Manyama, Table of n, a(n) for n = 0..450 (terms 0..100 from T. D. Noe)
M. A. Alekseyev, Weighted de Bruijn Graphs for the Menage Problem and Its Generalizations. Lecture Notes in Computer Science 9843 (2016), 151-162. doi:10.1007/978-3-319-44543-4_12 arXiv:1510.07926, [math.CO], 2015-2016.
Vladimir Baltic, On the number of certain types of strongly restricted permutations, Applicable Analysis and Discrete Mathematics Vol. 4, No 1 (April, 2010), 119-135.
Kenneth P. Bogart and Peter G. Doyle, Nonsexist solution of the ménage problem, Amer. Math. Monthly 93 (1986), no. 7, 514-519.
A. Cayley, On a problem of arrangements, Proceedings of the Royal Society of Edinburgh 9 (1878) 338-342.
A. Cayley, On a problem of arrangements, Proceedings of the Royal Society of Edinburgh 9 (1878) 338-342.
A. Cayley, On Mr Muir's discussion of Professor Tait's problem of arrangements, Proceedings of the Royal Society of Edinburgh 9 (1878) 388-391.
A. Cayley, On Mr Muir's discussion of Professor Tait's problem of arrangements, Proceedings of the Royal Society of Edinburgh 9 (1878) 388-391.
J. Dutka, On the Probleme des Menages, Mathem. Conversat. (2001) 277-287, reprinted from Math. Intell. 8 (1986) 18-25
A. de Gennaro, How many latin rectangles are there?, arXiv:0711.0527 [math.CO] (2007), see p. 2.
P. Flajolet and R. Sedgewick, Analytic Combinatorics, 2009; see page 372
Nick Hobson, Python program for this sequence.
Peter Kagey, Ranking and Unranking Restricted Permutations, arXiv:2210.17021 [math.CO], 2022.
Irving Kaplansky, Solution of the "Problème des ménages", Bull. Amer. Math. Soc. 49, (1943). 784-785.
Irving Kaplansky, Symbolic solution of certain problems in permutations, Bull. Amer. Math. Soc., 50 (1944), 906-914.
I. Kaplansky and J. Riordan, The problème des ménages, Scripta Math. 12, (1946), 113-124. [Scan of annotated copy]
S. M. Kerawala, Asymptotic solution of the "Probleme des menages, Bull. Calcutta Math. Soc., 39 (1947), 82-84. [Annotated scanned copy]
Vaclav Kotesovec, Non-attacking chess pieces, 6ed, 2013, p. 221.
D. E. Knuth, Comments on A000179, Nov 25 2018 - Nov 27 2018.
D. E. Knuth, Donald Knuth's 24th Annual Christmas Lecture: Dancing Links, Stanfordonline, Video published on YouTube, Dec 12, 2018.
A. R. Kräuter, Über die Permanente gewisser zirkulärer Matrizen..., Séminaire Lotharingien de Combinatoire, B11b (1984), 11 pp.
Yiting Li, Ménage Numbers and Ménage Permutations, J. Int. Seq. 18 (2015) 15.6.8
E. Lucas, Théorie des Nombres, Gauthier-Villars, Paris, 1891, Vol. 1, p. 495.
T. Muir, On Professor Tait's problem of arrangement, Proceedings of the Royal Society of Edinburgh 9 (1878) 382-387.
T. Muir, On Professor Tait's problem of arrangement, Proceedings of the Royal Society of Edinburgh 9 (1878) 382-387.
Vladimir Shevelev and Peter J. C. Moses, The ménage problem with a known mathematician, arXiv:1101.5321 [math.CO], 2011, 2015.
Vladimir Shevelev and Peter J. C. Moses, Alice and Bob go to dinner: A variation on menage, INTEGERS, Vol. 16 (2016), #A72.
R. J. Stones, S. Lin, X. Liu and G. Wang, On Computing the Number of Latin Rectangles, Graphs and Combinatorics (2016) 32:1187-1202; DOI 10.1007/s00373-015-1643-1.
H. M. Taylor, A problem on arrangements, Mess. Math., 32 (1902), 60ff. [Annotated scanned copy]
J. Touchard, Théorie des substitutions. Sur un problème de permutations, C. R. Acad. Sci. Paris 198 (1934), 631-633.
Eric Weisstein's World of Mathematics, Married Couples Problem.
Eric Weisstein's World of Mathematics, Rooks Problem.
Wikipedia, Menage problem.
M. Wyman and L. Moser, On the problème des ménages, Canad. J. Math., 10 (1958), 468-480.
D. Zeilberger, Automatic Enumeration of Generalized Menage Numbers, arXiv preprint arXiv:1401.1089 [math.CO], 2014.

