A337923 a(n) is the exponent of the highest power of 2 dividing the n-th Fibonacci number.
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
Keywords
Examples
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.
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. 234-239.
Crossrefs
Programs
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Mathematica
a[n_] := IntegerExponent[Fibonacci[n], 2]; Array[a, 100]
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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
Formula
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