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A376870
Reduced numerators of Newton's iteration for 1/sqrt(3), starting with 1/3.
3
1, 4, 130, 2739685, 21055737501685791580, 9337539302589041654242365815942422114384262970589593842110
OFFSET
0,2
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.
LINKS
X. Gourdon and P. Sebah, Pythagoras' Constant.
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
EXAMPLE
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)).
MATHEMATICA
Module[{n = 0}, NestList[#*(3^(3^n++ + 1) - #^2)/2 &, 1, 6]] (* Paolo Xausa, Oct 17 2024 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,frac,changed
AUTHOR
Steven Finch, Oct 07 2024
STATUS
approved