Diagonal of A058087. Also a diagonal of A008305.
Cf. A000904, A059375, A102761, A000186, A094047, A067998, A033999, A258664, A258665, A258666, A258667, A258673, A259212, A213234, A000023.
A000179, A102761, and A335700 are all essentially the same sequence but with different conventions for the initial terms a(0) and a(1). - N. J. A. Sloane, Aug 06 2020

import Data.List (zipWith5)
a000179 n = a000179_list !! n
a000179_list = 1 : -1 : 0 : 1 : zipWith5
   (\v w x y z -> (x * y + (v + 2) * z - w) `div` v) [2..] (cycle [4,-4])
   (drop 4 a067998_list) (drop 3 a000179_list) (drop 2 a000179_list)
-- Reinhard Zumkeller, Aug 26 2013

A000179:= n ->add ((-1)^k*(2*n)*binomial(2*n-k,k)*(n-k)!/(2*n-k), k=0..n); # for n >= 1
U:= proc(n) local k; add( (2*n/(2*n-k))*binomial(2*n-k,k)*(n-k)!*(x-1)^k, k=0..n); end; W := proc(r,s) coeff( U(r),x,s ); end; A000179 := n->W(n,0); # valid for n >= 1

a[n_] := 2*n*Sum[(-1)^k*Binomial[2*n - k, k]*(n - k)!/(2*n - k), {k, 0, n}]; a[0] = 1; Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Dec 05 2012, from 2nd formula *)

\\ 3 programs adapted to a(1) = -1 by Hugo Pfoertner, Aug 31 2020

{a(n) = my(A); if( n, A = vector(n,i,i-2); for(k=4, n, A[k] = (k * (k - 2) * A[k-1] + k * A[k-2] - 4 * (-1)^k) / (k-2)); A[n], 1)};/* Michael Somos, Jan 22 2008 */

a(n)=if(n>1, round(2*n*exp(-2)*besselk(n, 2)), 1-2*n) \\ Charles R Greathouse IV, Nov 03 2014

{a(n) = my(A); if( n, A = vector(n,i,i-2); for(k=5, n, A[k] = k * A[k-1] + 2 * A[k-2] + (4-k) * A[k-3] - A[k-4]); A[n], 1)} /* Michael Somos, May 02 2018 */

from math import comb, factorial
def A000179(n): return 1 if n == 0 else sum((-2*n if k & 1 else 2*n)*comb(m:=2*n-k,k)*factorial(n-k)//m for k in range(n+1)) # Chai Wah Wu, May 27 2022

