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A000183
Number of discordant permutations of length n.
(Formerly M2121 N0838)
13
0, 0, 0, 1, 2, 20, 144, 1265, 12072, 126565, 1445100, 17875140, 238282730, 3407118041, 52034548064, 845569542593, 14570246018686, 265397214435860, 5095853023109484, 102877234050493609, 2178674876680100744, 48296053720501168037, 1118480911876659396600
OFFSET
1,5
COMMENTS
Ways to reseat n diners at circular table, none in or next to original chair.
REFERENCES
J. Riordan, Discordant permutations, Scripta Math., 20 (1954), 14-23.
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).
R. P. Stanley, Enumerative Combinatorics I, Example 4.7.17.
K. Yamamoto, Structure polynomial of Latin rectangles and its application to a combinatorial problem, Memoirs of the Faculty of Science, Kyusyu University, Series A, 10 (1956), 1-13.
LINKS
J. Riordan, Discordant permutations, Scripta Math., 20 (1954), 14-23. [Annotated scanned copy]
Anthony C. Robin, 90.72 Circular Wife Swapping, The Mathematical Gazette, Vol. 90, No. 519 (Nov., 2006), pp. 471-478.
K. Yamamoto, Structure polynomial of Latin rectangles and its application to a combinatorial problem, Memoirs of the Faculty of Science, Kyusyu University, Series A, 10 (1956), 1-13. [Annotated scanned copy]
D. Zeilberger, Automatic Enumeration of Generalized Menage Numbers, arXiv preprint arXiv:1401.1089 [math.CO], 2014.
FORMULA
a(n) = Sum_{m=0..n} (-1)^m*(n-m)!*A061702(n, m), n>2.
From Vladimir Shevelev, Apr 17 2011: (Start)
Let f(n) = F(n-1) + F(n+1) + 2, where F(n) is the n-th Fibonacci number.
Then, for n>=7, we have the recursion:
a(n) = (-1)^n*(4*n+f(n)) + (n/(n-1))*((n+1)*a(n-1) + 2*(-1)^n*f(n-1)) - ((2*n)/(n-2))*((n-3)*a(n-2) + (-1)^n*f(n-2)) + (n/(n-3))*((n-5)*a(n-3) + 2*(-1)^(n-1)*f(n-3)) + (n/(n-4))*(a(n-4) + (-1)^(n-1)*f(n-4)).
This formula (in an equivalent form) is due to K. Yamamoto. (End)
a(n) ~ n!*exp(-3). - Vaclav Kotesovec, Aug 10 2013
EXAMPLE
a(5) = 2: [ 1 2 3 4 5 ] -> [ 3 4 5 1 2 ] or [ 4 5 1 2 3 ].
Let n=7. Then, using the previous values of a(n), we have a(7) = -(4*7+31) + (7/6)*(8*20-2*20) - (14/5)*(4*2-13) + (7/4)*(2*1+2*9) + (7/3)*6 = -59+140+14+35+14 = 144. - Vladimir Shevelev, Apr 17 2011
MAPLE
with(combinat): f:= n-> fibonacci(n-1) +fibonacci(n+1) +2:
a:= proc(n) option remember; `if` (n<7, [0$3, 1, 2, 20][n], (-1)^n*(4*n+f(n)) +(n/(n-1))*((n+1)*a(n-1) +2*(-1)^n*f(n-1)) -((2*n)/(n-2))*((n-3)*a(n-2) +(-1)^n*f(n-2)) +(n/(n-3))*((n-5)*a(n-3) +2*(-1)^(n-1)*f(n-3)) +(n/(n-4))*(a(n-4) +(-1)^(n-1)*f(n-4))) end:
seq(a(n), n=1..30); # Alois P. Heinz, Apr 19 2011
MATHEMATICA
max = 22; f[x_, y_] := y*(1 + 3*x - 4*x^2*y - 3*x^2*y^2 - 3*x^3*y^2 + 4*x^4*y^3)/((1 - y - 2*x*y - x*y^2 + x^3*y^3)*(1 - x*y)); se = Series[f[x, y], {x, 0, max}, {y, 0, max}]; coes = CoefficientList[se, {x, y}] ; t[n_, k_] := coes[[k, n]]; a[n_] := Sum[ (-1)^(k+1)*(n-k+1)!*t[n+1, k], {k, 1, n+1}]; a[1] = a[2] = a[3] = 0; Table[a[n], {n, 1, max}] (* Jean-François Alcover, Oct 24 2011 *)
Flatten[{0, 0, RecurrenceTable[{(382-1142 n+712 n^2-185 n^3+22 n^4-n^5) a[-7+n]+(-3776+11024 n-7689 n^2+2397 n^3-384 n^4+31 n^5-n^6) a[-6+n]+(7394-18064 n+12353 n^2-3937 n^3+661 n^4-57 n^5+2 n^6) a[-5+n]+(1452-10548 n+8254 n^2-2655 n^3+423 n^4-33 n^5+n^6) a[-4+n]+(-11046+26716 n-18588 n^2+6013 n^3-1015 n^4+87 n^5-3 n^6) a[-3+n]+(632+5546 n-3888 n^2+1007 n^3-116 n^4+5 n^5) a[-2+n]+(3966-4666 n+3655 n^2-1445 n^3+284 n^4-27 n^5+n^6) a[-1+n]+(2444-3214 n+1409 n^2-283 n^3+27 n^4-n^5) a[n]==0, a[8]==1265, a[9]==12072, a[3]==0, a[4]==1, a[5]==2, a[6]==20, a[7]==144}, a, {n, 3, 20}]}] (* Vaclav Kotesovec, Aug 10 2013 *)
CROSSREFS
KEYWORD
nonn,nice,easy
EXTENSIONS
More terms from Vladeta Jovovic, Jun 18 2001
STATUS
approved