|
| |
|
|
A145893
|
|
Triangle read by rows: T(n,k) is the number of permutations p of {1,2,...,n} such that j and p(j) are of opposite parities for k values of j (0<=k<=n).
|
|
1
| |
|
|
1, 1, 0, 1, 0, 1, 2, 0, 4, 0, 4, 0, 16, 0, 4, 12, 0, 72, 0, 36, 0, 36, 0, 324, 0, 324, 0, 36, 144, 0, 1728, 0, 2592, 0, 576, 0, 576, 0, 9216, 0, 20736, 0, 9216, 0, 576, 2880, 0, 57600, 0, 172800, 0, 115200, 0, 14400, 0, 14400, 0, 360000, 0, 1440000, 0, 1440000, 0, 360000
(list; table; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 0,7
|
|
|
COMMENTS
| Mirror image of A145894.
Without the 0's it is the triangle of A145891.
Sum of entries in row n = n! =A000142(n).
|
|
|
FORMULA
| T(2n,2k)=[n!*binom(n,k)]^2; T(2n+1,2k)=n!(n+1)!binom(n,k)binom(n+1,k); elsewhere T(n,k)=0.
|
|
|
EXAMPLE
| T(3,2)=4 because we have 132, 312, 213 and 231.
Triangle starts:
1;
1,0;
1,0,1;
2,0,4,0;
4,0,16,0,4;
12,0,72,0,36,0;
36,0,324,0,324,0,36;
|
|
|
MAPLE
| T:=proc(n, k) if `mod`(n, 2) = 0 and `mod`(k, 2) = 0 then factorial((1/2)*n)^2*binomial((1/2)*n, (1/2)*k)^2 elif `mod`(n, 2) = 1 and `mod`(k, 2) = 0 then factorial((1/2)*n-1/2)*factorial((1/2)*n+1/2)*binomial((1/2)*n-1/2, (1/2)*k)*binomial((1/2)*n+1/2, (1/2)*k) else 0 end if end proc: for n from 0 to 10 do seq(T(n, k), k = 0 .. n) end do; # yields sequence in triangular form
|
|
|
CROSSREFS
| A000142, A145891, A145894.
Sequence in context: A082519 A035688 A046769 * A137561 A082024 A114402
Adjacent sequences: A145890 A145891 A145892 * A145894 A145895 A145896
|
|
|
KEYWORD
| nonn,tabl
|
|
|
AUTHOR
| Emeric Deutsch (deutsch(AT)duke.poly.edu), Nov 30 2008
|
| |
|
|
