A292574 -id:A292574 - cp's OEIS frontend

A000271 Sums of ménage numbers.

1, 0, 0, 1, 3, 16, 96, 675, 5413, 48800, 488592, 5379333, 64595975, 840192288, 11767626752, 176574062535, 2825965531593, 48052401132800, 865108807357216, 16439727718351881, 328839946389605643, 6906458590966507696

Offset: 0

N. J. A. Sloane

Permanent of the (0,1)-matrix having (i,j)-th entry equal to 0 iff this is on the diagonal or the first upper-diagonal. - Simone Severini, Oct 14 2004
Equivalently, number of permutations p of {1,2,...,n} such that p(i)-i not in {0,1}. - Andrew Howroyd, Sep 19 2017
From Vladimir Shevelev, Jun 21 2015: (Start)
Let 2*n!*V(n)=A137886(n) be 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.
It is known [Riordan, Ch. 8, Th. 1, t=0] that, if 2*n!*U(n) is a solution of an analogous problem at a circular table, then U(n) = V(n) - V(n-1), n>=3, where U(n) = A000179(n). Thus V(n) = Sum_{i=3,...,n} A000179(i), n>=1, and comparing the initial conditions, we conclude that a(n) = V(n), n>=1. This gives a combinatorial interpretation for 2*n!*a(n).
(End)

			G.f. = 1 + x^3 + 3*x^4 + 16*x^5 + 96*x^6 + 675*x^7 + 5413*x^8 + ...

W. Ahrens, Mathematische Unterhaltungen und Spiele. Teubner, Leipzig, Vol. 1, 3rd ed., 1921; Vol. 2, 2nd ed., 1918. See Vol. 2, p. 79.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 198.
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.

Seiichi Manyama, Table of n, a(n) for n = 0..450 (terms 0..100 from T. D. Noe)
W. Ahrens, Mathematische Unterhaltungen und Spiele, Leipzig: B. G. Teubner, 1901.
J. D. H. Dickson, Discussion of two double series arising from the number of terms in determinants of certain forms, Proc. London Math. Soc., 10 (1879), 120-122.
J. D. H. Dickson, Discussion of two double series arising from the number of terms in determinants of certain forms, Proc. London Math. Soc., 10 (1879), 120-122. [Annotated scanned copy]
Dmitry Efimov, Determinants of generalized binary band matrices, arXiv:1702.05655 [math.RA], 2017.
V. Kotesovec, Non-attacking chess pieces, 6ed, 2013, p. 222.
Y. Li, Ménage Numbers and Ménage Permutations, J. Int. Seq. 18 (2015) 15.6.8
H. M. Taylor, A problem on arrangements, Mess. Math., 32 (1902), 60ff. [Annotated scanned copy]
M. Wyman and L. Moser, On the problème des ménages, Canad. J. Math., 10 (1958), 468-480.

Cf. A000179, A000904, A001883, A137886, A292574. A diagonal of A058057.

[ &+[(-1)^(n-k)*Binomial(n+k, 2*k)*Factorial(k): k in [0..n]]: n in [0..21]]; // Bruno Berselli, Apr 11 2011

V := proc(n) local k; add( binomial(2*n-k,k)*(n-k)!*(x-1)^k, k=0..n); end; W := proc(r,s) coeff( V(r),x,s ); end; A000271 := n->W(n-2,0);

Table[Sum[(-1)^(n - k) k! Binomial[n + k, 2 k], {k, 0, n}], {n, 0, 22}] (* Jean-François Alcover, Apr 11 2011, after Paul Barry *)
RecurrenceTable[{a[0] == 1, a[1] == a[2] == 0, a[n] == (n - 1) a[n - 2] + (n - 1) a[n - 1] +  a[n - 3]}, a, {n, 30}] (* Harvey P. Dale, Jun 01 2012 *)
Table[(-1)^n HypergeometricPFQ[{1, -n, n + 1}, {1/2}, 1/4], {n, 20}] (* Michael Somos, May 28 2014 *)

a(n) = if(n, round( 2*exp(-2)*(besselk(n+1,2) + besselk(n,2)) ), 1) \\ Charles R Greathouse IV, May 11 2016

a(n) = (n - 1) a(n - 2) + (n - 1) a(n - 1) + a(n - 3).
From Paul Barry, Feb 08 2009: (Start)
G.f.: 1/(1+x-x/(1+x-x/(1+x-2x/(1+x-2x/(1+x-3x/(1+x-3x/(1+x-4x/(1+... (continued fraction);
a(n) = Sum_{k=0..n} binomial(2n-k,k)*(n-k)!*(-1)^k. (End)
a(n) = (-1)^n*hypergeom([1, -n, n+1],[1/2],1/4). - Mark van Hoeij, Nov 12 2009
a(n) = round( 2*exp(-2)*(BesselK(1+n,2) + BesselK(n,2)) ) for n>0. - Mark van Hoeij, Nov 12 2009
a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n+k,2*k)*k!. - Paul Barry, Jun 23 2010
G.f.: Sum_{n>=0} n!*x^n/(1+x)^(2*n+1). - Ira M. Gessel, Jan 15 2013
a(n) ~ exp(-2)*n!. - Vaclav Kotesovec, Mar 10 2014
a(-1 - n) = -a(n) for all n in Z. - Michael Somos, May 28 2014
a(n) = Sum_{i=3..n} A000179(i), n>=1. - Vladimir Shevelev, Jun 21 2015
0 = a(n)*(-a(n+2) - a(n+3)) + a(n+1)*(+a(n+1) + 2*a(n+2) + a(n+3) - a(n+4)) + a(n+2)*(+a(n+2) + 2*a(n+3) - a(n+4)) + a(n+3)*(+a(n+3)) for all n in Z. - Michael Somos, Oct 16 2016

More terms from James Sellers, Aug 21 2000

More terms from Simone Severini, Oct 14 2004

A270174 a(n) is the number of different ways to seat a set of n married male-female couples at a straight table so that men and women alternate and every man is separated by at least two men from his wife.

Original entry on oeis.org

0, 0, 0, 0, 240, 8640, 584640, 40239360, 3493808640, 364941158400, 45683021260800, 6754660222464000, 1166167699041945600, 232618987254682828800, 53114643986227439616000, 13768242163527512973312000, 4021980517038414919532544000, 1315337131173516220415213568000

Offset: 1

Feng Jishe, Mar 12 2016

nonn

We assume that the chairs are uniform and indistinguishable.

First we arrange the women in alternating seats, in 2*n! ways. Second, we find the number, G_{n} say, of ways of arranging men in the remaining seats such that every husband cannot sit at the left or right next 1, 2, ..., h male's seats from his wife. Note that here h = 2. We give the board B4, where X denotes the seat cannot be set at, where there are h X's in first column, and h+1 X's in first row, ..., 2h X's in the h column, ..., other entries are 1's. Thus the number of different ways to seat a set of n married male-female couples at a straight table is a_{n}=2*n!*G_{n}.

Feng Jishe, The board B4
D. Zeilberger, Automatic Enumeration of Generalized Ménage Numbers
D. Zeilberger, Automatic Enumeration of Generalized Menage Numbers, arXiv preprint arXiv:1401.1089 [math.CO], 2014.

Cf. A267060, A292574.

a(n) = 2*n! * A292574(n). - Andrew Howroyd, Sep 19 2017

a(11)-a(18) from Andrew Howroyd, Sep 19 2017

cp's OEIS Frontend

A000271 Sums of ménage numbers.

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