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%I #14 Oct 30 2014 18:32:56
%S 1,1,3,1,6,3,1,9,15,3,1,12,36,24,3,1,15,66,90,33,3,1,18,105,228,171,
%T 42,3,1,21,153,465,579,279,51,3,1,24,210,828,1500,1200,414,60,3,1,27,
%U 276,1344,3258,3858,2172,576,69,3,1,30,351,2040,6258,10116,8430
%N Triangular array: row n gives the coefficients of the polynomial p(n,x) defined in Comments.
%C The polynomial p(n,x) is given by p(n,x) = (x + 1)*p(n-1,x) + 2x, where p(0,x) = 1.
%H Clark Kimberling, <a href="/A249250/b249250.txt">Rows 0..100, flattened</a>
%e f(0,x) = 1/1, so that p(0,x) = 1
%e f(1,x) = (1 + 3 x)/1, so that p(1,x) = 1 + 3 x;
%e f(2,x) = (1 + 6 x + 3 x^2)/(1 + 3 x), so that p(2,x) = 1 + 6 x + 3 x^2.
%e First 6 rows of the triangle of coefficients:
%e 1
%e 1 3
%e 1 6 3
%e 1 9 15 3
%e 1 12 36 24 3
%e 1 15 66 90 33 3
%t z = 14; f[x_, n_] := x + 1 + 2 x/f[x, n - 1]; f[x_, 1] = 1;
%t t = Table[Factor[f[x, n]], {n, 1, z}]
%t u = Numerator[t]
%t TableForm[Table[CoefficientList[u[[n]], x], {n, 1, z}]] (* A249250 array *)
%t Flatten[CoefficientList[u, x]] (* A249250 sequence *)
%o (PARI) rown(n) = if (n==0, 1, x + 1 + 2*x/rown(n-1));
%o tabl(nn) = for (n=0, nn, print(Vecrev(numerator(rown(n))))); \\ _Michel Marcus_, Oct 28 2014
%Y Cf. A249247.
%K nonn,tabl,easy
%O 0,3
%A _Clark Kimberling_, Oct 24 2014
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