

A143491


Unsigned 2Stirling numbers of the first kind.


14



1, 2, 1, 6, 5, 1, 24, 26, 9, 1, 120, 154, 71, 14, 1, 720, 1044, 580, 155, 20, 1, 5040, 8028, 5104, 1665, 295, 27, 1, 40320, 69264, 48860, 18424, 4025, 511, 35, 1, 362880, 663696, 509004, 214676, 54649, 8624, 826, 44, 1, 3628800, 6999840, 5753736, 2655764
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OFFSET

2,2


COMMENTS

Essentially the same as A136124 but with column numbers differing by one. See A049444 for a signed version of this array. The unsigned 2Stirling 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 and 2 belong to distinct cycles. This is the particular case r = 2 of the unsigned rStirling numbers of the first kind, which count the permutations of the set {1,2,...,n} into k disjoint cycles, with the restriction that the numbers 1, 2, ..., r belong to distinct cycles. The case r = 1 gives the usual unsigned Stirling numbers of the first kind, abs(A008275); for other cases see A143492 (r = 3) and A143493 (r = 4). The corresponding 2Stirling numbers of the second kind can be found in A143494.
In general, the lower unitriangular array of unsigned rStirling numbers of the first kind (with suitable offsets in the row and column indexing) equals the matrix product St1 * P^(r1), where St1 is the array of unsigned Stirling numbers of the first kind, abs(A008275) and P is Pascal's triangle, A007318. The theory of rStirling numbers of both kinds is developed in [Broder]. For details of the related rLah numbers see A143497.
This sequence also represents the number of permutations in the alternating group An of length k, where the length is taken with respect to the generators set {(12)(ij)}. For a bijective proof of the relation between these numbers and the 2Stirling numbers of the first kind see the Rotbart link.  Aviv Rotbart, May 05 2011
With offset n=0,k=0 : triangle T(n,k), read by rows, given by [2,1,3,2,4,3,5,4,6,5,...] DELTA [1,0,1,0,1,0,1,0,1,0,...] where DELTA is the operator defined in A084938.  Philippe Deléham, Sep 29 2011
With offset n=0 and k=0, this is the Sheffer triangle (1/(1x)^2,log(1x)) (in the umbral notation of S. Roman's book this would be called Sheffer for (exp(2*t),1exp(t))). See the e.g.f. given below. Compare also with the e.g.f. for the signed version A049444.  Wolfdieter Lang, Oct 10 2011
Reversed rows correspond to the Betti numbers of the moduli space M(0,n) of smooth Riemann surfaces (see Murri link).  Tom Copeland, Sep 19 2012


REFERENCES

Michael J. Schlosser and Meesue Yoo, Elliptic Rook and File Numbers, Electronic Journal of Combinatorics, 24(1) (2017), #P1.31


LINKS

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Broder, Andrei Z., The rStirling numbers, Discrete Math. 49, 241259 (1984)
A. Dzhumadildaev and D. Yeliussizov, Path decompositions of digraphs and their applications to Weyl algebra, arXiv preprint arXiv:1408.6764 [math.CO], 2014. [Version 1 contained many references to the OEIS, which were removed in Version 2.  N. J. A. Sloane, Mar 28 2015]
Askar Dzhumadil’daev and Damir Yeliussizov, Walks, partitions, and normal ordering, Electronic Journal of Combinatorics, 22(4) (2015), #P4.10.
Neuwirth Erich, Recursively defined combinatorial functions: Extending Galton's board, Discrete Math. 239 No. 13, 3351 (2001)
R. Murri, Fatgraph Algorithms and the Homology of the Kontsevich Complex, arXiv:1202.1820 [math.AG], 2012. (see Table 1, p. 3)
Aviv Rotbart, Generator Sets for the Alternating Group, Séminaire Lotharingien de Combinatoire 65 (2011), Article B65b, 16pp.
M. Shattuck, Generalized rLah numbers, arXiv:1412.8721 [math.CO], 2014


