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A337923
a(n) is the exponent of the highest power of 2 dividing the n-th Fibonacci number.
2
0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 4, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 5, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 4, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 6, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 4, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 5, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 4, 0, 0, 1
OFFSET
1,6
LINKS
Tamás Lengyel, The order of the Fibonacci and Lucas numbers, The Fibonacci Quarterly, Vol. 33, No. 3 (1995), pp. 234-239.
FORMULA
a(n) = A007814(A000045(n)).
The following 4 formulas completely specify the sequence (Lengyel, 1995):
1. a(n) = 0 if n == 1 (mod 3) or n == 2 (mod 3).
2. a(n) = 1 if n == 3 (mod 6).
3. a(n) = 3 if n == 6 (mod 12).
4. a(n) = A007814(n) + 2 if n == 0 (mod 12).
a(A001651(n)) = 0.
a(A016945(n)) = 1.
a(A017593(n)) = 3.
a(A073762(n)) = 4.
The image of this function is A184985, i.e., all the nonnegative integers excluding 2.
Asymptotic mean: lim_{n->oo} (1/n) * Sum_{k=1..n} a(k) = 5/6.
a(3*n) = A090740(n), a(3*n+1) = a(3*n+2) = 0. - Joerg Arndt, Mar 01 2023
EXAMPLE
a(1) = 0 since Fibonacci(1) = 1 is odd.
a(6) = 3 since Fibonacci(6) = 8 = 2^3.
a(12) = 4 since Fibonacci(12) = 144 = 2^4 * 3^2.
MATHEMATICA
a[n_] := IntegerExponent[Fibonacci[n], 2]; Array[a, 100]
PROG
(Python)
def A337923(n): return int(not n%3)+(int(not n%6)<<1) if n%12 else 2+(~n&n-1).bit_length() # Chai Wah Wu, Jul 10 2022
CROSSREFS
Cf. A090740 (sequence without zeros).
Sequence in context: A167163 A005890 A104515 * A363880 A324874 A324862
KEYWORD
nonn
AUTHOR
Amiram Eldar, Jan 29 2021
STATUS
approved