a(n) = ((n^2-2*n)*a(n-1) + n*a(n-2) - 4*(-1)^n)/(n-2) for n >= 3.
a(n) = A059375(n)/(2*n!) for n >= 2.
a(n) = Sum_{k=0..n} (-1)^k*(2*n)*binomial(2*n-k, k)*(n-k)!/(2*n-k) for n >= 1. - Touchard (1934)
G.f.: ((1-x)/(1+x))*Sum_{n>=0} n!*(x/(1+x)^2)^n. - Vladeta Jovovic, Jun 26 2007
a(2^k+2) == 0 (mod 2^k); for k >= 2, a(2^k) == 2(mod 2^k). - Vladimir Shevelev, Jan 14 2011
a(n) = round( 2*n*exp(-2)*BesselK(n,2) ) for n > 1. - Mark van Hoeij, Oct 25 2011
a(n) ~ (n/e)^n * sqrt(2*Pi*n)/e^2. - Charles R Greathouse IV, Jan 21 2016
0 = a(n)*(-a(n+2) +a(n+4)) +a(n+1)*(+a(n+1) +a(n+2) -3*a(n+3) -5*a(n+4) +a(n+5)) +a(n+2)*(+2*a(n+2) +3*a(n+3) -3*a(n+4)) +a(n+3)*(+2*a(n+3) +a(n+4) -a(n+5)) +a(n+4)*(+a(n+4)), for all n>1. If a(-2..1) = (0, -1, 2, -1) then also true for those values of n. - Michael Somos, Apr 29 2018
D-finite with recurrence: 0 = a(n) +n*a(n+1) -2*a(n+2) +(-n-4)*a(n+3) +a(n+4), for all n in Z where a(n) = a(-n) for all n in Z and a(0) = 2, a(1) = -1. - Michael Somos, May 02 2018
a(n) = Sum_{k=0..n} A213234(n,k) * A000023(n-2*k) = Sum_{k=0..n} (-1)^k * n/(n-k) * binomial(n-k, k) * (n-2*k)! Sum_{j=0..n-2*k} (-2)^j/j! for n >= 1. [Wyman and Moser (1958)]. - William P. Orrick, Jun 25 2020
a(k+4*p) - 2*a(k+2*p) + a(k) is divisible by p, for any k > 0 and any prime p. - Mark van Hoeij, Jan 11 2022

More terms from James Sellers, May 02 2000
Additional comments from David W. Wilson, Feb 18 2003
a(1) changed to -1 at the suggestion of Don Knuth. - N. J. A. Sloane, Nov 26 2018

A258664 A total of n married couples, including a mathematician M and his wife, are to be seated at the 2n chairs around a circular table, with no man seated next to his wife. After the ladies are seated at every other chair, M is the first man allowed to choose one of the remaining chairs. The sequence gives the number of ways of seating the other men, with no man seated next to his wife, if M chooses the chair that is 3 seats clockwise from his wife's chair.

Original entry on oeis.org

0, 0, 1, 1, 4, 20, 115, 787, 6184, 54888, 542805, 5916725, 70463900, 910167596, 12672415015, 189181881575, 3014307220880, 51054940726928, 915987174021609, 17352888926841897, 346144782915314740, 7251738265074465220, 159193007549552845339, 3654204694819144118651

Offset: 1

Vladimir Shevelev and Peter J. C. Moses, Jun 07 2015

nonn

This is a variation of the classic ménage problem (cf. A000179).
It is known [Riordan, ch. 8, ex. 7(b)] that, after the ladies are seated at every other chair, the number U_n of ways of seating the men in the ménage problem has asymptotic expansion U_n ~ e^(-2)*n!*(1 + Sum_{k>=1}(-1)^k/(k!(n-1)_k)), where (n)_k = n*(n-1)*...*(n-k+1).
Therefore, it is natural to conjecture that a(n) ~ e^(-2)*n!/(n-2)*(1 + Sum_{k>=1}(-1)^k/(k!(n-1)_k)).

I. Kaplansky and J. Riordan, The problème des ménages, Scripta Math. 12, (1946), 113-124.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, chs. 7, 8.

Peter J. C. Moses, Seatings for 6 couples
I. Kaplansky and J. Riordan, The problème des ménages, Scripta Math. 12, (1946), 113-124. [Scan of annotated copy]
E. Lucas, Sur le problème des ménages, Théorie des nombres, Paris, 1891, 491-496.
Vladimir Shevelev, Peter J. C. Moses, The ménage problem with a known mathematician, arXiv:1101.5321 [math.CO], 2011, 2015.
Vladimir Shevelev and Peter J. C. Moses, Alice and Bob go to dinner: A variation on menage, INTEGERS, Vol. 16 (2016), #A72.
J. Touchard, Sur un problème de permutations, C.R. Acad. Sci. Paris, 198 (1934), 631-633.

