A283151 Triangle read by rows: Riordan array (1/(1-9x)^(2/3), x/(9x-1)).
1, 6, -1, 45, -15, 1, 360, -180, 24, -1, 2970, -1980, 396, -33, 1, 24948, -20790, 5544, -693, 42, -1, 212058, -212058, 70686, -11781, 1071, -51, 1, 1817640, -2120580, 848232, -176715, 21420, -1530, 60, -1, 15677145, -20902860, 9754668, -2438667, 369495, -35190, 2070, -69, 1, 135868590, -203802885
Offset: 0
Examples
Triangle begins
1;
6, -1;
45, -15, 1;
360, -180, 24, -1;
2970, -1980, 396, -33, 1;
24948, -20790, 5544, -693, 42, -1;
212058, -212058, 70686, -11781, 1071, -51, 1;
1817640, -2120580, 848232, -176715, 21420, -1530, 60, -1;
15677145, -20902860, 9754668, -2438667, 369495, -35190, 2070, -69, 1;
Links
- 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.
Formula
a(m,n) = binomial(-n-2/3, m-n)*(-1)^m*9^(m-n).
G.f.: (1-9x)^(1/3)/(xy-9x+1).
Recurrence: a(m,n) = a(m,n-1)*(n-1-m)/(9*n-3) for 0 < n <= m; matrix inverse of a(m,n) is a(m,n). - 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-1/3) about 0. For example, for n = 4 we have (1 - 9*x)^(11/3) = 2970*x^4 - 1980*x^3 + 396*x^2 - 33*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)*(360 - 180*(9*x) + 24*(9*x)^2/2! - (9*x)^3/3!) = 360 - 1980*x + 5544*x^2/2! - 11781*x^3/3! + 21420*x^4/4! - ....
Let F(x) = (1 - ( 1 - 9*x)^(1/3))/(3*x). See A025748. 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-1/3) ), n = 0,1,2,.... Cf. A283151 and A046521. (End)
From Peter Bala, Aug 18 2021: (Start)
T(n,k) = (-1)^k*binomial(n-1/3, 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 = 2/3.
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)^(2/3)) * F(x/(1 - 9*b*x)) iff F(x) = (1/(1 + 9*b*x)^(2/3)) * G(x/(1 + 9*b*x)).
The infinitesimal generator of the unsigned array has the sequence (9*n+6) 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)
Extensions
Offset corrected by Werner Schulte, Aug 05 2017
Comments