OFFSET
0,3
COMMENTS
a(n+1)/a(n)->2 as n->infinity.
FORMULA
a(n) = 2^n * log_2 phi + O(1). - Charles R Greathouse IV, Jun 05 2013
MATHEMATICA
IntegerLength[Fibonacci[2^Range[0, 30]], 2] (* Harvey P. Dale, Apr 10 2019 *)
PROG
(Python)
TOP = 33
fib2m1 = [0]*TOP # Fibonacci(2^n-1)
fib2 = [1]*TOP # Fibonacci(2^n)
print(1, end=', ')
for n in range(1, TOP):
fib2[n] = (2*fib2m1[n-1] + fib2[n-1])*fib2[n-1]
fib2m1[n] = fib2m1[n-1]*fib2m1[n-1] + fib2[n-1]*fib2[n-1]
print(len(bin(fib2[n]))-2, end=', ')
(PARI) a(n) = #binary(fibonacci(2^n)) \\ Michel Marcus, Jun 05 2013
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Alex Ratushnyak, Aug 10 2012
STATUS
approved