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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
1, 2, 4, 8, 16, 64, 256, 1024, 8192, 131072, 1048576 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

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

Table of n, a(n) for n=1..11.

Blair Kelly, Lucas number factorizations

Index to divisibility sequences

MATHEMATICA

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

PROG

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

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

CROSSREFS

Cf. A000005, A000045, A063375, A074715.

Sequence in context: A318151 A076086 A246907 * A230863 A322037 A013133

Adjacent sequences: A074697 A074698 A074699 * A074701 A074702 A074703

KEYWORD

more,nonn,hard

AUTHOR

Benoit Cloitre, Sep 03 2002

EXTENSIONS

More terms from Sean A. Irvine, Jun 02 2005

STATUS

approved

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Last modified February 7 22:58 EST 2023. Contains 360132 sequences. (Running on oeis4.)