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.

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A143498 Triangle of unsigned 3-Lah numbers.

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, 34594560, 8648640, 1201200, 96096, 4368
Offset: 3

Views

Author

Peter Bala, Aug 25 2008

Keywords

Comments

For a signed version of this triangle see A062138. This is the case r = 3 of the unsigned r-Lah numbers L(r;n,k). The unsigned 3-Lah numbers count the partitions of the set {1,2,...,n} into k ordered lists with the restriction that the elements 1, 2 and 3 belong to different lists. For other cases see A105278 (r = 1), A143497 (r = 2 and comments on the general case) and A143499 (r = 4).
The unsigned 3-Lah numbers are related to the 3-Stirling numbers: the lower triangular array of unsigned 3-Lah numbers may be expressed as the matrix product St1(3) * St2(3), where St1(3) = A143492 and St2(3) = A143495 are the arrays of 3-Stirling numbers of the first and second kind respectively. An alternative factorization for the array is as St1 * P^4 * 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|......3......4......5......6......7......8
==============================================
3..|......1
4..|......6......1
5..|.....42.....14......1
6..|....336....168.....24......1
7..|...3024...2016....432.....36......1
8..|..30240..25200...7200....900.....50......1
...
T(4,3) = 6. The partitions of {1,2,3,4} into 3 ordered lists, such that the elements 1, 2 and 3 lie in different lists, are: {1}{2}{3,4} and {1}{2}{4,3}, {1}{3}{2,4} and {1}{3}{4,2}, {2}{3}{1,4} and {2}{3}{4,1}.
		

Crossrefs

Cf. A001725 (column 3), A007318, A008275, A008277, A062138, A062148 - A062152 (column 4 to column 8), A062191 (alt. row sums), A062192 (row sums), A105278 (unsigned Lah numbers), A143492, A143495, A143497, A143499.

Programs

  • Magma
    /* As triangle */ [[Factorial(n-3)/Factorial(k-3)*Binomial(n+2, k+2): k in [3..n]]: n in [3.. 15]]; // Vincenzo Librandi, Nov 27 2018
  • Maple
    with combinat: T := (n, k) -> (n-3)!/(k-3)!*binomial(n+2,k+2): for n from 3 to 12 do seq(T(n, k), k = 3..n) end do;
  • Mathematica
    T[n_, k_] := (n-3)!/(k-3)!*Binomial[n+2, k+2]; Table[T[n, k], {n, 3, 10}, {k, 3, n}] // Flatten (* Amiram Eldar, Nov 26 2018 *)

Formula

T(n,k) = (n-3)!/(k-3)!*binomial(n+2,k+2) for n,k >= 3.
Recurrence: T(n,k) = (n+k-1)*T(n-1,k) + T(n-1,k-1) for n,k >= 3, with the boundary conditions: T(n,k) = 0 if n < 3 or k < 3; T(3,3) = 1.
E.g.f. for column k: Sum_{n >= k} T(n,k)*t^n/(n-3)! = 1/(k-3)!*t^k/(1-t)^(k+3) for k >= 3.
E.g.f: Sum_{n = 3..inf} Sum_{k = 3..n} T(n,k)*x^k*t^n/(n-3)! = (x*t)^3/(1-t)^6*exp(x*t/(1-t)) = (x*t)^3*(1 + (6+x)*t +(42+14*x+x^2)*t^2/2! + ... ).
Generalized Lah identity: (x+5)*(x+6)*...*(x+n+1) = Sum_{k = 3..n} T(n,k)*(x-1)*(x-2)*...*(x-k+3).
The polynomials 1/n!*Sum_{k = 3..n+3} T(n+3,k)*(-x)^(k-3) for n >= 0 are the generalized Laguerre polynomials Laguerre(n,5,x). See A062138.
Array = A143492 * A143495 = abs(A008275) * ( A007318 )^4 * A008277 (apply Theorem 10 of [Neuwirth]). Array equals exp(D), where D is the array with the quadratic sequence (6,14,24,36, ... ) on the main subdiagonal and zeros everywhere else.

A343861 Coefficient triangle of generalized Laguerre polynomials n!*L(n,n,x) (rising powers of x).

Original entry on oeis.org

1, 2, -1, 12, -8, 1, 120, -90, 18, -1, 1680, -1344, 336, -32, 1, 30240, -25200, 7200, -900, 50, -1, 665280, -570240, 178200, -26400, 1980, -72, 1, 17297280, -15135120, 5045040, -840840, 76440, -3822, 98, -1, 518918400, -461260800, 161441280, -29352960, 3057600, -188160, 6720, -128, 1
Offset: 0

Views

Author

Seiichi Manyama, May 01 2021

Keywords

Examples

			The triangle begins:
       1;
       2,      -1;
      12,      -8,      1;
     120,     -90,     18,     -1;
    1680,   -1344,    336,    -32,    1;
   30240,  -25200,   7200,   -900,   50,  -1;
  665280, -570240, 178200, -26400, 1980, -72, 1;
		

Crossrefs

For k=0..1 the (unsigned) columns give A001813, A092956(n-1).
Row sums (signed) give A006902, row sums (unsigned) give A082545.

Programs

  • Magma
    [(-1)^k*Factorial(n-k)*Binomial(n,k)*Binomial(2*n, n+k): k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 11 2022
    
  • Mathematica
    T[n_, k_] := (-1)^k * (2*n)! * Binomial[n, k]/(k + n)!; Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Amiram Eldar, May 11 2021 *)
  • PARI
    T(n, k) = (-1)^k*(2*n)!*binomial(n,k)/(k+n)!;
    
  • PARI
    row(n) = Vecrev(n!*pollaguerre(n, n));
    
  • SageMath
    def A343861(n,k): return (-1)^k*factorial(n-k)*binomial(n,k)*binomial(2*n,n+k)
    flatten([[A343861(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Aug 11 2022

Formula

T(n, k) = (-1)^k * n! * binomial(2*n,n-k)/k! = (-1)^k * (2*n)! * binomial(n,k)/(k+n)!.
T(n, 0) = A001813(n).
T(n, 1) = -A092956(n-1).
Sum_{k=0..n} T(n, k) = A006902(n).
Sum_{k=0..n} (-1)^k * T(n, k) = A082545(n).
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