A114208 Number of permutations of [n] having exactly one fixed point and avoiding the patterns 123 and 231.

1, 0, 3, 2, 6, 6, 12, 10, 21, 16, 31, 24, 44, 32, 60, 42, 77, 54, 97, 66, 120, 80, 144, 96, 171, 112, 201, 130, 232, 150, 266, 170, 303, 192, 341, 216, 382, 240, 426, 266, 471, 294, 519, 322, 570, 352, 622, 384, 677, 416, 735, 450, 794, 486, 856, 522, 921, 560

Offset

1,3

Links

Harvey P. Dale, Table of n, a(n) for n = 1..1000

T. Mansour and A. Robertson, Refined restricted permutations avoiding subsets of patterns of length three, Annals of Combinatorics, 6, 2002, 407-418.

Index entries for linear recurrences with constant coefficients, signature (-1,2,3,0,-3,-2,1,1).

Formula

n^2/6 if n mod 6 = 0; (7*n^2-12*n+29)/24 if n mod 6 = 1 or 5; (n^2-4)/6 if n mod 6 = 2 or 4; (7*n^2-12*n+45)/24 if n mod 6 = 3.

a(n) = a(n-1)+ 2*a(n-2)+3*a(n-3)-3*a(n-5)-2*a(n-6)+a(n-7)+a(n-8). [Harvey P. Dale, Mar 05 2012]

G.f.: -x*(2*x^6+2*x^5+2*x^4+2*x^3+x^2+x+1) / ((x-1)^3*(x+1)^3*(x^2+x+1)). [Colin Barker, Aug 11 2013]

Example

a(2)=0 because none of the permutations 12 and 21 has exactly one fixed point.
a(3)=3 because we have 132, 213 and 321.
a(4)=2 because we have 4132 and 4213.

Maple

a:=proc(n) if n mod 6 = 0 then n^2/6 elif n mod 6 = 1 or n mod 6 = 5 then (7*n^2-12*n+29)/24 elif n mod 6 = 2 or n mod 6 = 4 then (n^2-4)/6 else (7*n^2-12*n+45)/24 fi end: seq(a(n), n=1..70);

Mathematica

npn[n_]:=Module[{c=Mod[n, 6]}, Which[c==0, n^2/6, c==1, (7n^2-12n+29)/24, c==2, (n^2-4)/6, c==3, (7n^2-12n+45)/24, c==4, (n^2-4)/6, c==5, (7n^2-12n+29)/24]]; Array[npn, 60] (* or *) LinearRecurrence[{-1, 2, 3, 0, -3, -2, 1, 1}, {1, 0, 3, 2, 6, 6, 12, 10}, 60] (* Harvey P. Dale, Mar 05 2012 *)

Crossrefs

Cf. A114209, A114210.

Sequence in context: A157793 A096375 A062200 * A014686 A053090 A264400

Adjacent sequences: A114205 A114206 A114207 * A114209 A114210 A114211

Keyword

nonn,easy

Author

Emeric Deutsch, Nov 17 2005

Status

approved