This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A258673 #68 Sep 03 2023 23:39:56
%S A258673 0,0,0,0,0,0,115,791,6204,55004,543597,5922929,70518904,910711192,
%T A258673 12678337943,189252400475,3015217932052,51067619064756,
%U A258673 916176426421297,17355904144767765,346195850534324608,7252654441500343712,159210363453691696379,3654550890669607979359
%N A258673 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 11 seats clockwise from his wife's chair.
%C A258673 This is a variation of the classic ménage problem (cf. A000179).
%C A258673 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).
%C A258673 Therefore, it is natural to conjecture that a(n) ~ e^(-2)*n!/(n-2)*(1 + Sum_{k>=1} (-1)^k/(k!(n-1)_k)).
%C A258673 In the general case, M chooses a chair at an odd distance d >= 3 clockwise from his wife. See the corresponding general formula below.
%D A258673 I. Kaplansky and J. Riordan, The problème des ménages, Scripta Math. 12, (1946), 113-124.
%D A258673 J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, chs. 7, 8.
%H A258673 I. Kaplansky and J. Riordan, <a href="/A000166/a000166_1.pdf">The problème des ménages</a>, Scripta Math. 12, (1946), 113-124. [Scan of annotated copy]
%H A258673 Peter J. C. Moses, <a href="/A258673/a258673_1.pdf">Seatings for 7 couples</a>.
%H A258673 E. Lucas, <a href="https://archive.org/details/thoriedesnombre00lucagoog/page/n495">Sur le problème des ménages</a>, Théorie des nombres, Paris, 1891, 491-496.
%H A258673 Vladimir Shevelev, Peter J. C. Moses, <a href="http://arxiv.org/abs/1101.5321">The ménage problem with a known mathematician</a>, arXiv:1101.5321 [math.CO], 2011, 2015.
%H A258673 Vladimir Shevelev and Peter J. C. Moses, <a href="http://www.emis.de/journals/INTEGERS/papers/q72/q72.Abstract.html">Alice and Bob go to dinner: A variation on menage</a>, INTEGERS, Vol. 16(2016), #A72.
%H A258673 J. Touchard, <a href="http://gallica.bnf.fr/ark:/12148/bpt6k31506/f631.image">Sur un problème de permutations</a>, C.R. Acad. Sci. Paris, 198 (1934), 631-633.
%F A258673 For n <= 6, a(n)=0; otherwise a(n) = Sum_{k=0..n-1} (-1)^k*(n-k-1)! * Sum_{j=max(k-n+6, 0)..min(k,5)} binomial(10-j, j)*binomial(2*n-k+j-12, k-j).
%F A258673 In the general case (see comment), let r=(d+3)/2 and denote the solution by A(r,n). Then A(r,n) is given by the formula
%F A258673 A(r,n)=0 for n <= (d+1)/2; otherwise A(r,n) = Sum_{k=0..n-1} ((-1)^k)*(n-k-1)! * Sum_{j=max(r+k-n-1, 0)..min(k,r-2)} binomial(2r-j-4, j)*binomial(2(n-r) - k + j + 2, k-j).
%F A258673 Note that, if n is even, then 2*Sum_{r=3..(n+2)/2} A(r,n) = A000179(n); if n is odd, then 2*Sum_{r=3..(n+1)/2} A(r,n) + A((n+3)/2, n) = A000179(n).
%t A258673 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];
%t A258673 Map[a[11,#]&,Range[20]] (* _Peter J. C. Moses_, Jun 07 2015 *)
%Y A258673 Cf. A000179, A258664, A258665, A258666, A258667, A259673.
%K A258673 nonn
%O A258673 1,7
%A A258673 _Vladimir Shevelev_ and _Peter J. C. Moses_, Jun 07 2015