|
| |
|
|
A143492
|
|
Unsigned 3-Stirling numbers of the first kind.
|
|
7
|
|
|
|
1, 3, 1, 12, 7, 1, 60, 47, 12, 1, 360, 342, 119, 18, 1, 2520, 2754, 1175, 245, 25, 1, 20160, 24552, 12154, 3135, 445, 33, 1, 181440, 241128, 133938, 40369, 7140, 742, 42, 1, 1814400, 2592720, 1580508, 537628, 111769, 14560, 1162, 52, 1, 19958400
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
3,2
|
|
|
COMMENTS
|
See A049458 for a signed version of this array. The unsigned 3-Stirling numbers of the first kind count the permutations of the set {1,2,...,n} into k disjoint cycles, with the restriction that the elements 1, 2 and 3 belong to distinct cycles. This is the case r = 3 of the unsigned r-Stirling numbers of the first kind. For other cases see abs(A008275) (r = 1), A143491 (r = 2) and A143493 (r = 4). See A143495 for the corresponding 3-Stirling numbers of the second kind. The theory of r-Stirling numbers of both kinds is developed in [Broder]. For details of the related 3-Lah numbers see A143498.
With offset n=0 and k=0, this is the Sheffer triangle (1/(1-x)^3, -log(1-x)) (in the umbral notation of S. Roman's book this would be called Sheffer for (exp(-3*t), 1-exp(-t))). See the e.g.f given below. Compare also with the e.g.f. for the signed version A049458. - Wolfdieter Lang, Oct 10 2011
With offset n=0 and k=0 : triangle T(n,k), read by rows, given by (3,1,4,2,5,3,6,4,7,5,8,6,...) DELTA (1,0,1,0,1,0,1,0,1,0,1,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 31 2011
|
|
|
LINKS
|
|
|
|
FORMULA
|
T(n,k) = (n-3)! * Sum_{j = k-3 .. n-3} C(n-j-1,2)*|Stirling1(j,k-3)|/j!.
Recurrence relation: T(n,k) = T(n-1,k-1) + (n-1)*T(n-1,k) for n > 3, with boundary conditions: T(n,2) = T(2,n) = 0, for all n; T(3,3) = 1; T(3,k) = 0 for k > 3.
Special cases:
T(n,3) = (n-1)!/2! for n >= 3.
T(n,4) = (n-1)!/2!*(1/3 + ... + 1/(n-1)) for n >= 3.
T(n,k) = Sum_{3 <= i_1 < ... < i_(n-k) < n} (i_1*i_2* ...*i_(n-k)). For example, T(6,4) = Sum_{3 <= i < j < 6} (i*j) = 3*4 + 3*5 + 4*5 = 47.
Row g.f.: Sum_{k = 3..n} T(n,k)*x^k = x^3*(x+3)*(x+4)* ... *(x+n-1).
E.g.f. for column (k+3): Sum_{n = k..inf} T(n+3,k+3)*x^n/n! = 1/k!*1/(1-x)^3 * (log(1/(1-x)))^k.
E.g.f.: (1/(1-t))^(x+3) = Sum_{n = 0..inf} Sum_{k = 0..n} T(n+3,k+3)*x^k*t^n/n! = 1 + (3+x)*t/1! + (12+7*x+x^2)*t^2/2! + ....
This array is the matrix product St1 * P^2, where St1 denotes the lower triangular array of unsigned Stirling numbers of the first kind, abs(A008275) and P denotes Pascal's triangle, A007318. The row sums are n!/3! ( A001715 ). The alternating row sums are (n-2)!.
If we define f(n,i,a) = sum(binomial(n,k)*Stirling1(n-k,i)*product(-a-j,j=0..k-1),k=0..n-i), then T(n,i) = |f(n,i,3)|, for n=1,2,...;i=0...n. - Milan Janjic, Dec 21 2008
|
|
|
EXAMPLE
|
Triangle begins
n\k|.....3.....4.....5.....6.....7.....8
========================================
3..|.....1
4..|.....3.....1
5..|....12.....7.....1
6..|....60....47....12.....1
7..|...360...342...119....18.....1
8..|..2520..2754..1175...245....25.....1
...
T(5,4) = 7. The permutations of {1,2,3,4,5} with 4 cycles such that 1, 2 and 3 belong to different cycles are: (14)(2)(3)(5), (15)(2)(3)(4), (24)(1)(3)(5), (25)(1)(3)(4), (34)(1)(2)(5), (35)(1)(2)(4) and (45)(1)(2)(3).
|
|
|
MAPLE
|
with combinat: T := (n, k) -> (n-3)! * add(binomial(n-j-1, 2)*abs(stirling1(j, k-3))/j!, j = k-3..n-3): for n from 3 to 12 do seq(T(n, k), k = 3..n) end do;
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
