|
|
A284816
|
|
Sum of entries in the first cycles of all permutations of [n].
|
|
6
|
|
|
1, 4, 21, 132, 960, 7920, 73080, 745920, 8346240, 101606400, 1337212800, 18920563200, 286442956800, 4620449433600, 79114299264000, 1433211107328000, 27387931963392000, 550604138692608000, 11617107089043456000, 256671161862635520000, 5926549291918295040000
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
Each cycle is written with the smallest element first and cycles are arranged in increasing order of their first elements.
Also, the number of colorings of n+1 given balls, two thereof identical, using n given colors (each color is used). - Ivaylo Kortezov, Jan 27 2024
|
|
LINKS
|
|
|
FORMULA
|
a(n) = n!*(n*(n+1) - (n-1)*(n+2)/2)/2.
E.g.f.: -x*(x^2-2*x+2)/(2*(x-1)^3).
a(n) = (n^2+n+2)*n*a(n-1)/(n^2-n+2) for n > 1, a(n) = n for n < 2.
|
|
EXAMPLE
|
a(3) = 21 because the sum of the entries in the first cycles of all permutations of [3] ((123), (132), (12)(3), (13)(2), (1)(23), (1)(2)(3)) is 6+6+3+4+1+1 = 21.
|
|
MAPLE
|
a:= n-> n!*(n*(n+1)-(n-1)*(n+2)/2)/2:
seq(a(n), n=1..25);
# second Maple program:
a:= proc(n) option remember; `if`(n<2, n,
(n^2+n+2)*n*a(n-1)/(n^2-n+2))
end:
seq(a(n), n=1..25);
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|