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A376867 Reduced numerators of Newton's iteration for 1/sqrt(2), starting with 1/2.

%I A376867 #47 Oct 24 2024 14:50:09
%S A376867 1,5,355,94852805,1709678476417571835487555,
%T A376867 9994796326591347130392203807311551183419838794447313956622219314498503205
%N A376867 Reduced numerators of Newton's iteration for 1/sqrt(2), starting with 1/2.
%C A376867 An explicit formula for a(n) is not known, although it arises from a recurrence and the corresponding denominators are simply 2^(3^n) = A023365(n+1).
%C A376867 Next term is too large to include.
%H A376867 Alois P. Heinz, <a href="/A376867/b376867.txt">Table of n, a(n) for n = 0..7</a>
%H A376867 X. Gourdon and P. Sebah, <a href="http://numbers.computation.free.fr/Constants/Sqrt2/sqrt2.html">Pythagoras' Constant</a>.
%F A376867 a(n) is the reduced numerator of b(n) = b(n-1)*(3/2 - b(n-1)^2); b(0) = 1/2.
%F A376867 Limit_{n -> oo} a(n)/A023365(n+1) = 1/sqrt(2) = A010503.
%F A376867 a(n+1) = a(n)*(3*2^(2*3^n-1)-a(n)^2). - _Chai Wah Wu_, Oct 11 2024
%e A376867 a(1) = 5 because b(1) = (1/2)*(3/2 - 1/4) = 5/8.
%e A376867 1/2, 5/8, 355/512, 94852805/134217728, ... = a(n)/A023365(n+1).
%p A376867 b:= proc(n) b(n):= `if`(n=0, 1/2, b(n-1)*(3/2-b(n-1)^2)) end:
%p A376867 a:= n-> numer(b(n)):
%p A376867 seq(a(n), n=0..5);  # _Alois P. Heinz_, Oct 07 2024
%t A376867 a[0]=1/2; a[n_]:=a[n-1](3/2-a[n-1]^2); Numerator[Array[a,6,0]] (* _Stefano Spezia_, Oct 15 2024 *)
%o A376867 (Python)
%o A376867 from itertools import count, islice
%o A376867 def A376867_gen(): # generator of terms
%o A376867     p = 1
%o A376867     for k in count(0):
%o A376867         yield p
%o A376867         p *= ((3<<((3**k<<1)-1))-p**2)
%o A376867 A376867_list = list(islice(A376867_gen(),6)) # _Chai Wah Wu_, Oct 11 2024
%Y A376867 Cf. A010503, A023365, A376870.
%K A376867 nonn,frac
%O A376867 0,2
%A A376867 _Steven Finch_, Oct 07 2024