Cf. A000179, A258665, A258666, A258667, A258673, A259212.

a[d_,n_]:=If[n<=#-1,0,Sum[((-1)^k)*(n-k-1)!Sum[Binomial[2#-j-4,j]*Binomial[2(n-#)-k+j+2,k-j],{j,Max[#+k-n-1,0],Min[k,#-2]}],{k,0,n-1}]]&[(d+3)/2];
Map[a[3,#]&,Range[25]] (* Peter J. C. Moses, Jun 07 2015 *)

a(n) = sum(k=0, n-1, (-1)^k*(n-k-1)!*sum(j=max(k-n+2, 0), min(k,1), binomial(2-j, j)*binomial(2*n-k+j-4, k-j))); \\ Michel Marcus, Jun 26 2015

a(n) = Sum_{0<=k<=n-1}(-1)^k*(n-k-1)! * Sum_{max(k-n+2, 0)<=j<=min(k,1)} binomial(2-j, j)*binomial(2*n-k+j-4, k-j).

A000904 a(n) = (n+1)a(n-1) + (n+2)a(n-2) + a(n-3); a(1)=0, a(2)=3, a(3)=13.

Original entry on oeis.org

0, 3, 13, 83, 592, 4821, 43979, 444613, 4934720, 59661255, 780531033, 10987095719, 165586966816, 2660378564777, 45392022568023, 819716784789193, 15620010933562688, 313219935456042955, 6593238655510464741, 145364470356686267259, 3349976056859294611696

Offset: 1

N. J. A. Sloane

Numbers connected with the ménage problem (cf. A000179).

			G.f. = 3 x^2 + 13 x^3 + 83 x^4 + 592 x^5 + 4821 x^6 + 43979 x^7 + 444613 x^8 + ...

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).

T. D. Noe, Table of n, a(n) for n=1..100
E. Lucas, Théorie des Nombres, Gauthier-Villars, Paris, 1891, Vol. 1, p. 495.
E. Lucas, Théorie des Nombres, Gauthier-Villars, Paris, 1891, Vol. 1. [Annotated scan of pages 488-499 only]

Cf. A000179, A000271.

a000904_list = 0 : 3 : 13 : zipWith (+) a000904_list
   (zipWith (+) (zipWith (*) [6..] $ drop 1 a000904_list)
                (zipWith (*) [5..] $ drop 2 a000904_list))
-- Reinhard Zumkeller, Nov 22 2011

max = 19; f[x_] = 1/(x+x^2)^2 * Sum[ n!*(x/(1+x)^2)^n, {n, 0, max+2}]; Series[ f[x]+1/x-1/x^2, {x, 0, max}] // CoefficientList[#, x]& // Rest (* Jean-François Alcover, May 16 2013, after Vladeta Jovovic *)
RecurrenceTable[{a[1]==0,a[2]==3,a[3]==13,a[n]==(n+1)a[n-1]+(n+2)a[n-2]+a[n-3]},a,{n,20}] (* Harvey P. Dale, Aug 03 2016 *)

