%I #44 Sep 03 2021 01:59:05

%S 1,6,-1,45,-15,1,360,-180,24,-1,2970,-1980,396,-33,1,24948,-20790,

%T 5544,-693,42,-1,212058,-212058,70686,-11781,1071,-51,1,1817640,

%U -2120580,848232,-176715,21420,-1530,60,-1,15677145,-20902860,9754668,-2438667,369495,-35190,2070,-69,1,135868590,-203802885

%N Triangle read by rows: Riordan array (1/(1-9x)^(2/3), x/(9x-1)).

%C This is an example of a Riordan group involution.

%C Dual Riordan array of A283150.

%C With A283150 and A248324, forms doubly infinite Riordan array. For b and c the sequences A283150 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(i,j), and d(-i,j) = 0 (except d(0,0)=1) is a doubly infinite Riordan array.

%H Peter Bala, <a href="/A264772/a264772_1.pdf">A 4-parameter family of embedded Riordan arrays</a>

%H Peter Bala, <a href="/A081577/a081577.pdf">A note on the diagonals of a proper Riordan Array</a>

%H H. Prodinger, <a href="https://www.fq.math.ca/Scanned/32-5/prodinger.pdf">Some information about the binomial transform</a>, The Fibonacci Quarterly, 32, 1994, 412-415.

%H Thomas M. Richardson, <a href="http://arxiv.org/abs/1609.01193">The three 'R's and Dual Riordan Arrays</a>, arXiv:1609.01193 [math.CO], 2016.

%F a(m,n) = binomial(-n-2/3, m-n)*(-1)^m*9^(m-n).

%F G.f.: (1-9x)^(1/3)/(xy-9x+1).

%F 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

%F From _Peter Bala_, Mar 05 2018 (Start):

%F 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).

%F 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! - ....

%F 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)

%F From _Peter Bala_, Aug 18 2021: (Start)

%F T(n,k) = (-1)^k*binomial(n-1/3, n-k)*9^(n-k).

%F 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.

%F 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

%F 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)).

%F 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)

%e Triangle begins

%e 1;

%e 6, -1;

%e 45, -15, 1;

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

%e 2970, -1980, 396, -33, 1;

%e 24948, -20790, 5544, -693, 42, -1;

%e 212058, -212058, 70686, -11781, 1071, -51, 1;

%e 1817640, -2120580, 848232, -176715, 21420, -1530, 60, -1;

%e 15677145, -20902860, 9754668, -2438667, 369495, -35190, 2070, -69, 1;

%Y Cf. A004988, A248324, A283150, A025748, A046521.

%K sign,tabl,easy

%O 0,2

%A _Tom Richardson_, Mar 01 2017

%E Offset corrected by _Werner Schulte_, Aug 05 2017