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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.
8
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
0,6
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
REFERENCES
J. Snell, Introduction to Probability, e-book, pp. 101 Q. 20.
LINKS
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 & J. Laurie Snell Introduction to Probability.
V. Kotesovec, Non-attacking chess pieces, 6ed, 2013, p. 626.
Art of Problem Solving forum, Random neighbors. [From Joel B. Lewis, Jan 28 2010]
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.
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/(15n^5) + 8428/(45n^6) + 40174/(63n^7)). Right asymptotic formula is a(n) ~ (n!/e^2)*(1 - 4/n + 20/(3n^3) + 58/(3n^4) + 796/(15n^5) + 7858/(45n^6) + 40324/(63n^7) + 140194/(63n^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
EXAMPLE
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.
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 *)
CROSSREFS
KEYWORD
nonn,nice
AUTHOR
Udi Hadad (somebody(AT)netvision.net.il), Dec 22 2003
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
STATUS
approved