A201461 Triangle read by rows: n-th row (n>=0) gives coefficients of the polynomial ((x+1)^(2^n) + (x-1)^(2^n))/2.

1, 1, 1, 1, 6, 1, 1, 28, 70, 28, 1, 1, 120, 1820, 8008, 12870, 8008, 1820, 120, 1, 1, 496, 35960, 906192, 10518300, 64512240, 225792840, 471435600, 601080390, 471435600, 225792840, 64512240, 10518300, 906192, 35960, 496, 1

Offset: 0

Author

N. J. A. Sloane, Dec 01 2011

Keywords

• nonn
• easy
• tabf

Comments

Wanted: reference for the fact that these polynomials are irreducible. Washington, Cyclotomic Fields, perhaps?

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

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

T(n,k) and A281122 are a bisection of row 2^n of Pascal's triangle A007318. - Martin Renner, Jan 15 2017

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

Examples

			The first few polynomials are:
1,
x^2 + 1,
x^4 + 6*x^2 + 1,
x^8 + 28*x^6 + 70*x^4 + 28*x^2 + 1,
x^16 + 120*x^14 + 1820*x^12 + 8008*x^10 + 12870*x^8 + 8008*x^6 + 1820*x^4 + 120*x^2 + 1.
The triangle of coefficients begins:
[0] [1]
[1] [1, 0, 1]
[2] [1, 0, 6, 0, 1]
[3] [1, 0, 28, 0, 70, 0, 28, 0, 1]
[4] [1, 0, 120, 0, 1820, 0, 8008, 0, 12870, 0, 8008, 0, 1820, 0, 120, 0, 1]
The triangle of nonzero coefficients begins:
[0] 1
[1] 1, 1
[2] 1, 6, 1
[3] 1, 28, 70, 28, 1
[4] 1, 120, 1820, 8008, 12870, 8008, 1820, 120, 1
[5] 1, 496, 35960, 906192, 10518300, 64512240, 225792840, 471435600, 601080390, 471435600, 225792840, 64512240, 10518300, 906192, 35960, 496, 1
...
		

Links

• Indranil Ghosh, Rows 0..11, flattened
• F. L. Bauer, Letters to the editor: An Infinite Product for Square-Rooting with Cubic Convergence, The Mathematical Intelligencer, Vol. 20, Issue 1, (1998), 12-14.

Crossrefs

Cf. A000984, A006516, A007318, A037293, A086645, A281122.

Programs

Mathematica

Flatten[Table[Binomial[2^n,2k],{n,0,6},{k,0,2^(n-1)}]] (* Indranil Ghosh, Feb 22 2017 *)

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

PARI

T(n,k)=binomial(2^n,2*k);
for(n=0,5,for(k=0,2^(n-1),print1(T(n,k),", "));print()); \\ Joerg Arndt, Jan 15 2017

SageMath

def A201461_polynomial(n): return expand(((x+1)^(2^n) + (x-1)^(2^n))/2)
for n in range(6): print(A201461_polynomial(n))
for n in range(6): print(A201461_polynomial(n).list()) # coefficients
for n in range(6): # depunched (not a mathematical operation)
    if n == 0: print([1])
    else: print(A201461_polynomial(n).list()[::2]) # Peter Luschny, Jan 11 2021

Formula

T(n,k) = binomial(2^n,2*k). - Joerg Arndt, Jan 15 2017