|
Search: id:A161133
|
|
|
| A161133 |
|
Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} having exactly k odd fixed points (0<=k<=ceil(n/2)). |
|
+0
2
|
|
| 1, 0, 1, 1, 1, 3, 2, 1, 14, 8, 2, 64, 42, 12, 2, 426, 234, 54, 6, 2790, 1704, 468, 72, 6, 24024, 12864, 3024, 384, 24, 205056, 120120, 32160, 5040, 480, 24, 2170680, 1145400, 272400, 37200, 3000, 120, 22852200, 13024080, 3436200, 544800, 55800
(list; graph; listen)
|
|
|
OFFSET
|
0,6
|
|
|
COMMENT
|
Row n contains 1 + ceil(n/2) entries.
Sum of row n is n! = A000142(n).
T(n,0)=A161131(n).
Sum(k*T(n,k), k>=0) = A052558(n-1).
|
|
FORMULA
|
T(n,k)=binom(ceil(n/2), k)*Sum[(-1)^j*(n-k-j)!*binom(ceil(n/2)-k, j), j=0..ceil(n/2)-k].
|
|
EXAMPLE
|
T(3,0)=3 because we have 312, 231, 321; T(3,2)=1 because we have 123.
Triangle starts:
1;
0,1;
1,1;
3,2,1;
14,8,2;
64,42,12,2;
426,234,54,6.
|
|
MAPLE
|
T := proc (n, k) options operator, arrow: binomial(ceil((1/2)*n), k)*add((-1)^j*binomial(ceil((1/2)*n)-k, j)*factorial(n-k-j), j = 0 .. ceil((1/2)*n)-k) end proc: for n from 0 to 12 do seq(T(n, k), k = 0 .. ceil((1/2)*n)) end do; # yields sequence in triangular form
|
|
CROSSREFS
|
A000142, A161131, A052558, A161134
Sequence in context: A129652 A154921 A127126 this_sequence A112911 A152405 A152400
Adjacent sequences: A161130 A161131 A161132 this_sequence A161134 A161135 A161136
|
|
KEYWORD
|
nonn,tabf
|
|
AUTHOR
|
Emeric Deutsch (deutsch(AT)duke.poly.edu), Jul 18 2009
|
|
|
Search completed in 0.002 seconds
|
