%I #59 Jan 27 2022 21:15:57
%S 1,1,1,1,6,1,1,28,70,28,1,1,120,1820,8008,12870,8008,1820,120,1,1,496,
%T 35960,906192,10518300,64512240,225792840,471435600,601080390,
%U 471435600,225792840,64512240,10518300,906192,35960,496,1
%N Triangle read by rows: n-th row (n>=0) gives coefficients of the polynomial ((x+1)^(2^n) + (x-1)^(2^n))/2.
%C Wanted: reference for the fact that these polynomials are irreducible. Washington, Cyclotomic Fields, perhaps?
%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+1)*(A^2 + 6*A + 1)), ... This is p(n,sqrt(A))/(2^n*Product_{k=1..n-1} p(k,sqrt(A))) with the given polynomial p(n,x) = ((x+1)^(2^n) + (x-1)^(2^n))/2. - _Martin Renner_, Jan 11 2017
%C The quadratic coefficient of this polynomial is A006516(n), the even-indexed coefficients are binomial(2^n,2*k) or A086645(2^(n-1),k) for 0 <= k <= 2^(n-1), in each row the maximum central coefficient for n>=2 is A037293(n) or A000984(2^(n-1)). - _Martin Renner_, Jan 14 2017
%C T(n,k) and A281122 are a bisection of row 2^n of Pascal's triangle A007318. - _Martin Renner_, Jan 15 2017
%C For nonnegative real x, sqrt(x) = (2*x/(1 + x)) * (2*(1 + x)^2/(1 + 6*x + x^2)) * (2*(1 + 6*x + x^2)^2/(1 + 28*x + 70*x^2 + 28*x^3 + x^4)) * .... See Bauer. - _Peter Bala_, Jan 18 2022
%H Indranil Ghosh, <a href="/A201461/b201461.txt">Rows 0..11, flattened</a>
%H F. L. Bauer, <a href="http://www.ega-math.narod.ru/Nquant/Bauer.htm">Letters to the editor: An Infinite Product for Square-Rooting with Cubic Convergence</a>, The Mathematical Intelligencer, Vol. 20, Issue 1, (1998), 12-14.
%F T(n,k) = binomial(2^n,2*k). - _Joerg Arndt_, Jan 15 2017
%e The first few polynomials are:
%e 1,
%e x^2 + 1,
%e x^4 + 6*x^2 + 1,
%e x^8 + 28*x^6 + 70*x^4 + 28*x^2 + 1,
%e x^16 + 120*x^14 + 1820*x^12 + 8008*x^10 + 12870*x^8 + 8008*x^6 + 1820*x^4 + 120*x^2 + 1.
%e The triangle of coefficients begins:
%e [0] [1]
%e [1] [1, 0, 1]
%e [2] [1, 0, 6, 0, 1]
%e [3] [1, 0, 28, 0, 70, 0, 28, 0, 1]
%e [4] [1, 0, 120, 0, 1820, 0, 8008, 0, 12870, 0, 8008, 0, 1820, 0, 120, 0, 1]
%e The triangle of nonzero coefficients begins:
%e [0] 1
%e [1] 1, 1
%e [2] 1, 6, 1
%e [3] 1, 28, 70, 28, 1
%e [4] 1, 120, 1820, 8008, 12870, 8008, 1820, 120, 1
%e [5] 1, 496, 35960, 906192, 10518300, 64512240, 225792840, 471435600, 601080390, 471435600, 225792840, 64512240, 10518300, 906192, 35960, 496, 1
%e ...
%t Flatten[Table[Binomial[2^n,2k],{n,0,6},{k,0,2^(n-1)}]] (* _Indranil Ghosh_, Feb 22 2017 *)
%o (PARI) row(n) = my(v = Vec(((x+1)^(2^n)+(x-1)^(2^n))/2)); vector(#v\2 + 1, k, v[2*k-1]); \\ _Michel Marcus_, Jan 14 2017
%o (PARI) T(n,k)=binomial(2^n,2*k);
%o for(n=0,5,for(k=0,2^(n-1),print1(T(n,k),", "));print()); \\ _Joerg Arndt_, Jan 15 2017
%o (SageMath)
%o def A201461_polynomial(n): return expand(((x+1)^(2^n) + (x-1)^(2^n))/2)
%o for n in range(6): print(A201461_polynomial(n))
%o for n in range(6): print(A201461_polynomial(n).list()) # coefficients
%o for n in range(6): # depunched (not a mathematical operation)
%o if n == 0: print([1])
%o else: print(A201461_polynomial(n).list()[::2]) # _Peter Luschny_, Jan 11 2021
%Y Cf. A000984, A006516, A007318, A037293, A086645, A281122.
%K nonn,easy,tabf
%O 0,5
%A _N. J. A. Sloane_, Dec 01 2011