A155579 -id:A155579 - cp's OEIS frontend

A004989 a(n) = (3^n/n!) * Product_{k=0..n-1} (3*k - 2).

1, -6, -9, -36, -189, -1134, -7371, -50544, -360126, -2640924, -19806930, -151252920, -1172210130, -9197341020, -72921775230, -583374201840, -4703454502335, -38180983607190, -311811366125385, -2560135427134740, -21121117273861605, -175003543126281870, -1455711290550435555

Offset: 0

Joe Keane (jgk(AT)jgk.org)

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A004988, A004990, A004991, A004992, A025748, A155579, A185047.

List([0..25], n-> 3^n*Product([0..n-1], k-> 3*k-2)/Factorial(n) ); # G. C. Greubel, Aug 22 2019

[1] cat [3^n*(&*[3*k-2: k in [0..n-1]])/Factorial(n): n in [1..25]]; // G. C. Greubel, Aug 22 2019

a:= n-> (3^n/n!)*product(3*k-2, k=0..n-1); seq(a(n), n=0..25); # G. C. Greubel, Aug 22 2019

Table[9^n*Pochhammer[-2/3, n]/n!, {n,0,25}] (* G. C. Greubel, Aug 22 2019 *)

a(n)=if(n<0,0,prod(k=0,n-1,3*k-2)*3^n/n!)

[9^n*rising_factorial(-2/3, n)/factorial(n) for n in (0..25)] # G. C. Greubel, Aug 22 2019

a(n) ~ -(2/3)*Gamma(1/3)^-1*n^(-5/3)*3^(2*n)*(1 + (5/9)*n^-1 + ...).
G.f.: (1-9*x)^(2/3).
D-finite with recurrence: n*a(n) +3*(-3*n+5)*a(n-1)=0. - R. J. Mathar, Jan 17 2020
Sum_{n>=0} 1/a(n) = 99/128 - 5*sqrt(3)*Pi/512 - 15*log(3)/512. - Amiram Eldar, Dec 02 2022
From Peter Bala, Oct 14 2024: (Start)
a(n) = -1/(sqrt(3)*Pi) * 9^n * Gamma(2/3)*Gamma(n-2/3)/Gamma(n+1).
For n >= 1, a(n) = - Integral_{x = 0..9} x^n * w(x) dx, where w(x) = sqrt(3)/(2*Pi) * x^(1/3)*(9 - x)^(2/3)/x^2. (End)

A283150 Riordan array (1/(1-9x)^(1/3), x/(9x-1)).

Original entry on oeis.org

1, 3, -1, 18, -12, 1, 126, -126, 21, -1, 945, -1260, 315, -30, 1, 7371, -12285, 4095, -585, 39, -1, 58968, -117936, 49140, -9360, 936, -48, 1, 480168, -1120392, 560196, -133380, 17784, -1368, 57, -1, 3961386, -10563696, 6162156, -1760616, 293436, -30096, 1881, -66, 1, 33011550, -99034650, 66023100

Offset: 0

Tom Richardson, Mar 01 2017

Triangle read by rows. This is an example of a Riordan group involution. Dual Riordan array of A283151. With A283151 and A248324, forms doubly infinite Riordan array. For b and c the sequences A283151 and A248324, respectively, and i,j >= 0, the doubly infinite array with d(i,j) = a(i,j), d(-j,-i) = b(i,j), d(i,-j) = c(j,i), and d(-i,j) = 0 (except d(0,0)=1) is a doubly infinite Riordan array.

Matrix inverse of a(m,n) is a(m,n). - Werner Schulte, Aug 05 2017

			The triangle begins
        1;
        3,        -1;
       18,       -12,       1;
      126,      -126,      21,       -1;
      945,     -1260,     315,      -30,      1;
     7371,    -12285,    4095,     -585,     39,     -1;
    58968,   -117936,   49140,    -9360,    936,    -48,    1;
   480168,  -1120392,  560196,  -133380,  17784,  -1368,   57,  -1;
  3961386, -10563696, 6162156, -1760616, 293436, -30096, 1881, -66, 1;

Peter Bala, A 4-parameter family of embedded Riordan arrays
Peter Bala, A note on the diagonals of a proper Riordan Array
H. Prodinger, Some information about the binomial transform, The Fibonacci Quarterly, 32, 1994, 412-415.
Thomas M. Richardson, The three 'R's and Dual Riordan Arrays, arXiv:1609.01193 [math.CO], 2016.