G.f.: (1/(x+x^2)^2)*Sum_{n>=0} n!*(x/(1+x)^2)^n. - Vladeta Jovovic, Jun 29 2007
G.f.: (1/E(0)-1+x-x^2)/x^2 where E(k) = (1+x)^2 + k*x - (k+1)*x*(1+x)^2/E(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Sep 17 2012
From Vladimir Shevelev, Jun 29 - Jul 22 2015: (Start)
Using formulas in [Lucas], we have
(i) a(1)=0, a(2)=3, for n>=3, a(n) = (n+2)*a(n-1) + a(n-2) + 2*(-1)^n
(Note that (i) yields a(3)=13 and, for n>=4,
a(n-1) = (n+1)*a(n-2) + a(n-3) - 2*(-1)^n. Summing this with (i), we see that (i) defines A000904);
(ii) for n>=1, a(n) = (A000179(n+3) + 2*(-1)^n)/(n+3);
(iii) for n>=3, a(n) = a(n-2) + A000179(n+2);
(iii)*  if n is even, then a(n) = Sum_{i=0..(n+2)/2} A000179(2*i);
(iii)** if n is odd, then a(n) = Sum_{i=2..(n+1)/2} A000179(2*i+1).
Also
(iv)  a(n+1) = A000271(n+3) - a(n);
(iv)*  a(n) = Sum_{i=0..n+2}(-1)^i*A000271(n-i+2);
(v) a(n-2) = Sum_{i=0..n}(-1)^i*binomial(2*n-i+1, i)*(n-i)!, n>=3.
Note that Lucas considered this sequence with other initials. His formulas (i) - (iii), which he proved on pp. 491-495 of his book [Lucas], we wrote for the current initials.
The other 5 formulas, including the explicit formula 5), are new and we give their proofs:
(iii)*,(iii)**. Formulas (iii)* and (iii)** are obtained by the direct summation of (iii) over even and odd values respectively, taking into account the initials.
(iv). To obtain (iv), write (iii) in the form
a(j+1) - a(j) = -(a(j) - a(j-1)) +  A000179(j+3), j>=2.
Summing Sum_{j=2..n}, we have
a(n+1) - a(2) = -a(n) + a(1) + Sum_{j=2..n}A000179(j+3). Since a(1)=0, a(2)=3, we find
a(n+1)-3 = -a(n) + Sum_{i=5..n+3}A000179(i). But A000179(3)=1, A000179(4)=2. Therefore, we have
a(n+1) = -a(n) + Sum_{i=3..n+3}A000179(i).
It is left to note that, by the definition,
A000271(n) = Sum{i=3..n}A000179(i). So
a(n+1) = A000271(n+3) - a(n).
(iv)*. In view of the first four initials 1,0,0,1 of A000271, we have
Sum_{i=0..n+2}(-1)^i*A000271(n-i+2) = Sum_{i=0..n-2}(-1)^i*A000271(n-i+2) and, by (iv), the last sum equals
Sum_{i=0..n-2}(-1)^i(a(n-i)+a(n-i-1)) = a(n) + (-1)^(n-2)*a(1) = a(n).
(v). We use induction for n>=3. In case n=3 (v) gives 6-6*2+10-4=0 = a(1). Set n:=n+2.
Suppose for some (now n>=1) we have
a(n) = Sum_{i=0..n+2} (-1)^i*binomial(2*n-i+5, i)*(n+2-i)! (*)
Further we use (iv). In A259212 we proved an explicit formula for A000271(n-1) such that we have
A000271(n+3) = Sum_{j=0..n+3}(-1)^j * binomial(2*n-j+6,j)*(n-j+3)!, n>3.
Now, by (iv) and (*) with changing summing j=i+1, we have
a(n+1) = Sum_{j=0..n+3} (-1)^j*binomial(2*n-j+6,j)*(n-j+3)! + Sum_{j=1..n+3} (-1)^j*binomial(2*n-j+6, j-1)*(n+3-j)!
Since binomial(n,-1)=0, then in the second sum the summing could begin with j=0.
So, we have
a(n+1) = Sum_{j=0..n+3} (-1)^j*binomial(2*n-j+7, j)*(n+3-j)! = Sum_{j=0..n+3} (-1)^j*binomial(2*(n+1)-j+5,j)*((n+1)+2-j)!
Thus this expression formally is obtained from (*) by replacing n with n+1.
QED (End)
a(n) ~ n! * n^2 / exp(2). - Vaclav Kotesovec, Jul 04 2015

More terms from Larry Reeves (larryr(AT)acm.org), Nov 27 2001

A258665 A total of n married couples, including a mathematician M and his wife, are to be seated at the 2n chairs around a circular table, with no man seated next to his wife. After the ladies are seated at every other chair, M is the first man allowed to choose one of the remaining chairs. The sequence gives the number of ways of seating the other men, with no man seated next to his wife, if M chooses the chair that is 5 seats clockwise from his wife's chair.

