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%I #23 Sep 08 2022 08:45:07
%S 1,2,4,8,16,64,256,1024,8192,131072,1048576
%N a(n) = tau(F(2^n)) where tau(x) is the number of divisors of x (A000005(x)) and F(k) the k-th Fibonacci number (A000045(k)).
%C Is there any pattern in this sequence? It seems also that tau(F(m^n)) is a power of 2 for any m, any n > 0.
%C F(2^n) = L(2)L(4)L(8)...L(2^(n-1)) where L(x) is the Lucas numbers. This greatly reduces the difficulty of factoring these numbers. To find a(9) one needs the factorization of F(512); this was done long ago: F(2^9) = 3 * 7 * 47 * 127 * 1087 * 2207 * 4481 * 34303 * 119809 * 73327699969 * 186812208641 * 4698167634523379875583 * 125960894984050328038716298487435392001; hence a(9) = 2^13 = 8192. Since L(512), L(1024) are completed factored the next few terms are also known. L(2048) has 1 known factor and a C411, thus the next term is at least 2^23. - _Sean A. Irvine_, Jun 02 2005
%C If no member of A037917 is a power of 2, then a(n) is a power of 2 for all n. - _Charles R Greathouse IV_, Apr 09 2012
%H Blair Kelly, <a href="http://mersennus.net/fibonacci//lucas.txt">Lucas number factorizations</a>
%H <a href="/index/Di#divseq">Index to divisibility sequences</a>
%t Table[DivisorSigma[0, Fibonacci[2^n]], {n, 11}] (* _Vincenzo Librandi_, Jun 18 2017 *)
%o (PARI) a(n)=numdiv(fibonacci(2^n)) \\ _Charles R Greathouse IV_, Apr 09 2012
%o (Magma) [NumberOfDivisors(Fibonacci(2^n)): n in [1..11]]; // _Vincenzo Librandi_, Jun 18 2017
%Y Cf. A000005, A000045, A063375, A074715.
%K more,nonn,hard
%O 1,2
%A _Benoit Cloitre_, Sep 03 2002
%E More terms from _Sean A. Irvine_, Jun 02 2005
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