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A114841 Indices of Fibonacci numbers with 3 distinct prime factors.

%I A114841 #44 Jan 05 2025 19:51:38
%S A114841 15,16,18,21,33,35,37,38,39,46,49,51,58,62,65,67,82,86,103,106,119,
%T A114841 122,125,139,142,145,158,166,179,181,226,233,235,241,257,263,274,281,
%U A114841 299,301,317,337,383,389,419,457,463,473,479,491,521,541,557,619,643,659,706,719,739,751,857,863,877,881,883,911,947,983,1021,1033,1061,1069,1109,1117,1123,1181,1187,1193,1213,1226
%N A114841 Indices of Fibonacci numbers with 3 distinct prime factors.
%H A114841 Amiram Eldar, <a href="/A114841/b114841.txt">Table of n, a(n) for n = 1..83</a>
%H A114841 Blair Kelly, <a href="http://mersennus.net/fibonacci/">Fibonacci and Lucas Factorizations</a>.
%H A114841 Prapanpong Pongsriiam, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Papers1/57-5/pongsriiam.pdf">Fibonacci and Lucas Numbers which have Exactly Three Prime Factors and Some Unique Properties of F18 and L18</a>, Fibonacci Quart. 57 (2019), no. 5, 130-144.
%e A114841 a(1) = 15 because 15th Fibonacci number has 3 distinct prime factors (i.e., 610 = 2 * 5 * 61).
%p A114841 with(numtheory): with(combinat):
%p A114841 a:=n->`if`(nops(factorset(fibonacci(n)))=3,n,NULL); [seq(a(n),n=1..300)]; # _Muniru A Asiru_, Mar 25 2018
%t A114841 Select[Range[500], PrimeNu[Fibonacci[#]]==3 &] (* _Vincenzo Librandi_, Mar 26 2018 *)
%o A114841 (PARI) n=1;while(n<340,if(omega(fibonacci(n))==3,print1(n,", "));n++)
%o A114841 (Magma) [n: n in [1..350] |(#(PrimeDivisors(Fibonacci(n)))) eq 3]; // _Vincenzo Librandi_, Mar 26 2018
%Y A114841 Cf. A114823, A114824, A114825, A114826, A114836, A114837, A114838, A114839, A114840.
%Y A114841 Column k=3 of A303217.
%K A114841 nonn
%O A114841 1,1
%A A114841 _Shyam Sunder Gupta_, Feb 19 2006
%E A114841 More terms from _Ryan Propper_, Apr 26 2006
%E A114841 a(57)-a(80) from _Max Alekseyev_, Aug 18 2013