%I #59 Jan 07 2023 10:15:22

%S 1,2,34,196418,37889062373143906,

%T 271964099255182923543922814194423915162591622175362

%N a(n+1) = 5*a(n)^3 - 3*a(n), a(0) = 1.

%C The next term, a(6), has 153 digits. - _Harvey P. Dale_, Oct 24 2011

%H Seiichi Manyama, <a href="/A045529/b045529.txt">Table of n, a(n) for n = 0..7</a>

%H Daniel Duverney and Takeshi Kurosawa, <a href="https://doi.org/10.1007/s40993-022-00366-1">Transcendence of infinite products involving Fibonacci and Lucas numbers</a>, Research in Number Theory, Vol. 8 (2002), Article 68.

%H Zalman Usiskin, <a href="https://fq.math.ca/Scanned/11-3/elementary11-3.pdf">Problem B-265</a>, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 11, No. 3 (1973), p. 333; <a href="https://www.fq.math.ca/Scanned/12-3/elementary12-3.pdf">Fibonacci Numbers for Powers of 3</a>, Solution to Problem B-265 by Ralph Garfield and David Zeitlin, ibid., Vol. 12, No. 3 (1974), p. 315.

%F The first example I know in which a(n) can be expressed as (4/5)^(1/2)*cosh(3^n*arccosh((5/4)^(1/2))).

%F a(n) = Fibonacci(3^n). - _Leroy Quet_, Mar 17 2002

%F a(n+1) = a(n)*A002814(n+1). - _Lekraj Beedassy_, Jun 16 2003

%F a(n) = (phi^(3^n) - (1 - phi)^(3^n))/sqrt(5), where phi is the golden ratio (A001622). - _Artur Jasinski_, Oct 05 2008

%F a(n) = Product_{k=0..n-1} (Lucas(2*3^k) - 1) (Usiskin, 1973). - _Amiram Eldar_, Jan 29 2022

%F From _Peter Bala_, Nov 24 2022: (Start)

%F a(2*n+2) == a(2*n) (mod 3^(2*n+1)); a(2*n+3) == a(2*n+1) (mod 3^(2*n+2));

%F a(2*n+1) + a(2*n) == 0 (mod 3^(2*n+1)).

%F a(2*n) == 1 (mod 3) and a(2*n+1) == 2 (mod 3).

%F 5*a(n)^2 == 2 (mod 3^(n+1)).

%F In the ring of 3-adic integers, the sequences {a(2*n)} and {a(2*n+1)} are both Cauchy sequences and converge to the pair of 3-adic roots of the quadratic equation 5*x^2 - 2 = 0. (End)

%F From _Amiram Eldar_, Jan 07 2023: (Start)

%F Product_{n>=1} (1 + 2/(sqrt(5)*a(n)-1)) = phi (A001622).

%F Product_{n>=1} (1 - 2/(sqrt(5)*a(n)+1)) = 1/phi (A094214).

%F Both formulas are from Duverney and Kurosawa (2022). (End)

%p a := proc(n) option remember; if n = 0 then 1 else 5*a(n-1)^3 - 3*a(n-1) end if; end:

%p seq(a(n), n = 0..5); # _Peter Bala_, Nov 24 2022

%t G = (1 + Sqrt[5])/2; Table[Expand[(G^(3^n) - (1 - G)^(3^n))/Sqrt[5]], {n, 1, 7}] (* _Artur Jasinski_, Oct 05 2008 *)

%t Table[Round[(4/5)^(1/2)*Cosh[3^n*ArcCosh[((5/4)^(1/2))]]], {n, 1, 4}] (* _Artur Jasinski_, Oct 05 2008 *)

%t RecurrenceTable[{a[0]==1,a[n]==5a[n-1]^3-3a[n-1]},a[n],{n,6}] (* _Harvey P. Dale_, Oct 24 2011 *)

%t NestList[5#^3-3#&,1,5] (* _Harvey P. Dale_, Dec 21 2014 *)

%o (Maxima) A045529(n):=fib(3^n)$

%o makelist(A045529(n),n,0,10); /* _Martin Ettl_, Nov 12 2012 */

%Y Cf. (k^n)-th Fibonacci number: A058635 (k=2), this sequence (k=3), A145231 (k=4), A145232 (k=5), A145233 (k=6), A145234 (k=7), A250487 (k=8), A250488 (k=9), A250489 (k=10).

%Y Cf. A000032, A000045, A001622, A001999, A002814, A006267, A094214.

%K nonn,easy

%O 0,2

%A _Jose Eduardo Blazek_