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%I #47 Jan 25 2022 17:10:27
%S 0,1,8,144
%N Fibonacci numbers which are perfect powers.
%C Also, Fibonacci numbers which are products of Fibonacci numbers (each greater than 1 when the product is greater than 1 - see A235383). - _Rick L. Shepherd_, Feb 19 2014
%C The terms of the subsequence (1, 8, 144) are the Fibonacci numbers that are powerful numbers. - _Robert C. Lyons_, Jul 12 2016
%C Also Fibonacci numbers without any primitive divisors. See [Heuberger & Wagner]. - _Michel Marcus_, Aug 21 2016
%C It was proved (Bugeaud, Mignotte, and Siksek, 2006, p. 971) that the only perfect powers among the Fibonacci numbers and Lucas numbers are {0, 1, 8, 144} and {1, 4}, respectively. - _Daniel Forgues_, Apr 09 2018
%H Vladica Andrejic, <a href="http://www.doiserbia.nb.rs/img/doi/0353-8893/2006/0353-88930617038A.pdf">On Fibonacci Powers</a>, Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. 17 (2006), 38-44.
%H Yann Bugeaud, Florian Luca, Maurice Mignotte, and Samir Siksek, <a href="http://www-irma.u-strasbg.fr/~bugeaud/travaux/OmegaDef.pdf">On Fibonacci numbers with few prime divisors</a>, Proc. Japan Acad., 81, Ser. A (2005), pp. 17-20.
%H Yann Bugeaud, Maurice Mignotte, and Samir Siksek, <a href="http://dx.doi.org/10.4007/annals.2006.163.969">Classical and modular approaches to exponential Diophantine equations I. Fibonacci and Lucas perfect powers</a>, Annals of Mathematics, 163 (2006), pp. 969-1018.
%H Clemens Heuberger, Stephan Wagner, <a href="https://arxiv.org/abs/1606.02639">On the monoid generated by a Lucas sequence</a>, arXiv:1606.02639 [math.NT], 2016. Gives the complement sequence w.r.t Fibonacci numbers.
%H J. Mc Laughlin, <a href="http://arxiv.org/abs/math/0110150">Small prime powers in the Fibonacci sequence</a>, arXiv:math/0110150 [math.NT] (2001).
%H Attila Pethõ, <a href="http://www.emis.de/journals/AMAPN/vol17_2/6.html">Diophantine properties of linear recursive sequences II</a>, Acta Mathematica Academiae Paedagogicae Nyíregyháziensis 17:2 (2001), pp. 81-96.
%t perfectPowerQ[0] = True; perfectPowerQ[1] = True; perfectPowerQ[n_] := GCD @@ FactorInteger[n][[All, 2]] > 1; Union[Select[Fibonacci /@ Range[0, 20], perfectPowerQ]]
%Y Cf. A000045, A001597, A072381, A114842.
%K nonn,bref,fini,full
%O 1,3
%A _Jean-François Alcover_, Oct 25 2013
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