A002000 a(n+1) = a(n)*(a(n)^2 - 3) with a(0) = 7. (Formerly M4463 N1892)

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

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.

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

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

Sequence in context: A220278 A163437 A109059 * A092588 A049686 A177019

Adjacent sequences: A001997 A001998 A001999 * A002001 A002002 A002003

Keyword

nonn,easy

Author

N. J. A. Sloane

Status

approved