A062139 - OEIS

%I #48 Mar 25 2024 19:20:49

%S 1,3,-1,12,-8,1,60,-60,15,-1,360,-480,180,-24,1,2520,-4200,2100,-420,

%T 35,-1,20160,-40320,25200,-6720,840,-48,1,181440,-423360,317520,

%U -105840,17640,-1512,63,-1,1814400,-4838400,4233600

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

%C 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).

%C 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

%C 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

%H Indranil Ghosh, <a href="/A062139/b062139.txt">Rows 0..125, flattened</a>

%H Robert S. Maier, <a href="https://arxiv.org/abs/2308.10332">Boson Operator Ordering Identities from Generalized Stirling and Eulerian Numbers</a>, arXiv:2308.10332 [math.CO], 2023. See p. 19.

%H <a href="/index/La#Laguerre">Index entries for sequences related to Laguerre polynomials</a>

%F T(n, m) = ((-1)^m)*n!*binomial(n+2, n-m)/m!.

%F E.g.f. for m-th column sequence: ((-x/(1-x))^m)/(m!*(1-x)^3), m >= 0.

%F n!*L(n,2,x) = (n+2)!*hypergeom([-n],[3],x)/2. - _Peter Luschny_, Apr 08 2015

%F From _Werner Schulte_, Mar 24 2024: (Start)

%F 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.

%F T = T^(-1), i.e., T is matrix inverse of T. (End)

%e Triangle begins:

%e      1;

%e      3,    -1;

%e     12,    -8,    1;

%e     60,   -60,   15,   -1;

%e    360,  -480,  180,  -24,  1;

%e   2520, -4200, 2100, -420, 35, -1;

%e   ...

%e 2!*L(2,2,x) = 12 - 8*x + x^2.

%e 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

%p with(PolynomialTools):

%p p := n -> (n+2)!*hypergeom([-n],[3],x)/2:

%p seq(CoefficientList(simplify(p(n)), x), n=0..9); # _Peter Luschny_, Apr 08 2015

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

%o (PARI) 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

%o (PARI) row(n) = Vecrev(n!*pollaguerre(n, 2)); \\ _Michel Marcus_, Feb 06 2021

%o (Python)

%o import math

%o f=math.factorial

%o def C(n,r):return f(n)//f(r)//f(n-r)

%o i=0

%o for n in range(16):

%o     for m in range(n+1):

%o         i += 1

%o         print(i,((-1)**m)*f(n)*C(n+2,n-m)//f(m)) # _Indranil Ghosh_, Feb 24 2017

%o (Python)

%o from functools import cache

%o @cache

%o def T(n, k):

%o     if k < 0 or k > n: return 0

%o     if k == n: return (-1)**n

%o     return (n + k + 2) * T(n-1, k) - T(n-1, k-1)

%o for n in range(7): print([T(n,k) for k in range(n + 1)])

%o # _Peter Luschny_, Mar 25 2024

%Y 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.

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

%K sign,easy,tabl

%O 0,2

%A _Wolfdieter Lang_, Jun 19 2001