cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A143492 Unsigned 3-Stirling numbers of the first kind.

Original entry on oeis.org

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
Offset: 3

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Author

Peter Bala, Aug 20 2008

Keywords

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

Examples

			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).
		

Crossrefs

Cf. A001710 - A001714 (column 3 - column 7), A001715 (row sums), A008275, A049458 (signed version), A143491, A143493, A143495, A143498.

Programs

  • 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;

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