

A114208


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


3



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
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OFFSET

1,3


REFERENCES

T. Mansour and A. Robertson, Refined restricted permutations avoiding subsets of patterns of length three, Annals of Combinatorics, 6, 2002, 407418.


LINKS

Harvey P. Dale, Table of n, a(n) for n = 1..1000
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^212*n+29)/24 if n mod 6 = 1 or 5; (n^24)/6 if n mod 6 = 2 or 4; (7*n^212*n+45)/24 if n mod 6 = 3.
a(1)=1, a(2)=0, a(3)=3, a(4)=2, a(5)=6, a(6)=6, a(7)=12, a(8)=10, a(n)= a(n1)+ 2*a(n2)+3*a(n3)3*a(n5)2*a(n6)+a(n7)+a(n8) [From 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) / ((x1)^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^212*n+29)/24 elif n mod 6 = 2 or n mod 6 = 4 then (n^24)/6 else (7*n^212*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^212n+29)/24, c==2, (n^24)/6, c==3, (7n^212n+45)/24, c==4, (n^24)/6, c==5, (7n^212n+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



