A002000 a(n+1) = a(n)*(a(n)^2 - 3) with a(0) = 7.

7, 322, 33385282, 37210469265847998489922, 51522323599677629496737990329528638956583548304378053615581043535682

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

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

• Vincenzo Librandi, Table of n, a(n) for n = 0..6
• E. B. Escott, Rapid method for extracting a square root, Amer. Math. Monthly, 44 (1937), 644-646.

Crossrefs

Cf. A000032, A001999, A006267, A219161, A271223.

Programs

Magma

[n eq 1 select 7 else Self(n-1)^3 - 3*Self(n-1): n in [1..6]]; // Vincenzo Librandi, Feb 09 2017

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 *)

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)