OFFSET
0,2
COMMENTS
a(6) has 209 digits and is too large to include.
Except for the first term, this is a subsequence of A175180.
The squares larger than 1 are in A076445.
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
LINKS
Wikipedia, Halley's method.
FORMULA
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
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
MATHEMATICA
RecurrenceTable[{a[0] == 1, a[n] == 2*a[n - 1]^3 + 3*a[n - 1]}, a[n], {n, 5}]
NestList[2#^3+3#&, 1, 5] (* Harvey P. Dale, Mar 22 2023 *)
PROG
(PARI) a=1; print1(a, ", "); for(n=1, 5, b=2*a^3+3*a; print1(b, ", "); a=b);
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Arkadiusz Wesolowski, Mar 05 2014
STATUS
approved