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%I #29 Nov 16 2017 15:52:24
%S 1,2,3,6,24,48
%N Numbers n >= 1 such that A000045(n)/A000005(A000045(n)) is an integer.
%C No further term between 48 and 320. - _R. J. Mathar_, Apr 01 2011
%C This sequence is complete. For proof see the Luca-Young paper in links section, pages 7-10. - _Altug Alkan_, Apr 26 2016
%H Florian Luca and Paul Thomas Young, <a href="https://web.math.pmf.unizg.hr/glasnik/vol_47/no2_05.html">On the number of divisors of n! and of the Fibonacci numbers</a>, Glasnik Matematicki, Vol. 47, No. 2 (2012), 285-293. DOI: 10.3336/gm.47.2.05.
%F {n: A063375(n) | A000045(n)} . - _R. J. Mathar_, Apr 01 2011
%p with(combinat):with(numtheory): A160683 := proc(n) option remember: local k: if(n=1)then return 1:fi: for k from procname(n-1)+1 do if(fibonacci(k) mod tau(fibonacci(k))=0)then return k:fi: od: end: seq(A160683(n), n=1..6); # _Nathaniel Johnston_, May 09 2011
%t Select[Range@ 120, IntegerQ[#/DivisorSigma[0, #]] &@ Fibonacci@ # &]
%o (PARI) isok(n) = my(f=fibonacci(n)); f % numdiv(f) == 0; \\ _Michel Marcus_, Jul 31 2015
%Y Cf. A000005, A000045, A063375, A160686.
%K nonn,fini,full
%O 1,2
%A _Ctibor O. Zizka_, May 23 2009
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