|
|
A068018
|
|
Number of fixed points in all 132- and 213-avoiding permutations of {1,2,...,n} (these are permutations with runs consisting of consecutive integers).
|
|
1
|
|
|
0, 1, 2, 4, 6, 12, 18, 40, 62, 148, 234, 576, 918, 2284, 3650, 9112, 14574, 36420, 58266, 145648, 233030, 582556, 932082, 2330184, 3728286, 9320692, 14913098, 37282720, 59652342, 149130828, 238609314, 596523256, 954437198, 2386092964, 3817748730, 9544371792
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,3
|
|
LINKS
|
|
|
FORMULA
|
a(n) = 2^n/4 - (-2)^n/36 + 2*n/3 - 2/9.
G.f.: z*(1 - 3*z^2)/((1 - 4*z^2)*(1 - z)^2).
E.g.f.: (cosh(x)*(5*sinh(x) + 6*x - 2) + 2*(cosh(2*x) + (3*x - 1)*sinh(x)))/9. - Stefano Spezia, Jun 12 2023
|
|
EXAMPLE
|
a(3) = 4 because the permutations 123, 231, 312, 321 of {1,2,3} contain 4 fixed points altogether (all three entries of the first permutations and the entry 2 in the last one.
|
|
MAPLE
|
seq(2^n/4-(-2)^n/36+2*n/3-2/9, n=0..40);
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|