%I
%S 1,1,1,1,2,2,1,3,8,4,1,4,20,24,8,1,5,40,84,64,16,1,6,70,
%T 224,288,160,32,1,7,112,504,960,880,384,64,1,8,168,1008,2640,
%U 3520,2496,896,128,1,9,240,1848,6336,11440,11648,6720,2048,256,1,10,330,3168,13728,32032,43680,35840,17408
%N Derived Shabat linear tree transform of A053120: Triangle of coefficients of transformed Chebyshev's T(n, x) polynomials (powers of x in increasing order) T(x,n)>c*T(c*x+d)+d: c=1;d=1; as substitution: 1x>y( here alternative starting polynomial of Q(y,1]=1y.
%C Row sums are:
%C {1, 0, 1, 2, 1, 0, 1, 2, 1, 0, 1}
%C integration is alternating on transformed {0,2} domain:
%C Table[Table[Integrate[Sqrt[1/(1  (1  y)^2)]*Q[y, n]*Q[y, m], {y, 0, 2}], {n, 0, 10}], {m, 0, 10}]
%D http://logic.pdmi.ras.ru/~yumat/personaljournal/chebyshev/chebysh.htm Quote: "... how many polynomials can generate given tree? It is easy to see that if P is a generalized Chebyshev Polynomial, then so is polynomial CP(cz+d)+D, moreover, it represents the same tree (of course, provided that both C and c are different from zero). In some natural sense these two linear transformations exhaust the variety of polynomials representing given tree. Namely, every drawing of a tree on the plane introduces an additional structurecircular order of edges around given vertex (say, clockwise). Dealing with Chebyshev polynomials, it is natural to speak about plane trees understanding by them trees with this additional structure. "
%F Q(y,0)=1;Q(y,1)=1y; Q(y, n) = (2 + 2 (1  y)  2 (1  y) Q(y, n  1) + Q(y, n  2))
%e {1},
%e {1, 1},
%e {1, 2, 2},
%e {1, 3, 8, 4},
%e {1, 4, 20, 24,8},
%e {1, 5, 40, 84, 64, 16},
%e {1, 6, 70, 224, 288, 160, 32},
%e {1, 7, 112, 504, 960, 880, 384, 64},
%e {1, 8, 168, 1008, 2640, 3520, 2496, 896, 128},
%e {1, 9, 240, 1848, 6336, 11440, 11648, 6720, 2048, 256},
%e {1, 10, 330, 3168, 13728, 32032, 43680, 35840, 17408, 4608, 512}
%t Clear[c, d, x0, x1, x2, P, Q, x, n, a] P[x, 0] = 1; P[x, 1] = x; P[x_, n_] := P[x, n] = 2*x*P[x, n  1]  P[x, n  2]; Solve[c*x0 + d  2*x*(c*x1 + d) + c*x2 + d == 0, x0] c = 1; d = 1; (* Transform : c*x + d > y*) Q[y, 1] = 0; Q[y, 0] = 1; Q[y, 1] =1 y; Q[y_, n_] := Q[y, n] = (2 + 2 (1  y)  2 (1  y) Q[y, n  1] + Q[y, n  2]); Table[ExpandAll[Q[y, n]], {n, 0, 10}]; a = Table[CoefficientList[Q[y, n], y], {n, 0, 10}]; Flatten[a]
%Y Cf. A053120.
%K uned,tabl,sign
%O 1,5
%A _Roger L. Bagula_, Mar 16 2008
