A376870 Reduced numerators of Newton's iteration for 1/sqrt(3), starting with 1/3.

1, 4, 130, 2739685, 21055737501685791580, 9337539302589041654242365815942422114384262970589593842110

Offset: 0

Author

Steven Finch, Oct 07 2024

Keywords

• nonn
• frac

Comments

An explicit formula for a(n) is not known, although it arises from a recurrence and the corresponding denominators are simply 3^((3^n + 1)/2) = 3*A134799(n).

Next term is too large to include.

Examples

			a(1) = 4 because b(1) = (3/2)*(1/3)*(1 - 1/9) = 4/9.
1/3, 4/9, 130/243, 2739685/4782969, ... = A376870(n)/(3*A134799(n)).
		

Links

• Paolo Xausa, Table of n, a(n) for n = 0..7
• X. Gourdon and P. Sebah, Pythagoras' Constant.

Crossrefs

Cf. A020760, A134799, A376867.

Programs

Mathematica

Module[{n = 0}, NestList[#*(3^(3^n++ + 1) - #^2)/2 &, 1, 6]] (* Paolo Xausa, Oct 17 2024 *)

Python

from itertools import count, islice
def A376870_gen(): # generator of terms
    p = 1
    for k in count(0):
        yield p
        p = p*(3**(3**k+1)-p**2)>>1
A376870_list = list(islice(A376870_gen(),6)) # Chai Wah Wu, Oct 11 2024

Formula

a(n) is the reduced numerator of b(n) = (3/2)*b(n-1)*(1 - b(n-1)^2); b(0) = 1/3.

Limit_{n -> oo} a(n)/(3*A134799(n)) = 1/sqrt(3) = A020760.

a(n+1) = a(n)*(3^(3^n+1)-a(n)^2)/2. - Chai Wah Wu, Oct 11 2024