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.

Showing 1-7 of 7 results.

A143495 Triangle read by rows: 3-Stirling numbers of the second kind.

Original entry on oeis.org

1, 3, 1, 9, 7, 1, 27, 37, 12, 1, 81, 175, 97, 18, 1, 243, 781, 660, 205, 25, 1, 729, 3367, 4081, 1890, 380, 33, 1, 2187, 14197, 23772, 15421, 4550, 644, 42, 1, 6561, 58975, 133057, 116298, 47271, 9702, 1022, 52, 1, 19683, 242461, 724260, 830845, 447195
Offset: 3

Views

Author

Peter Bala, Aug 20 2008

Keywords

Comments

This is the case r = 3 of the r-Stirling numbers of the second kind. The 3-Stirling numbers of the second kind give the number of ways of partitioning the set {1,2,...,n} into k nonempty disjoint subsets with the restriction that the elements 1, 2 and 3 belong to distinct subsets. For remarks on the general case see A143494 (r = 2). The corresponding array of 3-Stirling numbers of the first kind is A143492. The theory of r-Stirling numbers of both kinds is developed in [Broder]. For 3-Lah numbers refer to A143498.
From Peter Bala, Sep 19 2008: (Start)
Let D be the derivative operator d/dx and E the Euler operator x*d/dx. Then x^(-3)*E^n*x^3 = Sum_{k = 0..n} T(n+3,k+3)*x^k*D^k.
The row generating polynomials R_n(x) := Sum_{k= 3..n} T(n,k)*x^k satisfy the recurrence R_(n+1)(x) = x*R_n(x) + x*d/dx(R_n(x)) with R_3(x) = x^3. It follows that the polynomials R_n(x) have only real zeros (apply Corollary 1.2. of [Liu and Wang]).
Relation with the 3-Eulerian numbers E_3(n,j) := A144697(n,j): T(n,k) = (3!/k!)*Sum_{j = n-k..n-3} E_3(n,j)*binomial(j,n-k) for n >= k >= 3.
(End)
T(n,k) = S(n,k,3), n>=k>=3, in Mikhailov's first paper, eq.(28) or (A3). E.g.f. column k from (A20) with k->3, r->k. Therefore, with offset [0,0], this triangle is the Sheffer triangle (exp(3*x),exp(x)-1) with e.g.f. of column no. m>=0: exp(3*x)*((exp(x)-1)^m)/m!. See one of the formulas given below. For Sheffer matrices see the W. Lang link under A006232 with the S. Roman reference, also found in A132393. - Wolfdieter Lang, Sep 29 2011

Examples

			Triangle begins
  n\k|....3....4....5....6....7....8
  ==================================
  3..|....1
  4..|....3....1
  5..|....9....7....1
  6..|...27...37...12....1
  7..|...81..175...97...18....1
  8..|..243..781..660..205...25....1
  ...
T(5,4) = 7. The set {1,2,3,4,5} can be partitioned into four subsets such that 1, 2 and 3 belong to different subsets in 7 ways: {{1}{2}{3}{4,5}}, {{1}{2}{5}{3,4}}, {{1}{2}{4}{3,5}}, {{1}{3}{4}{2,5}}, {{1}{3}{5}{2,4}}, {{2}{3}{4}{1,5}} and {{2}{3}{5}{1,4}}.
From _Peter Bala_, Feb 23 2025: (Start)
The array factorizes as
/ 1               \       /1             \ /1             \ /1            \
| 3    1           |     | 3   1          ||0  1           ||0  1          |
| 9    7   1       |  =  | 9   4   1      ||0  3   1       ||0  0  1       | ...
|27   37  12   1   |     |27  13   5  1   ||0  9   4  1    ||0  0  3  1    |
|81  175  97  18  1|     |81  40  18  6  1||0 27  13  5  1 ||0  0  9  4  1 |
|...               |     |...             ||...            ||...           |
where, in the infinite product on the right-hand side, the first array is the Riordan array (1/(1 - 3*x), x/(1 - x)). See A106516. (End)
		

Crossrefs

Cf. A005061 (column 4), A005494 (row sums), A008277, A016753 (column 5), A028025 (column 6), A049458 (matrix inverse), A106516, A143492, A143494, A143496, A143498.