Original entry on oeis.org

0, 0, 0, 1, 5, 20, 116, 791, 6203, 55000, 543576, 5922813, 70518113, 910704988, 12678282940, 189251856883, 3015212009143, 51067548545968, 916175515710896, 17355891466436025, 346195661281979133, 7252651426282955236, 159210312386078554436, 3654549974493252076175

Offset: 1

Vladimir Shevelev and Peter J. C. Moses, Jun 07 2015

nonn

This is a variation of the classic ménage problem (cf. A000179).
It is known [Riordan, ch. 8, ex. 7(b)] that, after the ladies are seated at every other chair, the number U_n of ways of seating the men in the ménage problem has asymptotic expansion U_n ~ e^(-2)*n!*(1 + Sum_{k>=1}(-1)^k/(k!(n-1)_k)), where (n)_k = n*(n-1)*...*(n-k+1).
Therefore, it is natural to conjecture that a(n) ~ e^(-2)*n!/(n-2)*(1 + Sum_{k>=1} (-1)^k/(k!(n-1)_k)).

I. Kaplansky and J. Riordan, The problème des ménages, Scripta Math. 12, (1946), 113-124.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, chs. 7, 8.

Peter J. C. Moses, Seatings for 6 couples.
I. Kaplansky and J. Riordan, The problème des ménages, Scripta Math. 12, (1946), 113-124. [Scan of annotated copy]
E. Lucas, Sur le problème des ménages, Théorie des nombres, Paris, 1891, 491-496.
Vladimir Shevelev, Peter J. C. Moses, The ménage problem with a known mathematician, arXiv:1101.5321 [math.CO], 2011, 2015.
Vladimir Shevelev and Peter J. C. Moses, Alice and Bob go to dinner: A variation on menage, INTEGERS, Vol. 16(2016), #A72.
J. Touchard, Sur un problème de permutations, C.R. Acad. Sci. Paris, 198 (1934), 631-633.

Cf. A000179, A258664, A258666, A258667, A258673, A259212.

enumerateSeatings[pairs_,d_]:=If[d==1||d>=2pairs-1||EvenQ[d],{},
Map[#[[1]]&,DeleteCases[Map[{#,Differences[#]}&[Riffle[Range[pairs],#]]&,Map[Insert[#,1,(d+1)/2]&,Permutations[#,{Length[#]}]&[Rest[Range[pairs]]]]],{{_},{_,0,_}}]]];
enumerateSeatings[6,5]
a[pairs_,d_]:=If[pairs<=#-1||EvenQ[d]||d==1,0,Sum[((-1)^k)*(pairs-k-1)!Sum[Binomial[2#-j-4,j]*Binomial[2(pairs-#)-k+j+2,k-j],{j,Max[#+k-pairs-1,0],Min[k,#-2]}],{k,0,pairs-1}]]&[(d+3)/2];
Table[a[n,5],{n,15}] (* Peter J. C. Moses, Jun 13 2015 *)

a(n) = sum(k=0,n-1, (-1)^k*(n-k-1)! * sum(j=max(k-n+3, 0), min(k,2), binomial(4-j, j)*binomial(2*n-k+j-6, k-j))); \\ Michel Marcus, Jun 13 2015

a(n) = Sum_{0<=k<=n-1} (-1)^k*(n-k-1)! * Sum_{max(k-n+3, 0)<=j<=min(k,2)} binomial(4-j, j)*binomial(2*n-k+j-6, k-j).

