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%I #24 Feb 22 2017 10:44:10
%S 1,1,1,1,1,7,7,1,1,35,273,715,715,273,35,1,1,155,6293,105183,876525,
%T 4032015,10855425,17678835,17678835,10855425,4032015,876525,105183,
%U 6293,155,1,1,651,119133,9706503,430321633,11618684091,205263418941,2492484372855,21552658988805,136248095712855
%N Triangle T read by rows: n-th row (n>=0) gives the non-vanishing coefficients of the polynomial q(n,x) = 2^(-n)*((x+1)^(2^n) - (x-1)^(2^n))/2.
%C Row n has length 1 for n = 1 and 2^(n-1) = A000079(n-1) for n >= 1.
%C The triangle T gives the non-vanishing coefficients of the polynomial q(0,x) = 1 and q(n,x) = 2^(-n)*Sum_{k=0..2^(n-1)-1} A281122(n,k) * x^(2^n-1-2*k), n >= 1.
%C The polynomial q(n,x) = 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 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 + 1)), (A^4 + 28*A^3 + 70*A^2 + 28*A + 1)/(8*(A^3 + 7*A^2 + 7*A + 1), ... This is sqrt(A)*p(n,sqrt(A))/(2^n*q(n,sqrt(A))) with the given polynomials p(n,x) and q(n,x).
%H Indranil Ghosh, <a href="/A281123/b281123.txt">Rows 0..10, flattened</a>
%F T(n, k) = 1 for n=0, k=0, and T(n, k) = 2^(-n) * binomial(2^n,2*k+1) = A103328(2^(n-1),k) for k = 0..2^(n-1)-1 and n >= 1. - _Wolfdieter Lang_, Jan 20 2017
%e The triangle T(n, k) starts with
%e 1
%e 1
%e 1, 1
%e 1, 7, 7, 1
%e 1, 35, 273, 715, 715, 273, 35, 1
%e 1, 155, 6293, 105183, 876525, 4032015, 10855425, 17678835, 17678835, 10855425, 4032015, 876525, 105183, 6293, 155, 1
%e etc., since the first few polynomials are
%e q(0,x) = 1,
%e q(1,x) = x,
%e q(2,x) = x^3 + x = x*(x^2 + 1),
%e q(3,x) = x^7 + 7*x^5 + 7*x^3 + x = x*(x^2 + 1)*(x^4 + 6*x^2 + 1),
%e q(4,x) = x^15 + 35*x^13 + 273*x^11 + 715*x^9 + 715*x^7 + 273*x^5 + 35*x^3 + x = x*(x^2 + 1)*(x^4 + 6*x^2 + 1)*(x^8 + 28*x^6 + 70*x^4 + 28*x^2 + 1),
%e etc.
%t t={1};T[n_,k_]:=Table[2^(-n)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 Cf. A000079, A091043, A201461, A281122.
%K nonn,tabf,easy
%O 0,6
%A _Martin Renner_, Jan 15 2017
%E Edited. - _Wolfdieter Lang_, Jan 20 2017
%E More terms from _Indranil Ghosh_, Feb 22 2017
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