Programs

  • Maple
    A143495 := (n, k) -> (1/(k-3)!)*add((-1)^(k-i-1)*binomial(k-3,i)*(i+3)^(n-3), i = 0..k-3): for n from 3 to 12 do seq(A143495(n, k), k = 3..n) end do;
  • Mathematica
    nmax = 12; t[n_, k_] := 1/(k-3)!* Sum[ (-1)^(k-j-1)*Binomial[k-3, j]*(j+3)^(n-3), {j, 0, k-3}]; Flatten[ Table[ t[n, k], {n, 3, nmax}, {k, 3, n}]] (* Jean-François Alcover, Dec 07 2011, after Maple *)
  • Sage
    @CachedFunction
    def stirling2r(n, k, r) :
        if n < r: return 0
        if n == r: return 1 if k == r else 0
        return stirling2r(n-1, k-1, r) + k*stirling2r(n-1, k, r)
    A143495 = lambda n, k: stirling2r(n, k, 3)
    for n in (3..8): [A143495(n, k) for k in (3..n)] # Peter Luschny, Nov 19 2012

Formula

T(n+3,k+3) = (1/k!)*Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i)*(i+3)^n, n,k >= 0.
T(n,k) = Stirling2(n,k) - 3*Stirling2(n-1,k) + 2*Stirling2(n-2,k), n,k >= 3.
Recurrence relation: T(n,k) = T(n-1,k-1) + k*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) = 3^(n-3); T(n,4) = 4^(n-3) - 3^(n-3).
E.g.f. (k+3) column (with offset 3): (1/k!)*exp(3x)*(exp(x)-1)^k.
O.g.f. k-th column: Sum_{n >= k} T(n,k)*x^n = x^k/((1-3*x)*(1-4*x)*...*(1-k*x)).
E.g.f.: exp(3*t + x*(exp(t)-1)) = Sum_{n >= 0} Sum_{k = 0..n} T(n+3,k+3)*x^k*t^n/n! = Sum_{n >= 0} B_n(3;x)*t^n/n! = 1 + (3+x)*t/1! + (9+7*x+x^2)*t^2/2! + ..., where the row polynomials, B_n(3;x) := Sum_{k = 0..n} T(n+3,k+3)*x^k, may be called the 3-Bell polynomials.
Dobinski-type identities: Row polynomial B_n(3;x) = exp(-x)*Sum_{i >= 0} (i+3)^n*x^i/i!; Sum_{k = 0..n} k!*T(n+3,k+3)*x^k = Sum_{i >= 0} (i+3)^n*x^i/(1+x)^(i+1).
The T(n,k) are the connection coefficients between the falling factorials and the shifted monomials (x+3)^(n-3). For example, 9 + 7*x + x*(x-1) = (x+3)^2 and 27 + 37*x + 12x*(x-1) + x*(x-1)*(x-2) = (x+3)^3.
This array is the matrix product P^2 * S, where P denotes Pascal's triangle, A007318 and S denotes the lower triangular array of Stirling numbers of the second kind, A008277 (apply Theorem 10 of [Neuwirth]). The inverse array is A049458, the signed 3-Stirling numbers of the first kind.

A062138 Coefficient triangle of generalized Laguerre polynomials n!*L(n,5,x)(rising powers of x).

Original entry on oeis.org

1, 6, -1, 42, -14, 1, 336, -168, 24, -1, 3024, -2016, 432, -36, 1, 30240, -25200, 7200, -900, 50, -1, 332640, -332640, 118800, -19800, 1650, -66, 1, 3991680, -4656960, 1995840, -415800, 46200, -2772, 84, -1, 51891840, -69189120
Offset: 0

Views

Author

Wolfdieter Lang, Jun 19 2001

Keywords

Comments

The row polynomials s(n,x) := n!*L(n,5,x)= sum(a(n,m)*x^m,m=0..n) have e.g.f. exp(-z*x/(1-z))/(1-z)^6. They are Sheffer polynomials satisfying the binomial convolution identity s(n,x+y) = sum(binomial(n,k)*s(k,x)*p(n-k,y),k=0..n), with polynomials sum(|A008297(n,m)|*(-x)^m, m=1..n), n >= 1 and p(0,x)=1 (for Sheffer polynomials see A048854 for S. Roman reference).
These polynomials appear in the radial part of the l=2 (d-wave) eigen functions for the discrete energy levels of the H-atom. See Messiah reference.
For m=0..5 the (unsigned) column sequences (without leading zeros) are: A001725(n+5), A062148-A062152. Row sums (signed) give A062191; row sums (unsigned) give A062192.
The unsigned version of this triangle is the triangle of unsigned 3-Lah numbers A143498. - Peter Bala, Aug 25 2008

Examples

			Triangle begins:
  {1};
  {6, -1};
  {42, -14, 1};
  {336, -168, 24, -1};
  ...
2!*L(2, 5, x) = 42-14*x+x^2.
		

