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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).

%I A321044 #7 Jan 09 2024 13:12:17
%S A321044 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,
%T A321044 112,44,10,1,1,21,100,285,540,714,672,450,210,65,12,1,1,25,144,506,
%U A321044 1210,2079,2640,2508,1782,935,352,90,14,1,1,29,196,819,2366,5005,8008,9867,9438,7007
%N 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).
%C A321044 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.
%C A321044 The graph of the root structures of the polynomial are mostly on a circle with center at -1, see Mathematica.
%F A321044 For n > 0, the n-th row sum is 3*2^(2n-2) - 1. - _Charles R Greathouse IV_, Oct 30 2018
%e A321044 {{1},
%e A321044 {1, 1},
%e A321044 {1, 5, 4, 1},
%e A321044 {1, 9, 16, 14, 6, 1},
%e A321044 {1, 13, 36, 55, 50, 27, 8,  1},
%e A321044 {1, 17, 64, 140, 196, 182, 112, 44, 10, 1},
%e A321044 {1, 21, 100, 285, 540, 714, 672, 450, 210, 65, 12, 1},
%e A321044 {1, 25, 144, 506, 1210, 2079, 2640, 2508, 1782,  935, 352, 90, 14, 1},
%e A321044 {1, 29, 196, 819, 2366, 5005, 8008, 9867, 9438, 7007, 4004, 1729, 546, 119, 16, 1},
%e A321044 {1, 33, 256, 1240, 4200, 10556, 20384, 30888,37180, 35750, 27456, 16744, 8008, 2940, 800, 152, 18, 1},
%e A321044 {1, 37, 324, 1785,6936, 20196, 45696, 82212, 119340, 140998, 136136, 107406, 68952, 35700, 14688, 4692, 1122, 189, 20, 1}}
%t A321044 g[x_, n_] = If[n == 0, 1, (2 + x)*(1 + x)^(-1 + 2 n)/(1 + x) - 1]
%t A321044 Show[Table[ListPlot[{Re[x], Im[x]} /. NSolve[g[x, n] == 0, x], PlotStyle -> Red], {n, 1, 10}]]
%t A321044 a = Table[CoefficientList[g[x, n], x], {n, 0, 10}] (* _Roger L. Bagula_, Oct 26 2018 *)
%t A321044 row[n_] = If[n > 0, CoefficientList[(x+2)*(x+1)^(2n-2)-1, x], {1}]; (* _Charles R Greathouse IV_, Oct 30 2018 *)
%o A321044 (PARI) row(n)=if(n, Vec((x+2)*(x+1)^(2*n-2)-1), [1]) \\ _Charles R Greathouse IV_, Oct 30 2018
%K A321044 nonn,tabf,obsc
%O A321044 0,5
%A A321044 _Roger L. Bagula_, Oct 26 2018