|
Search: id:A136125
|
|
|
| A136125 |
|
Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} in which the size of the last cycle is k (the cycles are ordered by increasing smallest elements; 1 <= k <=n). |
|
+0 1
|
|
| 1, 1, 1, 3, 1, 2, 12, 4, 2, 6, 60, 20, 10, 6, 24, 360, 120, 60, 36, 24, 120, 2520, 840, 420, 252, 168, 120, 720, 20160, 6720, 3360, 2016, 1344, 960, 720, 5040, 181440, 60480, 30240, 18144, 12096, 8640, 6480, 5040, 40320
(list; table; graph; listen)
|
|
|
OFFSET
|
1,4
|
|
|
COMMENT
|
Row sums are the factorials (A000142). T(n,1)=n!/2 for n>=2; Sum(k*T(n,k),k=1..n)=s(n,2)=A000254(n) (Stirling numbers of the first kind).
|
|
FORMULA
|
T(n,k)=n!/[k(k+1)] if k<n; T(n,n)=(n-1)!. Rec. rel.: T(n,k)=(n-1-k)T(n-1,k) + (k-1)T(n-1,k-1) for 1 < k < n.
|
|
EXAMPLE
|
T(4,2)=4 because we have (1)(2)(34), (13)(24), (12)(34), and (14)(23).
Triangle starts:
1;
1,1;
3,1,2l
12,4,2,6;
60,20,10,6,24;
|
|
MAPLE
|
T:=proc(n, k) if k < n then factorial(n)/(k*(k+1)) elif k = n then factorial(n-1) else 0 end if end proc: for n to 9 do seq(T(n, k), k=1..n) end do; # yields sequence in triangular form
|
|
CROSSREFS
|
Cf. A000142, A000254.
Adjacent sequences: A136122 A136123 A136124 this_sequence A136126 A136127 A136128
Sequence in context: A116854 A016567 A109528 this_sequence A092580 A004468 A145463
|
|
KEYWORD
|
nonn,tabl
|
|
AUTHOR
|
Emeric Deutsch (deutsch(AT)duke.poly.edu), Jan 10 2008
|
|
|
Search completed in 0.002 seconds
|