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A290498 Numbers m such that the set of distinct prime divisors of the number of divisors of Fibonacci(m) is equal to the set of distinct prime divisors of m.

%I A290498 #44 Mar 01 2022 07:27:07
%S A290498 1,4,8,16,24,32,60,64,72,96,128,192,256,300,336,512,576,648,900,1008,
%T A290498 1024,1080,1250
%N A290498 Numbers m such that the set of distinct prime divisors of the number of divisors of Fibonacci(m) is equal to the set of distinct prime divisors of m.
%C A290498 Thanks to squarefree terms of A058635, numbers of the form 2^k appear in this sequence for k > 1. However it is not proven yet whether it is always true.
%C A290498 From _Jon E. Schoenfield_, Aug 05 2017: (Start)
%C A290498 The difficulty in extending this sequence is that it becomes hard to obtain the complete prime factorization of Fibonacci(m) as m increases. However, since every number having an odd number of divisors is a square, and the largest Fibonacci number that is also a square is Fibonacci(12) = 144, we can confine the search for terms > 12 to even numbers only.
%C A290498 Even for values of m for which we are unable to completely factorize Fibonacci(m), we can determine with a high degree of confidence whether m is in the sequence by considering only the multiplicities of the smaller primes in those factorizations, because multiplicities greater than 1 in the prime factorizations of Fibonacci numbers rarely occur among the larger prime factors. If, in place of the actual complete factorization of Fibonacci(m) for each examined value of m, we were to use only the multiplicities of the prime factors of Fibonacci(m) that are less than 10000 (which are quickly and easily counted using trial division), the terms we would obtain for this sequence would begin with 1, 4, 8, 16, 24, 32, 60, 64, 72, 96, 128, 192, 256, 300, 336, 512, 576, 648, 900, 1008, 1024, 1080, 1250, 1500, 1536, 1620, 1920, 2048, 2352, 2500, 2592, 2700, 4096, 4608, 5000, 5184, 5400, 5832, 7500, 8100, 8192, 8448, 8640, 9072, 9600, 10000, 13608, 15000, ...
%C A290498 Perhaps surprisingly, we would get the same terms (up through at least a(141) = 960000) if, instead of the multiplicities of prime factors <= 10000, we were to use the multiplicities of just the prime factors <= 13. (End)
%H A290498 Blair Kelly, <a href="http://mersennus.net/fibonacci/">Fibonacci and Lucas Factorizations</a>
%e A290498 72 is a term because d(Fibonacci(2^3*3^2)) = 2^9*3.
%e A290498 300 is a term because d(Fibonacci(2^2*3*5^2)) = 2^15*3^2*5.
%t A290498 Select[Range[12]~Join~Range[14, 300, 2], Apply[SameQ, Map[FactorInteger[#][[All, 1]] &, {#, DivisorSigma[0, Fibonacci@ #]}]] &] (* _Michael De Vlieger_, Aug 07 2017 *)
%o A290498 (PARI) is(n) = factor(numdiv(fibonacci(n)))[,1]==factor(n)[,1] \\ _David A. Corneth_, Aug 04 2017
%Y A290498 Cf. A000005, A000045, A022307, A038575, A058635, A063375, A081381, A086597, A135939.
%K A290498 nonn,more
%O A290498 1,2
%A A290498 _Altug Alkan_, Aug 04 2017
%E A290498 a(16)-a(19) from _David A. Corneth_, Aug 04 2017
%E A290498 a(20)-a(22) from _Jon E. Schoenfield_, Aug 05 2017
%E A290498 a(23) from _Amiram Eldar_, Oct 14 2019