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

%I A376870 #28 Oct 17 2024 14:46:06
%S A376870 1,4,130,2739685,21055737501685791580,
%T A376870 9337539302589041654242365815942422114384262970589593842110
%N A376870 Reduced numerators of Newton's iteration for 1/sqrt(3), starting with 1/3.
%C A376870 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).
%C A376870 Next term is too large to include.
%H A376870 Paolo Xausa, <a href="/A376870/b376870.txt">Table of n, a(n) for n = 0..7</a>
%H A376870 X. Gourdon and P. Sebah, <a href="http://numbers.computation.free.fr/Constants/Sqrt2/sqrt2.html">Pythagoras' Constant</a>.
%F A376870 a(n) is the reduced numerator of b(n) = (3/2)*b(n-1)*(1 - b(n-1)^2); b(0) = 1/3.
%F A376870 Limit_{n -> oo} a(n)/(3*A134799(n)) = 1/sqrt(3) = A020760.
%F A376870 a(n+1) = a(n)*(3^(3^n+1)-a(n)^2)/2. - _Chai Wah Wu_, Oct 11 2024
%e A376870 a(1) = 4 because b(1) = (3/2)*(1/3)*(1 - 1/9) = 4/9.
%e A376870 1/3, 4/9, 130/243, 2739685/4782969, ... = A376870(n)/(3*A134799(n)).
%t A376870 Module[{n = 0}, NestList[#*(3^(3^n++ + 1) - #^2)/2 &, 1, 6]] (* _Paolo Xausa_, Oct 17 2024 *)
%o A376870 (Python)
%o A376870 from itertools import count, islice
%o A376870 def A376870_gen(): # generator of terms
%o A376870     p = 1
%o A376870     for k in count(0):
%o A376870         yield p
%o A376870         p = p*(3**(3**k+1)-p**2)>>1
%o A376870 A376870_list = list(islice(A376870_gen(),6)) # _Chai Wah Wu_, Oct 11 2024
%Y A376870 Cf. A020760, A134799, A376867.
%K A376870 nonn,frac
%O A376870 0,2
%A A376870 _Steven Finch_, Oct 07 2024