A258666 A total of n married couples, including a mathematician M and his wife, are to be seated at the 2n chairs around a circular table, with no man seated next to his wife. After the ladies are seated at every other chair, M is the first man allowed to choose one of the remaining chairs. The sequence gives the number of ways of seating the other men, with no man seated next to his wife, if M chooses the chair that is 7 seats clockwise from his wife's chair.

Original entry on oeis.org

0, 0, 0, 0, 4, 20, 117, 791, 6204, 55004, 543595, 5922925, 70518884, 910711076, 12678337153, 189252394275, 3015217877068, 51067618521276, 916176420499159, 17355904074255065, 346195849623668420, 7252654428822549364, 159210363264445218829, 3654550887654460566191

Offset: 1

Vladimir Shevelev and Peter J. C. Moses, Jun 07 2015

nonn

This is a variation of the classic ménage problem (cf. A000179).
It is known [Riordan, ch. 8, ex. 7(b)] that, after the ladies are seated at every other chair, the number U_n of ways of seating the men in the ménage problem has asymptotic expansion U_n ~ e^(-2)*n!*(1 + Sum_{k>=1} (-1)^k/(k!(n-1)_k)), where (n)_k = n*(n-1)*...*(n-k+1).
Therefore, it is natural to conjecture that a(n) ~ e^(-2)*n!/(n-2)*(1 + Sum_{k>=1} (-1)^k/(k!(n-1)_k)).

I. Kaplansky and J. Riordan, The problème des ménages, Scripta Math. 12, (1946), 113-124.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, chs. 7, 8.

Peter J. C. Moses, Seatings for 5 couples
I. Kaplansky and J. Riordan, The problème des ménages, Scripta Math. 12, (1946), 113-124. [Scan of annotated copy]
E. Lucas, Sur le problème des ménages, Théorie des nombres, Paris, 1891, 491-496.
Vladimir Shevelev, Peter J. C. Moses, The ménage problem with a known mathematician, arXiv:1101.5321 [math.CO], 2011-2015.
Vladimir Shevelev and Peter J. C. Moses, Alice and Bob go to dinner: A variation on menage, INTEGERS, Vol. 16(2016), #A72.
J. Touchard, Sur un problème de permutations, C.R. Acad. Sci. Paris, 198 (1934), 631-633.

Cf. A000179, A258664, A258665, A258667, A258673, A259212.

a[n_] := If[n<5, 0, Sum[(-1)^k (n-k-1)! Sum[Binomial[6-j, j] Binomial[2n-k+j-8, k-j], {j, Max[k-n+4, 0], Min[k, 3]}], {k, 0, n-1}]];
Array[a, 24] (* Jean-François Alcover, Sep 19 2018 *)

vector(30, n, if (n<=4, 0, sum(k=0,n-1,(-1)^k*(n-k-1)!*sum(j=max(k-n+4, 0),min(k,3), binomial(6-j, j)*binomial(2*n-k+j-8, k-j))))) \\ Michel Marcus, Jun 17 2015

a(n)=0 for n <= 4; for n >= 5, a(n) = Sum_{k=0..n-1} (-1)^k*(n-k-1)! Sum_{j=max(k-n+4, 0)..min(k,3)} binomial(6-j, j)*binomial(2*n-k+j-8, k-j).

A258667 A total of n married couples, including a mathematician M and his wife, are to be seated at the 2n chairs around a circular table, with no man seated next to his wife. After the ladies are seated at every other chair, M is the first man allowed to choose one of the remaining chairs. The sequence gives the number of ways of seating the other men, with no man seated next to his wife, if M chooses the chair that is 9 seats clockwise from his wife's chair.

