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A281122 Triangle T read by rows: n-th row (n>=0) gives the non-vanishing coefficients of the polynomial q(n,x) = ((x+1)^(2^n) - (x-1)^(2^n))/2.

%I A281122 #28 Feb 04 2021 06:44:26
%S A281122 1,2,4,4,8,56,56,8,16,560,4368,11440,11440,4368,560,16,32,4960,201376,
%T A281122 3365856,28048800,129024480,347373600,565722720,565722720,347373600,
%U A281122 129024480,28048800,3365856,201376,4960,32
%N A281122 Triangle T read by rows: n-th row (n>=0) gives the non-vanishing coefficients of the polynomial q(n,x) = ((x+1)^(2^n) - (x-1)^(2^n))/2.
%C A281122 The length of row n is 1 for n = 0 and 2^(n-1) = A000079(n-1) for n >= 1.
%C A281122 The polynomial q(n,x) = 2^n*Product_{k=0..n-1} p(k,x) with polynomial p(n,x) = ((x+1)^(2^n) + (x-1)^(2^n))/2, whose coefficients are tabulated in A201461.
%C A281122 The row polynomials are q(n, x) = Sum_{k=0..2^(n-1)-1} T(n, k)*x^(2^n-1-2*k) for n >= 1, and q(0,x) = 1. - _Wolfdieter Lang_, Jan 19 2017
%C A281122 A201461 and T(n,k) are a bisection of row 2^n of Pascal's triangle A007318.
%C A281122 The algorithm r(n) = 1/2*(r(n-1) + A/r(n-1)), starting with r(0) = A, used for approximating sqrt(A), which is known as the Babylonian method or Hero's method after the first-century Greek mathematician Hero of Alexandria and which can be derived from Newton's method, generates fractions beginning with (A+1)/2, (A^2 + 6*A + 1)/(4*A + 4), (A^4 + 28*A^3 + 70*A^2 + 28*A + 1)/(8*A^3 + 56*A^2 + 56*A + 8), ... This is sqrt(A)*p(n,sqrt(A))/q(n,sqrt(A)) with the given polynomials p(n,x) and q(n,x).
%H A281122 Indranil Ghosh, <a href="/A281122/b281122.txt">Rows 0..11, flattened</a>
%F A281122 T(n, k) = 1 if n = 0 and k = 0, and T(n, k) = binomial(2^n,2*k+1) = A103328(2^(n-1),k) for k = 0..2^(n-1)-1 and n >= 1. - _Wolfdieter Lang_, Jan 19 2017
%e A281122 The triangle of non-vanishing coefficients starts with
%e A281122 1
%e A281122 2
%e A281122 4, 4
%e A281122 8, 56, 56, 8
%e A281122 16, 560, 4368, 11440, 11440, 4368, 560, 16
%e A281122 etc., since the first few polynomials are
%e A281122 q(0,x) = 1,
%e A281122 q(1,x) = 2*x,
%e A281122 q(2,x) = 4*x^3 + 4*x = 4*x*(x^2 + 1),
%e A281122 q(3,x) = 8*x^7 + 56*x^5 + 56*x^3 + 8*x = 8*x*(x^2 + 1)*(x^4 + 6*x^2 + 1),
%e A281122 q(4,x) = 16*x^15 + 560*x^13 + 4368*x^11 + 11440*x^9 + 11440*x^7 + 4368*x^5 + 560*x^3 + 16*x = 16*x*(x^2 + 1)*(x^4 + 6*x^2 + 1)*(x^8 + 28*x^6 + 70*x^4 + 28*x^2 + 1),
%e A281122 etc.
%p A281122 CoeffList := p -> remove(n->n=0, [op(PolynomialTools:-CoefficientList(p, x))]):
%p A281122 Tpoly := n -> (1/2)*((x+1)^(2^n) - (x-1)^(2^n));
%p A281122 seq(print(CoeffList(Tpoly(n))), n=0..5); # _Peter Luschny_, Feb 04 2021
%t A281122 q[n_] := DeleteCases[ CoefficientList[ Expand[((x +1)^(2^n) - (x -1)^(2^n))/2], x], 0]; Array[q, 7, 0] // Flatten (* _Robert G. Wilson v_, Jan 16 2017 *)
%t A281122 t={1};T[n_,k_]:=Table[Binomial[2^n,2k+1],{n,1,6},{k,0,2^(n-1)-1}];Do[AppendTo[t,T[n,k]]];Flatten[t] (* _Indranil Ghosh_, Feb 22 2017 *)
%Y A281122 Cf. A000079, A007318, A103328, A201461, A281123.
%K A281122 nonn,tabf
%O A281122 0,2
%A A281122 _Martin Renner_, Jan 15 2017