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A249100 Triangular array read by rows: row n gives the coefficients of the polynomial p(n,x) defined in Comments.

%I A249100 #19 Feb 28 2025 07:45:32
%S A249100 1,3,1,5,3,1,21,12,3,1,45,48,21,3,1,231,177,81,32,3,1,585,855,450,120,
%T A249100 45,3,1,3465,3240,2070,930,165,60,3,1,9945,18000,10890,4110,1695,216,
%U A249100 77,3,1,65835,71505,57330,28560,7245,2835,273,96,3,1,208845,443835,300195,143640,64155,11781,4452,336,117,3,1
%N A249100 Triangular array read by rows: row n gives the coefficients of the polynomial p(n,x) defined in Comments.
%C A249100 The polynomial p(n,x) is the numerator of the rational function given by f(n,x) = x + (2*n+1)/f(n-1,x), where f(0,x) = 1.
%C A249100 (Sum of numbers in row n) = A249101(n) for n >= 0.
%C A249100 (n-th term of column 1) = A235136(n) for n >= 1.
%H A249100 Clark Kimberling, <a href="/A249100/b249100.txt">Rows 0..100, flattened</a>
%F A249100 T(n, k) = T(n-1, k-1) + (2*n-1)*T(n-2, k). - _Michael Somos_, Oct 27 2022
%e A249100 f(0,x) = 1/1, so that p(0,x) = 1;
%e A249100 f(1,x) = (3 + x)/1, so that p(1,x) = 3 + x;
%e A249100 f(2,x) = (5 + 3*x + x^2)/(3 + x), so that p(2,x) = 5 + 3*x + x^2.
%e A249100 First 6 rows of the triangle of coefficients:
%e A249100     1;
%e A249100     3,   1;
%e A249100     5,   3,   1;
%e A249100    21,  12,   3,   1;
%e A249100    45,  48,  21,   3,   1;
%e A249100   231, 177,  81,  32,   3,   1;
%t A249100 z = 11; p[x_, n_] := x + (2 n - 1)/p[x, n - 1]; p[x_, 1] = 1;
%t A249100 t = Table[Factor[p[x, n]], {n, 1, z}]
%t A249100 u = Numerator[t]
%t A249100 TableForm[Table[CoefficientList[u[[n]], x], {n, 1, z}]] (* A249100 array *)
%t A249100 Flatten[CoefficientList[u, x]] (* A249100 sequence *)
%t A249100 v = u /. x -> 1  (* A249101 *)
%t A249100 u /. x -> 0  (* A235136 *)
%t A249100 T[ n_Integer, k_Integer] := (T[n, k] = If[n<2, Boole[0==k], T[n-1, k-1] + (2*n-1)*T[n-2 ,k] ]); Join @@ Table[T[n, k], {n, 10}, {k, 0, n-1}] (* _Michael Somos_, Oct 27 2022 *)
%Y A249100 Cf. A249101, A245136, A087299.
%K A249100 nonn,tabl,easy
%O A249100 0,2
%A A249100 _Clark Kimberling_, Oct 21 2014