%I #27 Nov 27 2023 03:04:14

%S 2,8,2,16,2,8,2,32,2,8,2,16,2,8,2,64,2,8,2,16,2,8,2,32,2,8,2,16,2,8,2,

%T 128,2,8,2,16,2,8,2,32,2,8,2,16,2,8,2,64,2,8,2,16,2,8,2,32,2,8,2,16,2,

%U 8,2,256,2,8,2,16,2,8,2,32,2,8,2,16,2,8,2,64,2,8,2,16,2,8,2,32,2,8,2

%N Largest power of 2 dividing F(3n) where F(k) is the k-th Fibonacci number.

%C If m == 1 or 2 (mod 3) then F(m) is odd.

%H Robert Israel, <a href="/A074723/b074723.txt">Table of n, a(n) for n = 1..10000</a>

%F It appears that 4 never appears and : a(2k+1)=2 a(2^m*(2k+1))=2^(m+2) for k>=0 and m >=1.

%F From _Robert Israel_, Oct 10 2016: (Start)

%F a(2k+1)=2 follows from F(n+6) = 5 F(n) + 8 F(n+1) == F(n) mod 4.

%F a(2*(2k+1))=8 follows from F(n+12) = 89 F(n) + 144 F(n+1) == 9 F(n) mod 16.

%F a(2^m*(2k+1)) = 2^(m+2) for m > 2 follows from F(2n) = F(n) (2 F(n-1) + F(n)).

%F G.f. 2*x/(1-x^2) + Sum_{m>=1} 2^(m+2)*x^(2^m)/(1 - x^(2^(m+1))). (End)

%F a(n) = A006519(A014445(n)). - _Michel Marcus_, Oct 10 2016

%F As proved above, for m > 0, a(2m-1) = 2 and a(2m) = 2^(k+2) where k is the exponent of the even prime in the prime factorization of 2m. Also a(n) = 2*A297402(n). - _Frank M Jackson_, Jul 28 2018

%F Sum_{k=1..n} a(k) ~ (2*n/log(2)) * (log(n) + gamma + log(2) - 1), where gamma is Euler's constant (A001620). - _Amiram Eldar_, Nov 27 2023

%p seq(`if`(n::odd,2,2^(2+padic:-ordp(n,2))),n=1..100); # _Robert Israel_, Oct 10 2016

%t Table[2^(Length@ NestWhileList[#/2 &, Fibonacci[3 n], IntegerQ] - 2), {n, 120}] (* _Michael De Vlieger_, Oct 10 2016 *)

%t a[n_] := If[EvenQ[n], 2^(FactorInteger[n][[1]][[2]] + 2), 2]; Array[a, 90] (* _Frank M Jackson_, Jul 28 2018 *)

%o (PARI) a(n) = 2^valuation(fibonacci(3*n), 2); \\ _Michel Marcus_, Oct 10 2016

%Y Cf. A000045, A001620, A006519, A014445, A297402.

%K nonn

%O 1,1

%A _Benoit Cloitre_, Sep 04 2002