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 A045529 a(n+1) = 5*a(n)^3 - 3*a(n), a(0) = 1. 31
 1, 2, 34, 196418, 37889062373143906, 271964099255182923543922814194423915162591622175362 (list; graph; refs; listen; history; text; internal format)
 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: A155205 A303444 A230244 * A293245 A077747 A041012 Adjacent sequences: A045526 A045527 A045528 * A045530 A045531 A045532 KEYWORD nonn,easy AUTHOR STATUS approved

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Last modified March 20 21:07 EDT 2023. Contains 361391 sequences. (Running on oeis4.)