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a(n) = tau(Fibonacci(24*2^n))/(24*2^n) where tau(x) is the number of divisors of x (A000005(x)).
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%I #27 Sep 08 2022 08:45:07

%S 3,7,32,144,5120,180224,3145728,3489660928

%N a(n) = tau(Fibonacci(24*2^n))/(24*2^n) where tau(x) is the number of divisors of x (A000005(x)).

%C Are terms always integers?

%p with(numtheory): with(combinat): a:=n->tau(fibonacci(24*2^n))/(24*2^n): seq(a(n),n=0..4); # _Emeric Deutsch_, Jan 30 2006

%t Table[DivisorSigma[0, Fibonacci[24 2^n]] / (24 2^n), {n, 0, 5}] (* _Vincenzo Librandi_, Sep 11 2017 *)

%o (PARI) a(n) = numdiv(fibonacci(24*2^n))/(24*2^n); \\ _Michel Marcus_, Sep 10 2017

%o (Magma) [NumberOfDivisors(Fibonacci(24*2^n))/(24*2^n): n in [0..5]]; // _Vincenzo Librandi_, Sep 11 2017

%Y Cf. A063375, A074698.

%K more,nonn

%O 0,1

%A _Benoit Cloitre_, Sep 03 2002

%E a(5) from _Eric Rowland_, Jun 18 2017

%E a(6)-a(7) from _Amiram Eldar_, Sep 03 2019 (using FactorDB)