A074700 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)).

1, 2, 4, 8, 16, 64, 256, 1024, 8192, 131072, 1048576

Offset: 1

Author

Benoit Cloitre, Sep 03 2002

Keywords

• more
• nonn
• hard

Comments

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.

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

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

Links

• Blair Kelly, Lucas number factorizations
• Index to divisibility sequences

Crossrefs

Cf. A000005, A000045, A063375, A074715.

Programs

Magma

[NumberOfDivisors(Fibonacci(2^n)): n in [1..11]]; // Vincenzo Librandi, Jun 18 2017

Mathematica

Table[DivisorSigma[0, Fibonacci[2^n]], {n, 11}] (* Vincenzo Librandi, Jun 18 2017 *)

PARI

a(n)=numdiv(fibonacci(2^n)) \\ Charles R Greathouse IV, Apr 09 2012

Extensions

More terms from Sean A. Irvine, Jun 02 2005