Original entry on oeis.org

0, 0, 0, 0, 0, 20, 116, 791, 6205, 55004, 543596, 5922929, 70518903, 910711188, 12678337924, 189252400363, 3015217931281, 51067619058668, 916176426367084, 17355904144230373, 346195850528456683, 7252654441430368404, 159210363452786908116, 3654550890657000160319

Offset: 1

Vladimir Shevelev and Peter J. C. Moses, Jun 07 2015

nonn

This is a variation of the classic ménage problem (cf. A000179).
It is known [Riordan, ch. 8, ex. 7(b)] that, after the ladies are seated at every other chair, the number U_n of ways of seating the men in the ménage problem has asymptotic expansion U_n ~ e^(-2)*n!*(1 + Sum_{k>=1} (-1)^k/(k!(n-1)_k)), where (n)_k = n*(n-1)*...*(n-k+1).
Therefore, it is natural to conjecture that a(n) ~ e^(-2)*n!/(n-2)*(1 + Sum_{k>=1} (-1)^k/(k!(n-1)_k)).

I. Kaplansky and J. Riordan, The problème des ménages, Scripta Math. 12, (1946), 113-124.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, chs. 7, 8.

I. Kaplansky and J. Riordan, The problème des ménages, Scripta Math. 12, (1946), 113-124. [Scan of annotated copy]
Peter J. C. Moses, Seatings for 6 couples
E. Lucas, Sur le problème des ménages, Théorie des nombres, Paris, 1891, 491-496.
Vladimir Shevelev, Peter J. C. Moses, The ménage problem with a known mathematician, arXiv:1101.5321 [math.CO], 2011-2015.
Vladimir Shevelev and Peter J. C. Moses, Alice and Bob go to dinner: A variation on menage, INTEGERS, Vol. 16(2016), #A72.
J. Touchard, Sur un problème de permutations, C.R. Acad. Sci. Paris, 198 (1934), 631-633.

Cf. A000179, A258664, A258665, A258666, A258673, A259212.

a[n_] := If[n<6, 0, Sum[(-1)^k (n-k-1)! Sum[Binomial[8-j, j] Binomial[2n-k+j-10, k-j], {j, Max[k-n+5, 0], Min[k, 4]}], {k, 0, n-1}]];
Array[a, 24] (* Jean-François Alcover, Sep 19 2018 *)

a(n) = if (n<=5, 0, sum(k=0, n-1, (-1)^k*(n-k-1)!*sum(j=max(k-n+5, 0), min(k,4), binomial(8-j, j)*binomial(2*n-k+j-10, k-j)))); \\ Michel Marcus, Jun 26 2015

For n <= 5, a(n)=0; otherwise a(n) = Sum_{0<=k<=n-1}(-1)^k*(n-k-1)! Sum_{max(k-n+5, 0)<=j<=min(k,4)} binomial(8-j, j)*binomial(2*n-k+j-10, k-j).

The "probleme des menages" asks for the number of gender-alternating seating arrangements for n couples around a circular table with the condition that no two spouses are seated adjacently. - Paul C. Kainen and Michael Somos, Mar 11 2011

The reference to A094047 arises in the formula because that sequence is also the number of directed Hamiltonian cycles in the n-crown graph. (Each cycle can be broken in 2n ways to give a path.) - Andrew Howroyd, Feb 21 2016

Also, the number of ways of seating n married couples at 2*n chairs arranged side-by-side in a straight line, men and women in alternate positions, so that no husband is next to his wife. - Andrew Howroyd, Sep 19 2017

cp's OEIS Frontend

A000179 Ménage numbers: a(0) = 1, a(1) = -1, and for n >= 2, a(n) = number of permutations s of [0, ..., n-1] such that s(i) != i and s(i) != i+1 (mod n) for all i.

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A000904 a(n) = (n+1)*a(n-1) + (n+2)*a(n-2) + a(n-3); a(1)=0, a(2)=3, a(3)=13.

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A059375 Number of seating arrangements for the ménage problem.

A000904 a(n) = (n+1)a(n-1) + (n+2)a(n-2) + a(n-3); a(1)=0, a(2)=3, a(3)=13.