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A283150
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Riordan array (1/(1-9x)^(1/3), x/(9x-1)).
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3
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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
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OFFSET
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0,2
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COMMENTS
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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.
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LINKS
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FORMULA
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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
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)
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)
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EXAMPLE
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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;
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MAPLE
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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
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PROG
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(PARI) 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
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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