Cf. A004987, A248324, A283151, A046521, A155579.

T := (n, k) -> (-1)^k*binomial(n - 2/3, n - k)*9^(n - k):
for n from 0 to 6 do seq(T(n, k), k = 0..n) od; # Peter Luschny, Sep 03 2021

a(m,n) = binomial(-n-1/3, m-n)*(-1)^m*9^(m-n);
tabl(nn) = for(n=0, nn, for (k=0, n, print1(a(n, k), ", ")); print); \\ Michel Marcus, Aug 07 2017

a(m,n) = binomial(-n-1/3, m-n)*(-1)^m*9^(m-n).
G.f.: (1-9x)^(2/3)/(xy-9x+1).
Recurrence: a(m,n) = a(m, n-1)*(n-1-m)/(9*n-6) for 0 < n <= m. - Werner Schulte, Aug 05 2017
From Peter Bala, Mar 05 2018 (Start):
Let P(n,x) = Sum_{k = 0..n} T(n,k)*x^(n-k) denote the n-th row polynomial in descending powers of x. Then (-1)^n*P(n,x) is the n-th degree Taylor polynomial of (1 - 9*x)^(n-2/3) about 0. For example, for n = 4 we have (1 - 9*x)^(10/3) = 945*x^4 - 1260*x^3 + 315*x^2 - 30*x + 1 + O(x^5).
Let R(n,x) denote the n-th row polynomial of this triangle. The polynomial R(n,9*x) has the e.g.f. Sum_{k = 0..n} T(n,k)*(9*x)^k/k!. The e.g.f. for the n-th diagonal of the triangle (starting at n = 0 for the main diagonal) equals exp(-x) * the e.g.f. for the polynomial R(n,9*x). For example, when n = 3 we have exp(-x)*(126 - 126*(9*x) + 21*(9*x)^2/2! - (9*x)^3/3!) = 126 - 1260*x + 4095*x^2/2! - 9360*x^3/3! + 17784*x^4/4! - ....
Let F(x) = (1 - ( 1 - 9*x)^(2/3))/(3*x) denote the o.g.f. of A155579. The derivatives of F(x) are related to the row polynomials P(n,x) by the identity x^n/n! * (d/dx)^n(F(x)) = 1/(3*x)*( (-1)^n - P(n,x)/(1 - 9*x)^(n-2/3) ), n = 0,1,2,.... Cf. A283151 and A046521. (End)
From Peter Bala, Aug 18 2021: (Start)
T(n,k) = (-1)^k*binomial(n-2/3, n-k)*9^(n-k).
Analogous to the binomial transform we have the following sequence transformation formula: g(n) = Sum_{k = 0..n} T(n,k)*b^(n-k)*f(k) iff f(n) = Sum_{k = 0..n} T(n,k)*b^(n-k)*g(k). See Prodinger, bottom of p. 413, with b replaced with 9*b, c = -1 and d = 1/3.
Equivalently, if F(x) = Sum_{n >= 0} f(n)*x^n and G(x) = Sum_{n >= 0} g(n)*x^n are a pair of formal power series then
G(x) = (1/(1 - 9*b*x)^(1/3)) * F(x/(1 - 9*b*x)) iff F(x) = (1/(1 + 9*b*x)^(1/3)) * G(x/(1 + 9*b*x)).
The infinitesimal generator of the unsigned array has the sequence (9*n+3) n>=0 on the main subdiagonal and zeros elsewhere. The m-th power of the unsigned array has entries m^(n-k)*|T(n,k)|. (End)

Offset corrected by Werner Schulte, Aug 05 2017

cp's OEIS Frontend

A004989 a(n) = (3^n/n!) * Product_{k=0..n-1} (3*k - 2).

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