A089222 Number of ways of seating n people around a table for the second time without anyone sitting next to the same person as they did the first time.

1, 0, 0, 0, 0, 10, 36, 322, 2832, 27954, 299260, 3474482, 43546872, 586722162, 8463487844, 130214368530, 2129319003680, 36889393903794, 675098760648204, 13015877566642418, 263726707757115400, 5603148830577775218, 124568968969991162100, 2892414672938546871250

Offset: 0

Author

Udi Hadad (somebody(AT)netvision.net.il), Dec 22 2003

Keywords

• nonn
• nice

Comments

A078603 counts these arrangements up to circular symmetry (i.e., two arrangements are the same if one can be rotated to give the other). A002816 counts them up to dihedral symmetry (i.e., two arrangements are the same if one can be rotated or reflected to give the other). - Joel B. Lewis, Jan 28 2010

Examples

			a(4)=0 because trying to arrange 1,2,3,4 around a table will always give a couple who is sitting next to each other and differ by 1.
		

References

• J. Snell, Introduction to Probability, e-book, pp. 101 Q. 20.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• Art of Problem Solving forum, Random neighbors. - _Joel B. Lewis_, Jan 28 2010
• B. Aspvall and F. M. Liang, The dinner table problem, Technical Report CS-TR-80-829, Computer Science Department, Stanford, California, 1980.
• Charles M. Grinstead and J. Laurie Snell, Introduction to Probability.
• Vaclav Kotesovec, Non-attacking chess pieces, 6ed, 2013, p. 626.
• Roberto Tauraso, The Dinner Table Problem: The Rectangular Case, INTEGERS: Electronic Journal of Combinatorial Number Theory, Vol. 6 (2006), #A11. Note that in this paper a(1) = 1. See Column 2 in the table on page 3.

Crossrefs

Cf. A002464, A002816, A078603, A078630, A078631.

Programs

Mathematica

Same[cperm_, n_] := ( For[same = False; i = 2, (i <= n) && ! same, i++, same = ((Mod[cperm[[i - 1]], n] + 1) == cperm[[i]]) || ((Mod[cperm[[ i]], n] + 1) == cperm[[i - 1]])]; same = same || ((Mod[cperm[[n]], n] + 1) == cperm[[1]]) || ((Mod[ cperm[[1]], n] + 1) == cperm[[n]]); Return[same]); CntSame[n_] := (allPerms = Permutations[Range[n]]; count = 0; For[j = 1, j <= n!, j++, perm = allPerms[[j]]; If[ ! Same[perm, n], count++ ]]; Return[count]);
(* or direct computation of terms *)
Table[If[n<3, 0, n! + (-1)^n*2n + Sum[(-1)^r*(n/(n-r))^2 * (n-r)! * Sum[2^c * Binomial[r-1,c-1] * Binomial[n-r,c], {c,1,r}], {r,1,n-1}]], {n,1,25}] (* Vaclav Kotesovec, Apr 06 2012 *)

Formula

Inclusion-exclusion gives that for n > 2, we have a(n) = n! + 2*n*(-1)^n + Sum_{1 <= k <= m < n} (-1)^m * (n/k) * binomial(n-m-1, k-1) * binomial(m-1, k-1) * 2^k * n * (n-m-1)!. - Joel B. Lewis, Jan 28 2010

a(n) = (3*n-30)*a(n-11) + (6*n-45)*a(n-10) + (5*n+18)*a(n-9) - (8*n-139)*a(n-8) - (26*n-204)*a(n-7) - (4*n-30)*a(n-6) + (26*n-148)*a(n-5) + (8*n-74)*a(n-4) - (9*n-18)*a(n-3) - (2*n-15)*a(n-2) + (n+2)*a(n-1), n >= 14. - Vaclav Kotesovec, Apr 13 2010

The asymptotic expansion from article by Aspvall and Liang (also cited in article by Tauraso) is wrong. Bad terms are 736/(15*n^5) + 8428/(45*n^6) + 40174/(63*n^7). Right asymptotic formula is a(n) ~ (n!/e^2)*(1 - 4/n + 20/(3*n^3) + 58/(3*n^4) + 796/(15*n^5) + 7858/(45*n^6) + 40324/(63*n^7) + 140194/(63*n^8) + ...). Verified also numerically. For example, for n=200, exact/asymptotic results are 1.0000000000125542243 (Aspvall + Liang), 1.0000000000000008990 (Kotesovec 7 terms) or 1.0000000000000000121 (Kotesovec 8 terms). - Vaclav Kotesovec, Apr 06 2012

a(n) = 2*n*A002816(n) for n > 1. - Martin Renner, Apr 01 2022

Extensions

Tauraso reference from Parthasarathy Nambi, Dec 21 2006

More terms from Vladeta Jovovic, Nov 29 2009

a(0)=1 prepended by Alois P. Heinz, Jul 31 2019