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A045529
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a(n+1) = 5*a(n)^3 - 3*a(n), a(0) = 1.
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31
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OFFSET
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0,2
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COMMENTS
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The next term, a(6), has 153 digits. - Harvey P. Dale, Oct 24 2011
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LINKS
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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.
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FORMULA
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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)
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MAPLE
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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
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MATHEMATICA
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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 *)
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PROG
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(Maxima) A045529(n):=fib(3^n)$
makelist(A045529(n), n, 0, 10); /* Martin Ettl, Nov 12 2012 */
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CROSSREFS
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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
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KEYWORD
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nonn,easy
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AUTHOR
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Jose Eduardo Blazek
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STATUS
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approved
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