Svoboda | Graniru | BBC Russia | Golosameriki | Facebook
login
A002000
a(n+1) = a(n)*(a(n)^2 - 3) with a(0) = 7.
(Formerly M4463 N1892)
2
7, 322, 33385282, 37210469265847998489922, 51522323599677629496737990329528638956583548304378053615581043535682
OFFSET
0,1
REFERENCES
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
E. B. Escott, Rapid method for extracting a square root, Amer. Math. Monthly, 44 (1937), 644-646.
FORMULA
From Peter Bala, Feb 01 2017: (Start)
a(n) = ((7 + sqrt(45))/2)^(3^n) + ((7 - sqrt(45))/2)^(3^n).
a(n) = 2*T(3^n,7/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
Product_{n >= 0} (1 + 2/(a(n) - 1)) = 3*sqrt(5)/5.
Cf. A001999 and A219161. (End)
From Peter Bala, Nov 15 2022: (Start)
a(n) = Lucas(4*(3^n)).
a(n+1) == a(n) (mod 3^(n+1)) (a particular case of the Gauss congruences for the Lucas numbers).
Conjecture: a(n+1) == a(n) (mod 3^(n+r+2)) for n >= r.
The least positive residue of a(n) mod(3^n) = 3^n - 2 = A058481(n). In the ring of 3-adic integers the limit_{n -> oo} a(n) exists and is equal to -2.
Product_{k = 0..n} (a(k) - 1) = (1/3)*Lucas(6*(3^n)). (End)
MAPLE
a := proc(n) option remember; if n = 0 then 7 else a(n-1)^3 - 3*a(n-1) end if; end;
seq(a(n), n = 0..4); # Peter Bala, Nov 15 2022
MATHEMATICA
RecurrenceTable[{a[0] == 7, a[n] == a[n - 1]^3 - 3 a[n - 1]}, a, {n, 0, 8}]
(* Vincenzo Librandi, Feb 09 2017 *)
NestList[#(#^2-3)&, 7, 4] (* Harvey P. Dale, Aug 11 2021 *)
PROG
(Magma) [n eq 1 select 7 else Self(n-1)^3 - 3*Self(n-1): n in [1..6]]; // Vincenzo Librandi, Feb 09 2017
CROSSREFS
KEYWORD
nonn,easy
STATUS
approved