References

  • A. Messiah, Quantum mechanics, vol. 1, p. 419, eq.(XI.18a), North Holland, 1969.

Crossrefs

For m=0..5 the (unsigned) column sequences (without leading zeros) are: A001725(n+5), A062148, A062149, A062150, A062151, A062152.
Row sums (signed) give A062191, row sums (unsigned) give A062192.
Cf. A143498.

Programs

  • Mathematica
    Flatten[Table[((-1)^m)*n!*Binomial[n+5,n-m]/m!,{n,0,8},{m,0,n}]] (* Indranil Ghosh, Feb 24 2017 *)
  • PARI
    tabl(nn) = {for (n=0, nn, for (m=0, n, print1(((-1)^m)*n!*binomial(n+5, n-m)/m!, ", "); ); print(); ); } \\ Indranil Ghosh, Feb 24 2017
    
  • PARI
    row(n) = Vecrev(n!*pollaguerre(n, 5)); \\ Michel Marcus, Feb 06 2021
    
  • Python
    import math
    f=math.factorial
    def C(n, r):return f(n)//f(r)//f(n-r)
    i=-1
    for n in range(26):
        for m in range(n+1):
            i += 1
            print(str(i)+" "+str(((-1)**m)*f(n)*C(n+5, n-m)//f(m))) # Indranil Ghosh, Feb 24 2017

Formula

T(n, m) = ((-1)^m)*n!*binomial(n+5, n-m)/m!.
E.g.f. for m-th column: ((-x/(1-x))^m)/(m!*(1-x)^6), m >= 0.

A143497 Triangle of unsigned 2-Lah numbers.

Original entry on oeis.org

1, 4, 1, 20, 10, 1, 120, 90, 18, 1, 840, 840, 252, 28, 1, 6720, 8400, 3360, 560, 40, 1, 60480, 90720, 45360, 10080, 1080, 54, 1, 604800, 1058400, 635040, 176400, 25200, 1890, 70, 1, 6652800, 13305600, 9313920, 3104640, 554400, 55440, 3080, 88, 1
Offset: 2