FORMULA

T(n,k) = (n2)! * Sum_{j = k2 .. n2} (nj1)*stirling1(j,k2)/j!.
Recurrence relation: T(n,k) = T(n1,k1) + (n1)*T(n1,k) for n > 2, with boundary conditions: T(n,1) = T(1,n) = 0, for all n; T(2,2) = 1; T(2,k) = 0 for k > 2.
Special cases: T(n,2) = (n1)!; T(n,3) = (n1)!*(1/2 + 1/3 + ... + 1/(n1)).
T(n,k) = Sum_{2 <= i_1 < ...< i_(nk) < n} (i_1*i_2* ...*i_(nk)). For example, T(6,4) = Sum_{2 <= i < j < 6} (i*j) = 2*3 + 2*4 + 2*5 + 3*4 + 3*5 + 4*5 = 71.
Row g.f.: Sum_{k = 2..n} T(n,k)*x^k = x^2*(x+2)*(x+3)* ... *(x+n1).
E.g.f. for column (k+2): Sum_{n = k..inf} T(n+2,k+2)*x^n/n! = 1/k!*1/(1x)^2* (log(1/(1x)))^k.
E.g.f.: (1/(1t))^(x+2) = Sum_{n = 0..inf} Sum_{k = 0..n} T(n+2,k+2)*x^k*t^n/n! = 1 + (2+x)*t/1! + (6+5*x+x^2)*t^2/2! + ... .
This array is the matrix product St1 * P, 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!/2 ( A001710 ). The alternating row sums are (n2)!.
If we define f(n,i,a)=sum(binomial(n,k)*stirling1(nk,i)*product(aj,j=0..k1),k=0..ni), then T(n1,i) = f(n,i,2), for n=1,2,...;i=0...n.  Milan Janjic, Dec 21 2008
From Gary W. Adamson, Jul 19 2011: (Start)
nth row of the triangle = top row of M^(n2), M = a reversed variant of the (1,2) Pascal triangle (Cf. A029635); as follows:
2, 1, 0, 0, 0, 0, ...
2, 3, 1, 0, 0, 0, ...
2, 5, 4, 1, 0, 0, ...
2, 7, 9, 5, 1, 0, ...
... (End)
The reversed, row polynomials of this entry multiplied by (1+x) are the row polynomials of A094638. E.g., (1+x)(1+5x+6x^2) = (1+6x+11x^2+6x^3).  Tom Copeland, Dec 11 2016


EXAMPLE

Triangle begins
n\k.....2.....3.....4.....5.....6.....7
========================================
2.......1
3.......2.....1
4.......6.....5.....1
5......24....26.....9.....1
6.....120...154....71....14.....1
7.....720..1044...580...155....20.....1
...
T(4,3) = 5. The permutations of {1,2,3,4} with 3 cycles such that 1 and 2 belong to different cycles are: (1)(2)(3 4), (1)(3)(24), (1)(4)(23), (2)(3)(14) and (2)(4)(13). The remaining possibility (3)(4)(12) is not allowed.
From Aviv Rotbart, May 05 2011: (Start)
Example of the alternating group permutations numbers:
Triangle begins
n\k.....0.....1.....2.....3.....4.....5.....6.....7
====================================================
2.......1
3.......1.....2
4.......1.....5.....6
5.......1.....9....26....24
6.......1....14....71...154...120
7.......1....20...155...580..1044..720
A(n,k) = number of permutations in An of length k, with respect to the generators set {(12)(ij)}. For example, A(2,0)=1 (only the identity is there), for A4, the generators are {(12)(13),(12)(14),(12,23),(12)(24),(12)(34)}, thus we have A(4,1)=5 (exactly 5 generators), the permutations of length 2 are:
(12)(13)(12)(13) = (312)
(12)(13)(12)(14) = (41)(23)
(12)(13)(12)(24) = (432)(1)
(12)(13)(12)(34) = (342)(1)
(12)(23)(12)(24) = (13)(24)
(12)(14)(12)(14) = (412)(3)
Namely, A(4,2)=6. Together with the identity [=(12)(12), of length 0. therefore A(4,0)=1] we have 12 permutations, comprising all A4 (4!/2=12). (End)


MAPLE

with combinat: T := (n, k) > (n2)! * add((nj1)*abs(stirling1(j, k2))/j!, j = k2..n2): for n from 2 to 10 do seq(T(n, k), k = 2..n) end do;


MATHEMATICA

t[n_, k_] := (n2)!*Sum[(nj1)*Abs[StirlingS1[j, k2]]/j!, {j, k2, n2}]; Table[t[n, k], {n, 2, 11}, {k, 2, n}] // Flatten (* JeanFrançois Alcover, Apr 16 2013, after Maple *)


CROSSREFS

Cf. A001705  A001709 (column 3  column 7), A001710 (row sums), A008275, A049444 (signed version), A136124, A143492, A143493, A143494, A143497.
Cf. A094638.
Sequence in context: A121575 A049444 A136124 * A295517 A070918 A113381
Adjacent sequences: A143488 A143489 A143490 * A143492 A143493 A143494


KEYWORD

easy,nonn,tabl


AUTHOR

Peter Bala, Aug 20 2008


STATUS

approved



