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 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. (Formerly M2062 N0815) 62
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 COMMENTS 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) REFERENCES 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. LINKS 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. FORMULA 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 EXAMPLE 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. MAPLE 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 MATHEMATICA 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 *) PROG (PARI) \\ 3 programs adapted to a(1) = -1 by Hugo Pfoertner, Aug 31 2020 (PARI) {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 */ (PARI) a(n)=if(n>1, round(2*n*exp(-2)*besselk(n, 2)), 1-2*n) \\ Charles R Greathouse IV, Nov 03 2014 (PARI) {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 */ (Haskell) 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 (Python) 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 CROSSREFS 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 Sequence in context: A037571 A179237 A216316 * A335700 A246383 A189087 Adjacent sequences: A000176 A000177 A000178 * A000180 A000181 A000182 KEYWORD sign,nice,easy AUTHOR N. J. A. Sloane EXTENSIONS More terms from James A. 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 STATUS approved

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Last modified September 22 16:46 EDT 2023. Contains 365531 sequences. (Running on oeis4.)