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A337923 a(n) is the exponent of the highest power of 2 dividing the n-th Fibonacci number.

%I A337923 #24 Jun 30 2025 22:49:38
%S A337923 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,
%T A337923 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,
%U A337923 1,0,0,5,0,0,1,0,0,3,0,0,1,0,0,4,0,0,1
%N A337923 a(n) is the exponent of the highest power of 2 dividing the n-th Fibonacci number.
%H A337923 Amiram Eldar, <a href="/A337923/b337923.txt">Table of n, a(n) for n = 1..10000</a>
%H A337923 Tamás Lengyel, <a href="https://www.fq.math.ca/Scanned/33-3/lengyel.pdf">The order of the Fibonacci and Lucas numbers</a>, The Fibonacci Quarterly, Vol. 33, No. 3 (1995), pp. 234-239.
%F A337923 a(n) = A007814(A000045(n)).
%F A337923 The following 4 formulas completely specify the sequence (Lengyel, 1995):
%F A337923 1. a(n) = 0 if n == 1 (mod 3) or n == 2 (mod 3).
%F A337923 2. a(n) = 1 if n == 3 (mod 6).
%F A337923 3. a(n) = 3 if n == 6 (mod 12).
%F A337923 4. a(n) = A007814(n) + 2 if n == 0 (mod 12).
%F A337923 a(A001651(n)) = 0.
%F A337923 a(A016945(n)) = 1.
%F A337923 a(A017593(n)) = 3.
%F A337923 a(A073762(n)) = 4.
%F A337923 The image of this function is A184985, i.e., all the nonnegative integers excluding 2.
%F A337923 Asymptotic mean: lim_{n->oo} (1/n) * Sum_{k=1..n} a(k) = 5/6.
%F A337923 a(3*n) = A090740(n), a(3*n+1) = a(3*n+2) = 0. - _Joerg Arndt_, Mar 01 2023
%e A337923 a(1) = 0 since Fibonacci(1) = 1 is odd.
%e A337923 a(6) = 3 since Fibonacci(6) = 8 = 2^3.
%e A337923 a(12) = 4 since Fibonacci(12) = 144 = 2^4 * 3^2.
%t A337923 a[n_] := IntegerExponent[Fibonacci[n], 2]; Array[a, 100]
%o A337923 (Python)
%o A337923 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
%Y A337923 Cf. A000045, A007814, A248174.
%Y A337923 Cf. A001651, A016945, A017593, A073762, A184985.
%Y A337923 Cf. A090740 (sequence without zeros).
%K A337923 nonn
%O A337923 1,6
%A A337923 _Amiram Eldar_, Jan 29 2021