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Triangle read by rows: n-th row (n>=0) gives coefficients of the polynomial ((x+1)^(2^n) + (x-1)^(2^n))/2.
5

%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