A249100 Triangular array read by rows: row n gives the coefficients of the polynomial p(n,x) defined in Comments.
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, 45, 3, 1, 3465, 3240, 2070, 930, 165, 60, 3, 1, 9945, 18000, 10890, 4110, 1695, 216, 77, 3, 1, 65835, 71505, 57330, 28560, 7245, 2835, 273, 96, 3, 1, 208845, 443835, 300195, 143640, 64155, 11781, 4452, 336, 117, 3, 1
Offset: 0
Examples
f(0,x) = 1/1, so that p(0,x) = 1; f(1,x) = (3 + x)/1, so that p(1,x) = 3 + x; f(2,x) = (5 + 3*x + x^2)/(3 + x), so that p(2,x) = 5 + 3*x + x^2. First 6 rows of the triangle of coefficients: 1; 3, 1; 5, 3, 1; 21, 12, 3, 1; 45, 48, 21, 3, 1; 231, 177, 81, 32, 3, 1;
Links
- Clark Kimberling, Rows 0..100, flattened
Programs
-
Mathematica
z = 11; p[x_, n_] := x + (2 n - 1)/p[x, n - 1]; p[x_, 1] = 1; t = Table[Factor[p[x, n]], {n, 1, z}] u = Numerator[t] TableForm[Table[CoefficientList[u[[n]], x], {n, 1, z}]] (* A249100 array *) Flatten[CoefficientList[u, x]] (* A249100 sequence *) v = u /. x -> 1 (* A249101 *) u /. x -> 0 (* A235136 *) 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 *)
Formula
T(n, k) = T(n-1, k-1) + (2*n-1)*T(n-2, k). - Michael Somos, Oct 27 2022
Comments