Views

Author

Peter Bala, Aug 25 2008

Keywords

Comments

For a signed version of this triangle see A062137. The unsigned 2-Lah number L(2; n,k) gives the number of partitions of the set {1, 2, ..., n} into k ordered lists with the restriction that the elements 1 and 2 must belong to different lists. More generally, the unsigned r-Lah number L(r; n, k) gives the number of partitions of the set {1, 2, ..., n} into k ordered lists with the restriction that the elements 1, 2, ..., r belong to different lists. If r = 1 there is no restriction and we obtain the unsigned Lah numbers A105278. For other cases see A143498 (r=3) and A143499 (r=4). We make some remarks on the general case.
The unsigned r-Lah numbers occur as connection constants in the generalized Lah identity (x + 2*r - 1)*(x + 2*r)*...*(x + 2*r + n - r - 2) = Sum_{k=r..n} L(r; n, k)*(x - 1)*(x - 2)*...*(x - k + r) for n >= r and where any empty products are taken equal to 1 (for a bijective proof of the identity, follow the proof of [Petkovsek and Pisanski] but restrict the first r of the Argonauts to different paths).
The unsigned r-Lah numbers satisfy the same recurrence as the unsigned Lah numbers, namely, L(r; n, k) = (n + k - 1)*L(r; n - 1,k) + L(r; n - 1,k - 1), but with the boundary conditions: L(r; n, k) = 0 if n < r or if k < r; L(r; r, r) = 1. The recurrence has the explicit solution L(r; n, k) = ((n - r)!/(k - r)!)*binomial(n + r - 1, k + r - 1) for n, k >= r. It follows that the unsigned r-Lah numbers have 'vertical' generating functions for k >= r of the form Sum_{n>=k} L(r; n, k)*t^n/(n -r)! = 1/(k - r)!*t^k/(1 - t)^(k + r). This yields the e.g.f. for the array of unsigned r-restricted Lah numbers in the form: Sum_{n,k>=r} L(r; n, k)*x^k*t^n/(n-r)! = (x*t)^r * 1/(1 - t)^(2*r) * exp(x*t/(1 - t)) = (x*t)^r (1 + (2*r + x)*t + (2r*(2*r + 1) + 2*(2*r + 1)*x + x^2)*t^2/2! + ...).
The array of unsigned r-Lah numbers begins
1
2r 1
2r*(2r+1) 2*(2r+1) 1
2r*(2r+1)*(2r+2) 3*(2r+1)*(2r+2) 3*(2r+2) 1
...
and equals exp(D(r)), where D(r) is the array with the sequence (2*r, 2*(2*r + 1), 3*(2*r + 2), 4*(2*r + 3), ...) on the main subdiagonal and zeros everywhere else.
The unsigned r-Lah numbers are related to the r-Stirling numbers: the lower triangular array of unsigned r-Lah numbers may be expressed as the matrix product St1(r) * St2(r), where St1(r) and St2(r) denote the arrays of r-Stirling numbers of the first and second kind respectively. The theory of r-Stirling numbers is developed in [Broder]. See A143491 - A143496 for tables of r-Stirling numbers. An alternative factorization for the array is as St1 * P^(2r - 2) * St2, where P denotes Pascal's triangle, A007318, St1 is the triangle of unsigned Stirling numbers of the first kind, abs(A008275) and St2 denotes the triangle of Stirling numbers of the second kind, A008277 (apply Theorem 10 of [Neuwirth]).
The array of unsigned r-Lah numbers is an example of the fundamental matrices sketched in A133314. So redefining the offset as n=0, given matrices A and B with A(n, k) = T(n, k)*a(n - k) and B(n, k) = T(n, k)*b(n - k), then A*B = C where C(n, k) = T(n,k)*[a(.) + b(.)]^(n - k), umbrally. An e.g.f. for the row polynomials of A is exp(x*t) exp{-x*t*[a*t/(a*t - 1)]}/(1 - a*t)^4 = exp(x*t) exp[(.)!*Laguerre(., 3, -x*t)* a(.)*t)], umbrally. - Tom Copeland, Sep 19 2008

Examples

			Triangle begins:
=========================================
n\k |     2     3     4     5     6     7
----+------------------------------------
  2 |     1
  3 |     4     1
  4 |    20    10     1
  5 |   120    90    18     1
  6 |   840   840   252    28     1
  7 |  6720  8400  3360   560    40     1
 ...
T(4,3) = 10. The ten partitions of {1,2,3,4} into 3 ordered lists such that the elements 1 and 2 lie in different lists are: {1}{2}{3,4} and {1}{2}{4,3}, {1}{3}{2,4} and {1}{3}{4,2}, {1}{4}{2,3} and {1}{4}{3,2}, {2}{3}{1,4} and {2}{3}{4,1}, {2}{4}{1,3} and {2}{4}{3,1}. The remaining two partitions {3}{4}{1,2} and {3}{4}{2,1} are not allowed because the elements 1 and 2 belong to the same block.
		

Crossrefs

Cf. A001715 (column 2), A007318, A008275, A008277, A061206 (column 3), A062137, A062141 - A062144 ( column 4 to column 7), A062146 (alt. row sums), A062147 (row sums), A105278 (unsigned Lah numbers), A143491, A143494, A143498, A143499.

