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%I M4463 N1892 #38 Nov 19 2022 20:36:45
%S 7,322,33385282,37210469265847998489922,
%T 51522323599677629496737990329528638956583548304378053615581043535682
%N a(n+1) = a(n)*(a(n)^2 - 3) with a(0) = 7.
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H Vincenzo Librandi, <a href="/A002000/b002000.txt">Table of n, a(n) for n = 0..6</a>
%H E. B. Escott, <a href="http://www.jstor.org/stable/2301484">Rapid method for extracting a square root</a>, Amer. Math. Monthly, 44 (1937), 644-646.
%F From _Peter Bala_, Feb 01 2017: (Start)
%F a(n) = ((7 + sqrt(45))/2)^(3^n) + ((7 - sqrt(45))/2)^(3^n).
%F a(n) = 2*T(3^n,7/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
%F Product_{n >= 0} (1 + 2/(a(n) - 1)) = 3*sqrt(5)/5.
%F Cf. A001999 and A219161. (End)
%F From _Peter Bala_, Nov 15 2022: (Start)
%F a(n) = Lucas(4*(3^n)).
%F a(n+1) == a(n) (mod 3^(n+1)) (a particular case of the Gauss congruences for the Lucas numbers).
%F Conjecture: a(n+1) == a(n) (mod 3^(n+r+2)) for n >= r.
%F 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.
%F Product_{k = 0..n} (a(k) - 1) = (1/3)*Lucas(6*(3^n)). (End)
%p a := proc(n) option remember; if n = 0 then 7 else a(n-1)^3 - 3*a(n-1) end if; end;
%p seq(a(n), n = 0..4); # _Peter Bala_, Nov 15 2022
%t RecurrenceTable[{a[0] == 7, a[n] == a[n - 1]^3 - 3 a[n - 1]}, a, {n, 0, 8}]
%t (* _Vincenzo Librandi_, Feb 09 2017 *)
%t NestList[#(#^2-3)&,7,4] (* _Harvey P. Dale_, Aug 11 2021 *)
%o (Magma) [n eq 1 select 7 else Self(n-1)^3 - 3*Self(n-1): n in [1..6]]; // _Vincenzo Librandi_, Feb 09 2017
%Y Cf. A000032, A001999, A006267, A219161, A271223.
%K nonn,easy
%O 0,1
%A _N. J. A. Sloane_
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