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%I #37 Nov 12 2022 02:13:11
%S 1,4,1,13,7,1,40,34,10,1,121,142,64,13,1,364,547,334,103,16,1,1093,
%T 2005,1549,643,151,19,1,3280,7108,6652,3478,1096,208,22,1,9841,24604,
%U 27064,17086,6766,1720,274,25,1,29524,83653,105796,78322,37384,11926
%N Mirror image of the triangle A193842.
%C A193843 is obtained by reversing the rows of the triangle A193842.
%C Consider the transformation 1 + x + x^2 + x^3 + ... + x^n = A_0*(x-3)^0 + A_1*(x-3)^1 + A_2*(x-3)^2 + ... + A_n*(x-3)^n. This sequence gives A_0, ... A_n as the entries in the n-th row of this triangle, starting at n = 0. - _Derek Orr_, Oct 14 2014
%H E. Neuwirth, <a href="https://doi.org/10.1016/S0012-365X(00)00373-3">Recursively defined combinatorial functions: Extending Galton's board,</a> Discrete Math. 239 (2001) 33-51.
%F Write w(n,k) for the triangle at A193842. The triangle at A193843 is then given by w(n,n-k).
%F From _Peter Bala_, Jul 31 2012: (Start)
%F Matrix product of the shifted Pascal triangle {C(n+1,k+1)}n,k>=0 and the square of the Pascal triangle {2^(n-k)*C(n,k)}n,k>=0. Thus the triangle is the product of two triangular Galton arrays and so is also a Galton array (Neuwirth, Theorem 10).
%F T(n,k) = sum {i = 0..n} C(n+1,i+1)*2^(i-k)*C(i,k).
%F Riordan array [1/((1 - x)*(1 - 3*x)), x/(1 - 3*x)].
%F O.g.f.: 1/((1 - x)*(1 - (3+t)*x)) = 1 + (4+t)*x + (13+7*t+t^2)*x^2 + ....
%F First column A003462. Row sums A002450. Alternating row sums A000225. Antidiagonal sums (Sum_{k} T(n-k,k)) A082574. Weighted sums (Sum_{k} k*T(n,k)) A014916.
%F (End)
%F T(n,k) = 4*T(n-1,k) + T(n-1,k-1) - 3*T(n-2,k) - T(n-2,k-1), T(0,0) = T(1,1) = 1, T(1,0) = 4, T(n,k) = 0 if k < 0 or if k > n. - _Philippe Deléham_, Jan 17 2014
%e First six rows:
%e 1;
%e 4, 1;
%e 13, 7, 1;
%e 40, 34, 10, 1;
%e 121, 142, 64, 13, 1;
%e 364, 547, 334, 103, 16, 1;
%p T := proc(n,k) option remember;
%p if k<0 or k>n then 0 elif n=k then 1 elif n=1 and k=0 then 4
%p else 4*T(n-1,k) + T(n-1,k-1) -3*T(n-2,k) - T(n-2,k-1) fi end;
%p seq(seq(T(n,k), k=0..n), n=0..9); # _Peter Luschny_, Jan 18 2014
%t z = 10;
%t p[n_, x_] := (x + 1)^n;
%t q[n_, x_] := (x + 2)^n
%t p1[n_, k_] := Coefficient[p[n, x], x^k];
%t p1[n_, 0] := p[n, x] /. x -> 0;
%t d[n_, x_] := Sum[p1[n, k]*q[n - 1 - k, x], {k, 0, n - 1}]
%t h[n_] := CoefficientList[d[n, x], {x}]
%t TableForm[Table[Reverse[h[n]], {n, 0, z}]]
%t Flatten[Table[Reverse[h[n]], {n, -1, z}]] (* A193842 *)
%t TableForm[Table[h[n], {n, 0, z}]] (* A193843 *)
%t Flatten[Table[h[n], {n, -1, z}]]
%o (PARI) for(n=0,20,for(k=0,n,print1(1/k!*sum(i=0,n,(3^(i-k)*prod(j=0,k-1,i-j))),", "))) \\ _Derek Orr_, Oct 14 2014
%Y Cf. A193842.
%Y Cf. A000225 (alt.row sums), A002450 (row sums), A014916 (weighted sums).
%Y Cf. A003462 (first col.), A082574 (anti-diag.sums).
%K nonn,tabl
%O 0,2
%A _Clark Kimberling_, Aug 07 2011
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