Programs

  • GAP
    T:=Flat(List([2..10],n->List([2..n],k->(Factorial(n-2)/Factorial(k-2))*Binomial(n+1,k+1)))); # Muniru A Asiru, Nov 27 2018
  • Maple
    T := (n, k) -> ((n-2)!/(k-2)!)*binomial(n+1, k+1):
    for n from 2 to 11 do seq(T(n, k), k = 2..n) od;
  • Mathematica
    T[n_, k_] := (n-2)!/(k-2)!*Binomial[n+1, k+1]; Table[T[n, k], {n,2,10}, {k,2,n}] // Flatten (* Amiram Eldar, Nov 27 2018 *)
  • Maxima
    create_list((n - 2)!/(k - 2)!*binomial(n + 1, k + 1), n, 2, 12, k, 2, n); /* Franck Maminirina Ramaharo, Nov 27 2018 */
    

Formula

T(n, k) = ((n - 2)!/(k - 2)!)*C(n+1, k+1), for n, k >= 2.
Recurrence: T(n, k) = (n + k - 1)*T(n-1, k) + T(n-1, k-1) for n, k >= 2, with the boundary conditions: T(n, k) = 0 if n < 2 or k < 2; T(2, 2) = 1.
E.g.f. for column k: Sum_{n>=k} T(n, k)*t^n/(n - 2)! = 1/(k - 2)!*t^k/(1 - t)^(k+2) for k >= 2.
E.g.f: Sum_{n=2..inf} Sum_{k=2..n} T(n, k)*x^k*t^n/(n - 2)! = (x*t)^2/(1 - t)^4* exp(x*t/(1 - t)) = (x*t)^2*(1 + (4 + x)*t + (20 + 10*x + x^2)*t^2/2! + ... ).
Generalized Lah identity: (x + 3)*(x + 4)*...*(x + n) = Sum_{k = 2..n} T(n, k)*(x - 1)*(x - 2)*...*(x - k + 2).
The polynomials 1/n!*Sum_{k=2..n+2} T(n+2, k)*(-x)^(k - 2) for n >= 0 are the generalized Laguerre polynomials Laguerre(n,3,x). See A062137.
Array = A143491 * A143494 = abs(A008275) * (A007318)^2 * A008277 (apply Theorem 10 of [Neuwirth]). Array equals exp(D), where D is the array with the quadratic sequence (4, 10, 18, 28, ...) on the main subdiagonal and zeros elsewhere.
After adding 1 to the head of the main diagonal and a zero to each of the subdiagonals, the n-th diagonal may be generated as coefficients of (1/n!) [D^(-1) tDt t^(-3)D t^3]^n exp(x*t), where D is the derivative w.r.t. t and D^(-1) t^j/j! = t^(j + 1)/(j + 1)!. E.g., n = 2 generates 20*x*t^3/3! + 90*x^2*t^4/4! + 252*x^3* t^5/5! + ... . For the general unsigned r-Lah number array, replace the threes by (2*r - 1) in the operator. The e.g.f. of the row polynomials is then exp[D^(-1) tDt t^(-(2*r-1))D t^(2*r - 1)] exp(x*t), with offset n = 0. - Tom Copeland, Sep 21 2008

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

Views

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

A143499 Triangle of unsigned 4-Lah numbers.

Original entry on oeis.org

1, 8, 1, 72, 18, 1, 720, 270, 30, 1, 7920, 3960, 660, 44, 1, 95040, 59400, 13200, 1320, 60, 1, 1235520, 926640, 257400, 34320, 2340, 78, 1, 17297280, 15135120, 5045040, 840840, 76440, 3822, 98, 1, 259459200, 259459200, 100900800, 20180160, 2293200
Offset: 4

Views

Author

Peter Bala, Aug 25 2008

Keywords

Comments

This is the case r = 4 of the unsigned r-Lah numbers L(r;n,k). The unsigned 4-Lah numbers count the partitions of the set {1,2,...,n} into k ordered lists with the restriction that the elements 1, 2, 3 and 4 belong to different lists. For other cases see A105278 (r = 1), A143497 (r = 2 and comments on the general case) and A143498 (r = 3).
The unsigned 4-Lah numbers are related to the 4-Stirling numbers: the lower triangular array of unsigned 4-Lah numbers may be expressed as the matrix product St1(4) * St2(4), where St1(4) = A143493 and St2(4) = A143496 are the arrays of 4-restricted Stirling numbers of the first and second kind respectively. An alternative factorization for the array is as St1 * P^6 * St2, where P denotes Pascal's triangle, A007318, St1 is the triangle of unsigned Stirling numbers of the first kind, abs(A008275) and St2 denotes the triangle of Stirling numbers of the second kind, A008277.

