A321044 Irregular table related to f[(a*x+b)/(c*x+d)]=(c*x+d)^(2*n)*f[x], and f[x]=1/(x+1), f[x]=(a*x+b)/(c*x+d).

1, 1, 1, 1, 5, 4, 1, 1, 9, 16, 14, 6, 1, 1, 13, 36, 55, 50, 27, 8, 1, 1, 17, 64, 140, 196, 182, 112, 44, 10, 1, 1, 21, 100, 285, 540, 714, 672, 450, 210, 65, 12, 1, 1, 25, 144, 506, 1210, 2079, 2640, 2508, 1782, 935, 352, 90, 14, 1, 1, 29, 196, 819, 2366, 5005, 8008, 9867, 9438, 7007

Offset: 0

Author

Roger L. Bagula, Oct 26 2018

Keywords

• nonn
• tabf
• obsc

Comments

The author writes: This derivation is an interesting Fibonacci modular form. The modular form function is: f[(a*x+b)/(c*x+d)] = (c*x+d)^(2*n)*f[x] and I have used f[x]=1/(x+1) and f[x]=(a*x+b)/(c*x+d) for Möbius function matrix: {{0,1},{1,1}}. The polynomial is solved as a zero based form.

The graph of the root structures of the polynomial are mostly on a circle with center at -1, see Mathematica.

Examples

			{{1},
{1, 1},
{1, 5, 4, 1},
{1, 9, 16, 14, 6, 1},
{1, 13, 36, 55, 50, 27, 8,  1},
{1, 17, 64, 140, 196, 182, 112, 44, 10, 1},
{1, 21, 100, 285, 540, 714, 672, 450, 210, 65, 12, 1},
{1, 25, 144, 506, 1210, 2079, 2640, 2508, 1782,  935, 352, 90, 14, 1},
{1, 29, 196, 819, 2366, 5005, 8008, 9867, 9438, 7007, 4004, 1729, 546, 119, 16, 1},
{1, 33, 256, 1240, 4200, 10556, 20384, 30888,37180, 35750, 27456, 16744, 8008, 2940, 800, 152, 18, 1},
{1, 37, 324, 1785,6936, 20196, 45696, 82212, 119340, 140998, 136136, 107406, 68952, 35700, 14688, 4692, 1122, 189, 20, 1}}
		

Programs

Mathematica

g[x_, n_] = If[n == 0, 1, (2 + x)*(1 + x)^(-1 + 2 n)/(1 + x) - 1]
Show[Table[ListPlot[{Re[x], Im[x]} /. NSolve[g[x, n] == 0, x], PlotStyle -> Red], {n, 1, 10}]]
a = Table[CoefficientList[g[x, n], x], {n, 0, 10}] (* Roger L. Bagula, Oct 26 2018 *)
row[n_] = If[n > 0, CoefficientList[(x+2)*(x+1)^(2n-2)-1, x], {1}]; (* Charles R Greathouse IV, Oct 30 2018 *)

PARI

row(n)=if(n, Vec((x+2)*(x+1)^(2*n-2)-1), [1]) \\ Charles R Greathouse IV, Oct 30 2018

Formula

For n > 0, the n-th row sum is 3*2^(2n-2) - 1. - Charles R Greathouse IV, Oct 30 2018