A062195 -id:A062195 - cp's OEIS frontend

1, 3, -1, 12, -8, 1, 60, -60, 15, -1, 360, -480, 180, -24, 1, 2520, -4200, 2100, -420, 35, -1, 20160, -40320, 25200, -6720, 840, -48, 1, 181440, -423360, 317520, -105840, 17640, -1512, 63, -1, 1814400, -4838400, 4233600

Offset: 0

Wolfdieter Lang, Jun 19 2001

The row polynomials s(n,x) := n!*L(n,2,x) = Sum_{m=0..n} a(n,m)*x^m have e.g.f. exp(-z*x/(1-z))/(1-z)^3. They are Sheffer polynomials satisfying the binomial convolution identity s(n,x+y) = Sum_{k=0..n} binomial(n,k)*s(k,x)*p(n-k,y), with polynomials p(n,x) = Sum_{m=1..n} |A008297(n,m)|*(-x)^m, n >= 1 and p(0,x)=1 (for Sheffer polynomials see A048854 for S. Roman reference).
This unsigned matrix is embedded in the matrix for n!*L(n,-2,-x). Introduce 0,0 to each unsigned row and then add 1,-1,1 to the array as the first two rows to generate n!*L(n,-2,-x). - Tom Copeland, Apr 20 2014
The unsigned n-th row reverse polynomial equals the numerator polynomial of the finite continued fraction 1 - x/(1 + (n+1)*x/(1 + n*x/(1 + n*x/(1 + ... + 2*x/(1 + 2*x/(1 + x/(1 + x/(1)))))))). Cf. A089231. The denominator polynomial of the continued fraction is the (n+1)-th row polynomial of A144084. An example is given below. - Peter Bala, Oct 06 2019

			Triangle begins:
     1;
     3,    -1;
    12,    -8,    1;
    60,   -60,   15,   -1;
   360,  -480,  180,  -24,  1;
  2520, -4200, 2100, -420, 35, -1;
  ...
2!*L(2,2,x) = 12 - 8*x + x^2.
Unsigned row 3 polynomial in reverse form as the numerator of a continued fraction: 1 - x/(1 + 4*x/(1 + 3*x/(1 + 3*x/(1 + 2*x/(1 + 2*x/(1 + x/(1 + x))))))) = (60*x^3 + 60*x^2 + 15*x + 1)/(24*x^4 + 96*x^3 + 72*x^2 + 16*x + 1). - _Peter Bala_, Oct 06 2019

Indranil Ghosh, Rows 0..125, flattened
Robert S. Maier, Boson Operator Ordering Identities from Generalized Stirling and Eulerian Numbers, arXiv:2308.10332 [math.CO], 2023. See p. 19.
Index entries for sequences related to Laguerre polynomials

For m=0..5 the (unsigned) columns give A001710, A005990, A005461, A062193-A062195. The row sums (signed) give A062197, the row sums (unsigned) give A052852.

Cf. A021009, A062137-A062140, A066667, A089231, A144084.

with(PolynomialTools):
p := n -> (n+2)!*hypergeom([-n],[3],x)/2:
seq(CoefficientList(simplify(p(n)), x), n=0..9); # Peter Luschny, Apr 08 2015

Flatten[Table[((-1)^m)*n!*Binomial[n+2,n-m]/m!,{n,0,8},{m,0,n}]] (* Indranil Ghosh, Feb 24 2017 *)

tabl(nn) = {for (n=0, nn, for (k=0, n, print1(((-1)^k)*n!*binomial(n+2, n-k)/k!, ", ");); print(););} \\ Michel Marcus, May 06 2014

row(n) = Vecrev(n!*pollaguerre(n, 2)); \\ Michel Marcus, Feb 06 2021

import math
f=math.factorial
def C(n,r):return f(n)//f(r)//f(n-r)
i=0
for n in range(16):
    for m in range(n+1):
        i += 1
        print(i,((-1)**m)*f(n)*C(n+2,n-m)//f(m)) # Indranil Ghosh, Feb 24 2017

from functools import cache
@cache
def T(n, k):
    if k < 0 or k > n: return 0
    if k == n: return (-1)**n
    return (n + k + 2) * T(n-1, k) - T(n-1, k-1)
for n in range(7): print([T(n,k) for k in range(n + 1)])
# Peter Luschny, Mar 25 2024

T(n, m) = ((-1)^m)*n!*binomial(n+2, n-m)/m!.
E.g.f. for m-th column sequence: ((-x/(1-x))^m)/(m!*(1-x)^3), m >= 0.
n!*L(n,2,x) = (n+2)!*hypergeom([-n],[3],x)/2. - Peter Luschny, Apr 08 2015
From Werner Schulte, Mar 24 2024: (Start)
T(n, k) = (n+k+2) * T(n-1, k) - T(n-1, k-1) with initial values T(0, 0) = 1 and T(i, j) = 0 if j < 0 or j > i.
T = T^(-1), i.e., T is matrix inverse of T. (End)

cp's OEIS Frontend

A062139 Coefficient triangle of generalized Laguerre polynomials n!*L(n,2,x) (rising powers of x).

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