Examples

			Triangle begins
n\k|......4......5......6......7......8......9
==============================================
4..|......1
5..|......8......1
6..|.....72.....18......1
7..|....720....270.....30......1
8..|...7920...3960....660.....44......1
9..|..95040..59400..13200...1320.....60......1
...
T(5,4) = 8. The partitions of {1,2,3,4,5} into 4 ordered lists, such that the elements 1, 2, 3 and 4 lie in different lists, are: {1}{2}{3}{4,5} and {1}{2}{3}{5,4}, {1}{2}{4}{3,5} and {1}{2}{4}{5,3}, {1}{3}{4}{2,5} and {1}{3}{4}{5,2}, {2}{3}{4}{1,5} and {2}{3}{4}{5,1}.
		

Crossrefs

Cf. A007318, A008275, A008277, A049388 (column 4), A105278 (unsigned Lah numbers), A143493, A143496, A143497, A143498.

Programs

  • Maple
    with combinat: T := (n, k) -> (n-4)!/(k-4)!*binomial(n+3,k+3): for n from 4 to 13 do seq(T(n, k), k = 4..n) end do;
  • Mathematica
    T[n_, k_] := (n-4)!/(k-4)!*Binomial[n+3, k+3]; Table[T[n, k], {n, 4, 10}, {k, 4, n}] // Flatten (* Amiram Eldar, Nov 26 2018 *)

Formula

T(n,k) = (n-4)!/(k-4)!*binomial(n+3,k+3), n,k >= 4.
Recurrence: T(n,k) = (n+k-1)*T(n-1,k) + T(n-1,k-1) for n,k >= 4, with the boundary conditions: T(n,k) = 0 if n < 4 or k < 4; T(4,4) = 1.
E.g.f. for column k: Sum_{n >= k} T(n,k)*t^n/(n-4)! = 1/(k-4)!*t^k/(1-t)^(k+4) for k >= 4.
E.g.f: Sum_{n = 4..inf} Sum_{k = 4..n} T(n,k)*x^k*t^n/(n-4)! = (x*t)^4/(1-t)^8*exp(x*t/(1-t)) = (x*t)^4*(1 + (8+x)*t +(72+18*x+x^2)*t^2/2! + ...).
Generalized Lah identity: (x+7)*(x+8)*...*(x+n+2) = Sum_{k = 4..n} T(n,k)*(x-1)*(x-2)*...*(x-k+4).
The polynomials 1/n!*Sum_{k = 4..n+4} T(n+4,k)*(-x)^(k-4) for n >= 0 are the generalized Laguerre polynomials Laguerre(n,7,x).
Array = A132493* A143496 = abs(A008275) * ( A007318 )^6 * A008277 (apply Theorem 10 of [Neuwirth]). Array equals exp(D), where D is the array with the quadratic sequence (8,18,30,44, ... ) on the main subdiagonal and zeros everywhere else.

A371259 Triangle read by rows, (2, 3)-Lah numbers.

Original entry on oeis.org

1, 36, 1, 1764, 100, 1, 112896, 9864, 200, 1, 9144576, 1099296, 34064, 344, 1, 914457600, 142159392, 6004512, 92200, 540, 1, 110649369600, 21385410048, 1156921920, 24075712, 213700, 796, 1, 15933509222400, 3724783667712, 248142106368, 6573957120, 78782912, 443744, 1120, 1
Offset: 3

Views

Author

Aleks Zigon Tankosic, Mar 16 2024

Keywords

Comments

The (2, 3)-Lah numbers T(n, k) count ordered 2-tuples (pi(1), pi(2)) of partitions of the set {1, ..., n} into k linearly ordered blocks (lists, for short) such that the numbers 1, 2, 3 are in distinct lists, and bl(pi(1)) = bl(pi(2)) where for i = {1, 2} and pi(i) = b(1)^i, b(2)^i, ..., b(k)^i, where b(1)^i, b(2)^i, ..., b(k)^i are the blocks of partition pi(i), bl(pi(i)) = {min(b(1))^i, min(b(2))^i, ..., min(b(k))^i} is the set of block leaders, i.e., of minima of the lists in partition pi(i).
The (2, 3)-Lah numbers T(n, k) are the (m, r)-Lah numbers for m=2 and r=3. More generally, the (m, r)-Lah numbers count ordered m-tuples (pi(1), pi(2), ..., pi(m)) of partitions of the set {1, 2, ..., n} into k linearly ordered blocks (lists, for short) such that the numbers 1, 2, ..., r are in distinct lists, and bl(pi(1)) = bl(pi(2)) = ... = bl(pi(m)) where for i = {1, 2, ..., m} and pi(i) = {b(1)^i, b(2)^i, ..., b(k)^i}, where b(1)^i, b(2)^i,..., b(k)^i are the blocks of partition pi(i), bl(pi(i)) = {min(b(1))^i, min(b(2))^i, ..., min (b(k))^i} is the set of block leaders, i.e., of minima of the lists in partition pi(i).

