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%I #22 Dec 29 2023 13:27:09
%S 1,3,1,11,6,1,45,30,9,1,195,144,58,12,1,873,685,330,95,15,1,3989,3258,
%T 1770,630,141,18,1,18483,15533,9198,3801,1071,196,21,1,86515,74280,
%U 46928,21672,7210,1680,260,24,1,408105,356283,236736,119154,44982,12510,2484,333,27,1
%N Riordan array (1/sqrt(1-6x+5x^2),(1-3x-sqrt(1-6x+5x^2))/(2x)).
%C Columns include A026375, A026376 and A026377. Inverse is A110168. Rows sums are A110166. Diagonal sums are A110167.
%C From _Peter Bala_, Jan 09 2022: (Start)
%C This Riordan array has the form ( x*h'(x)/h(x), h(x) ) with h(x) = (1 - 3*x - sqrt(1 - 6*x + 5*x^2))/(2*x) and so belongs to the hitting time subgroup H of the Riordan group (see Peart and Woan).
%C T(n,k) = [x^(n-k)] f(x)^n with f(x) = 1 + 3*x + x^2. In general the (n,k)-th entry of the hitting time array ( x*h'(x)/h(x), h(x) ) has the form [x^(n-k)] f(x)^n, where f(x) = x/( series reversion of h(x) ). (End)
%H P. Peart and W.-J. Woan, <a href="https://doi.org/10.1016/S0166-218X(99)00166-3">A divisibility property for a subgroup of Riordan matrices</a>, Discrete Applied Mathematics, Vol. 98, Issue 3, Jan 2000, 255-263.
%F Number triangle T(n, k) = Sum_{j = 0..n} C(n, j)C(2j, j+k).
%F T(n,0) = 3*T(n-1,0) + 2*T(n-1,1), T(n,k) = T(n-1,k-1) + 3*T(n-1,k) + T(n-1,k+1) for k > 0, T(0,0) = 1, T(n,k) = 0 if k < 0 or if k > n. - _Philippe Deléham_, Jan 24 2014
%e Rows begin
%e 1;
%e 3, 1;
%e 11, 6, 1;
%e 45, 30, 9, 1;
%e 195, 144, 58, 12, 1;
%e 873, 685, 330, 95, 15, 1;
%e Production array begins:
%e 3, 1;
%e 2, 3, 1;
%e 0, 1, 3, 1;
%e 0, 0, 1, 3, 1;
%e 0, 0, 0, 1, 3, 1;
%e 0, 0, 0, 0, 1, 3, 1;
%e 0, 0, 0, 0, 0, 1, 3, 1;
%e ... - _Philippe Deléham_, Feb 08 2014
%p seq(seq( coeff((x^2 + 3*x + 1)^n, x, n-k), k = 0..n ), n = 0..10); # _Peter Bala_, Jan 09 2022
%t (* The function RiordanArray is defined in A256893. *)
%t RiordanArray[1/Sqrt[1-6#+5#^2]&, (1-3#-Sqrt[1-6#+5#^2])/(2#)&, 10] // Flatten (* _Jean-François Alcover_, Jul 19 2019 *)
%K easy,nonn,tabl
%O 0,2
%A _Paul Barry_, Jul 14 2005
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