%I #51 May 05 2024 19:19:25
%S 120,200,220,228,260,368,392,405,414,434,472,492,512,536,584,585,595,
%T 610,615,618,645,654,693,741,762,777,830,867,894,904,931,942,957,962,
%U 978,1045,1066,1070,1074,1102,1106,1108,1147,1194,1209,1266,1268,1309,1310,1317
%N Indices of Fibonacci numbers with 13 distinct prime factors.
%C From _Robert Israel_, Aug 18 2015: (Start)
%C Numbers n such that A022307(n) = 13.
%C If n is in the sequence, then k*n is not in the sequence for k > 1.
%C This is because A000045(n) divides A000045(k*n) while Carmichael's theorem says A000045(k*n) has at least one primitive prime factor. (End)
%H Amiram Eldar, <a href="/A114823/b114823.txt">Table of n, a(n) for n = 1..54</a>
%H Blair Kelly, <a href="http://mersennus.net/fibonacci/">Fibonacci and Lucas Factorizations</a>.
%e a(1)=120 because the 120th Fibonacci number consists of 13 distinct prime factors (i.e., 5358359254990966640871840 = 2^5 * 3^2 * 5 * 7 * 11 * 23 * 31 * 41 * 61 * 241 * 2161 * 2521 * 20641).
%p select(t -> nops(numtheory:-factorset(combinat:-fibonacci(t)))=13, [$1..1000]); # _Robert Israel_, Aug 10 2015
%t Select[Range[1250], PrimeNu[Fibonacci[#]]==13&] (* _Harvey P. Dale_, Apr 30 2015 *)
%o (PARI) n=1;while(n<265,if(omega(fibonacci(n))==13,print1(n,", "));n++)
%o (SageMath)
%o for n in range(1,3*10^2):
%o if len(prime_factors(fibonacci(n)))==13:
%o print(n) # _Manfred Scheucher_, Aug 04 2015
%o (Magma) [n: n in [1..3*10^2] |(#(PrimeDivisors(Fibonacci(n)))) eq 13]; // _Vincenzo Librandi_, Aug 05 2015
%Y Cf. A000045, A001221, A022307.
%Y Column k=13 of A303217.
%K hard,nonn
%O 1,1
%A _Shyam Sunder Gupta_, Feb 19 2006
%E More terms from _Ryan Propper_, Apr 26 2006
%E a(36)-a(45) from _Max Alekseyev_, Aug 18 2013
%E a(46)-a(50) from _Amiram Eldar_, Oct 14 2019