Examples

			Triangle begins:
           1;
          36,           1;
        1764,         100,          1;
      112896,        9864,        200,        1;
     9144576,     1099296,      34064,      344,     1;
   914457600,   142159392,    6004512,    92200,   540,  1;
110649369600, 21385410048, 1156921920, 24075712, 213700, 796, 1.
  ...
An example for T(4, 3). The corresponding partitions are
pi(1) = {(1),(2),(3,4)},
pi(2) = {(1),(2),(4,3)},
pi(3) = {(1),(3),(2,4)},
pi(4) = {(1),(3),(4,2)},
pi(5) = {(1,4),(2),(3)},
pi(6) = {(4,1),(2),(3)}, since A143498 for n=4, k=3 equals 6. Sets of their block leaders are bl(pi(1)) = bl(pi(2)) = bl(pi(3)) = bl(pi(4)) = bl(pi(5)) = bl(pi(6)) = {1,2,3}.
Compute the number of ordered 2-tuples (i.e., ordered pairs) of partitions pi(1), pi(2), ..., pi(6) such that partitions in the same pair share the same set of block leaders. As there are six partitions with the set of block leaders equal to {1,2,3}, T(4, 3) = 6^2 = 36.
		

Crossrefs

Programs

  • Maple
    T:= proc(n, k) option remember; `if`(k<3 or k>n, 0,
          `if`(n=k, 1, T(n-1, k-1)+(n+k-1)^2*T(n-1, k)))
        end:
    seq(seq(T(n, k), k=3..n), n=3..10);
  • Mathematica
    A371259[n_, k_] := A371259[n, k] = Which[n < k || k < 3, 0, n == k, 1, True, A371259[n-1, k-1] + (n+k-1)^2*A371259[n-1, k]];
    Table[A371259[n, k], {n, 3, 10},{k, 3, n}] (* Paolo Xausa, Jun 11 2024 *)
  • Python
    A371259 = lambda n, k: 0 if (k < 3 or k > n) else (1 if (n == 3 and k == 3) else (A371259(n-1, k-1) + ((n + k - 1)**2) * A371259(n-1, k)))
    print([A371259(n, k) for n in range(3, 11) for k in range(3, n+1)])

Formula

Recurrence relation: T(n, k) = T(n-1, k-1) + (n+k-1)^2*T(n-1, k).
Explicit formula: T(n, k) = Sum_{4 <= j(1) < j(2) < ... < j(n-k) <= n} (2j(1)-2)^2 * (2j(2)-3)^2 * ... * (2j(n-k)-(n-k+1))^2.
Special cases:
T(n, k) = 0 for n < k or k < 3, [corrected by Paolo Xausa, Jun 11 2024]
T(n, n) = 1,
T(n, 3) = (A143498(n, 3))^2 = ((n+2)!)^2/14400,
T(n, n-1) = 2^2 * Sum_{j=3..n-1} j^2.

Extensions

More terms from Michel Marcus, Jun 12 2025

A372208 Triangle read by rows, (3, 3)-Lah numbers.

Original entry on oeis.org

1, 216, 1, 74088, 728, 1, 37933056, 604800, 1728, 1, 27653197824, 642733056, 2904768, 3456, 1, 27653197824000, 883130895360, 5662172160, 10497600, 6200, 1, 36806406303744000, 1553703385006080, 13322923130880, 34467586560, 31422600, 10296, 1, 63601470092869632000, 3450292743162101760, 38111804456140800, 129651027770880, 163174556160, 82006848, 16128, 1
Offset: 3

