This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A238799 #41 Dec 29 2024 14:08:08
%S A238799 1,5,265,37220045,103124220135120334842385,
%T A238799 2193370648451279691104497113491599222165108730278225579497595691360405
%N A238799 a(0) = 1, a(n+1) = 2*a(n)^3 + 3*a(n).
%C A238799 a(6) has 209 digits and is too large to include.
%C A238799 Except for the first term, this is a subsequence of A175180.
%C A238799 The squares larger than 1 are in A076445.
%C A238799 If we define u(0) = 1 , u(n+1) = (u(n)/3)*(u(n)^2+9) / (u(n)^2 + 1), then u(n) = a(n) / A378683(n) ; this is Halley's method to calculate sqrt(3). - _Robert FERREOL_, Dec 21 2024
%H A238799 Wikipedia, <a href="https://en.wikipedia.org/wiki/Halley's_method">Halley's method</a>.
%F A238799 a(n) = sqrt(2) * sinh( 3^n * arcsinh(1/sqrt(2)) ) = (1+sqrt(3))/2 * (2+sqrt(3))^((3^n-1)/2) + (1-sqrt(3))/2 * (2-sqrt(3))^((3^n-1)/2). - _Max Alekseyev_, Sep 04 2018
%F A238799 a(n) = ((1 + sqrt(3))^(3^n) + (1 - sqrt(3))^(3^n))/2^((3^n+1)/2) = A002531(3^n) = A080040(3^n)/2^((3^n+1)/2). - _Robert FERREOL_, Nov 19 2024
%t A238799 RecurrenceTable[{a[0] == 1, a[n] == 2*a[n - 1]^3 + 3*a[n - 1]}, a[n], {n, 5}]
%t A238799 NestList[2#^3+3#&,1,5] (* _Harvey P. Dale_, Mar 22 2023 *)
%o A238799 (PARI) a=1; print1(a, ", "); for(n=1, 5, b=2*a^3+3*a; print1(b, ", "); a=b);
%o A238799 (PARI) { A238799(n) = my(q=Mod(x,x^2-3)); lift( (1+q)*(2+q)^((3^n-1)/2) + (1-q)*(2-q)^((3^n-1)/2) )/2; } \\ _Max Alekseyev_, Sep 04 2018
%Y A238799 Cf. A175180, A378683.
%K A238799 nonn,easy
%O A238799 0,2
%A A238799 _Arkadiusz Wesolowski_, Mar 05 2014