

A337923


a(n) is the exponent of the highest power of 2 dividing the nth 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
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OFFSET

1,6


LINKS

Amiram Eldar, Table of n, a(n) for n = 1..10000
Tamás Lengyel, The order of the Fibonacci and Lucas numbers, The Fibonacci Quarterly, Vol. 33, No. 3 (1995), pp. 234239.


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.


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&n1).bit_length() # Chai Wah Wu, Jul 10 2022


CROSSREFS

Cf. A000045, A007814, A248174.
Cf. A001651, A016945, A017593, A073762, A184985.
Sequence in context: A167163 A005890 A104515 * A324874 A324862 A324864
Adjacent sequences: A337920 A337921 A337922 * A337924 A337925 A337926


KEYWORD

nonn


AUTHOR

Amiram Eldar, Jan 29 2021


STATUS

approved