Views

Author

Aleks Zigon Tankosic, Apr 22 2024

Keywords

Comments

The (3, 3)-Lah numbers T(n, k) count ordered 3-tuples (pi(1), pi(2), pi(3)) of partitions of the set {1, ..., n} into k linearly ordered blocks (lists, for short) such that the numbers 1, 2, 3 are in distinct lists, and bl(pi(1)) = bl(pi(2)) = bl(pi(3)) where for i = {1, 2, 3} and pi(i) = b(1)^i, b(2)^i, ..., b(k)^i, where b(1)^i, b(2)^i, ..., b(k)^i are the blocks of partition pi(i), bl(pi(i)) = {min(b(1))^i, min(b(2))^i, ..., min(b(k))^i} is the set of block leaders, i.e., of minima of the lists in partition pi(i).
The (3, 3)-Lah numbers T(n, k) are the (m, r)-Lah numbers for m=3 and r=3. More generally, the (m, r)-Lah numbers count ordered m-tuples (pi(1), pi(2), ..., pi(m)) of partitions of the set {1, 2, ..., n} into k linearly ordered blocks (lists, for short) such that the numbers 1, 2, ..., r are in distinct lists, and bl(pi(1)) = bl(pi(2)) = ... = bl(pi(m)) where for i = {1, 2, ..., m} and pi(i) = {b(1)^i, b(2)^i, ..., b(k)^i}, where b(1)^i, b(2)^i, ..., b(k)^i are the blocks of partition pi(i), bl(pi(i)) = {min(b(1))^i, min(b(2))^i, ..., min (b(k))^i} is the set of block leaders, i.e., of minima of the lists in partition pi(i).

Examples

			Triangle begins:
              1;
            216,             1;
          74088,           728,           1;
       37933056,        604800,        1728,         1;
    27653197824,     642733056,     2904768,      3456,     1;
 27653197824000,  883130895360,  5662172160,  10497600,  6200,  1.
  ...
An example for T(4, 3). The corresponding partitions are
pi(1) = {(1),(2),(3,4)},
pi(2) = {(1),(2),(4,3)},
pi(3) = {(1),(3),(2,4)},
pi(4) = {(1),(3),(4,2)},
pi(5) = {(1,4),(2),(3)},
pi(6) = {(4,1),(2),(3)}, since A143498 for n=4, k=3 equals 6. Sets of their block leaders are bl(pi(1)) = bl(pi(2)) = bl(pi(3)) = bl(pi(4)) = bl(pi(5)) = bl(pi(6)) = {1,2,3}. Compute the number of ordered 3-tuples (i.e., ordered pairs) of partitions pi(1), pi(2), ..., pi(6) such that partitions in the same pair share the same set of block leaders. As there are six partitions with the set of block leaders equal to {1,2,3}, T(4, 3) = 6^3 = 216.
		

Crossrefs

Programs

  • Maple
    T:= proc(n, k) option remember; `if`(k<3 or k>n, 0,
          `if`(n=k, 1, T(n-1, k-1)+(n+k-1)^3*T(n-1, k)))
        end:
    seq(seq(T(n, k), k=3..n), n=3..10);
  • Mathematica
    A372208[n_, k_] := A372208[n, k] = Which[n < k || k < 3, 0, n == k, 1, True, A372208[n-1, k-1] + (n+k-1)^3*A372208[n-1, k]];
    Table[A372208[n, k], {n, 3, 10}, {k, 3, n}] (* Paolo Xausa, Jun 11 2024 *)
  • Python
    A372208 = lambda n, k: 0 if (k < 3 or k > n) else (1 if (n == 3 and k == 3) else (A372208(n-1, k-1) + ((n + k - 1)**3) * A372208(n-1, k)))
    print([A372208(n, k) for n in range(3, 11) for k in range(3, n+1)])

Formula

Recurrence relation: T(n, k) = T(n-1, k-1) + (n+k-1)^3*T(n-1, k).
Explicit formula: T(n, k) = Sum_{4 <= j(1) < j(2) < ... < j(n-k) <= n} (2j(1)-2)^3 * (2j(2)-3)^3 * ... * (2j(n-k)-(n-k+1))^3.
Special cases:
T(n, k) = 0 for n < k or k < 3, [corrected by Paolo Xausa, Jun 11 2024]
T(n, n) = 1,
T(n, 3) = (A143498(n, 3))^3 = ((n+2)!)^3/1728000,
T(n, n-1) = 2^3 * Sum_{j=3..n-1} j^3.
Showing 1-7 of 7 results.