A045529 a(n+1) = 5*a(n)^3 - 3*a(n), a(0) = 1.

1, 2, 34, 196418, 37889062373143906, 271964099255182923543922814194423915162591622175362

Offset

0,2

Comments

The next term, a(6), has 153 digits. - Harvey P. Dale, Oct 24 2011

Links

Seiichi Manyama, Table of n, a(n) for n = 0..7

Daniel Duverney and Takeshi Kurosawa, Transcendence of infinite products involving Fibonacci and Lucas numbers, Research in Number Theory, Vol. 8 (2002), Article 68.

Zalman Usiskin, Problem B-265, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 11, No. 3 (1973), p. 333; Fibonacci Numbers for Powers of 3, Solution to Problem B-265 by Ralph Garfield and David Zeitlin, ibid., Vol. 12, No. 3 (1974), p. 315.

Formula

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))).

a(n) = Fibonacci(3^n). - Leroy Quet, Mar 17 2002

a(n+1) = a(n)*A002814(n+1). - Lekraj Beedassy, Jun 16 2003

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

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

From Peter Bala, Nov 24 2022: (Start)

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));

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

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

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

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)

From Amiram Eldar, Jan 07 2023: (Start)

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

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

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

Maple

a := proc(n) option remember; if n = 0 then 1 else 5*a(n-1)^3 - 3*a(n-1) end if; end:
seq(a(n), n = 0..5); # Peter Bala, Nov 24 2022

Mathematica

G = (1 + Sqrt[5])/2; Table[Expand[(G^(3^n) - (1 - G)^(3^n))/Sqrt[5]], {n, 1, 7}] (* Artur Jasinski, Oct 05 2008 *)
Table[Round[(4/5)^(1/2)*Cosh[3^n*ArcCosh[((5/4)^(1/2))]]], {n, 1, 4}] (* Artur Jasinski, Oct 05 2008 *)
RecurrenceTable[{a[0]==1, a[n]==5a[n-1]^3-3a[n-1]}, a[n], {n, 6}] (* Harvey P. Dale, Oct 24 2011 *)
NestList[5#^3-3#&, 1, 5] (* Harvey P. Dale, Dec 21 2014 *)

Prog

(Maxima) A045529(n):=fib(3^n)$
makelist(A045529(n), n, 0, 10); /* Martin Ettl, Nov 12 2012 */

Crossrefs

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).

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

Sequence in context: A303444 A230244 A365881 * A293245 A077747 A041012

Adjacent sequences: A045526 A045527 A045528 * A045530 A045531 A045532

Keyword

nonn,easy

Author

Jose Eduardo Blazek

Status

approved