A074723 Largest power of 2 dividing F(3n) where F(k) is the k-th Fibonacci number.

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, 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, 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

Offset

1,1

Comments

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

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Formula

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.

From Robert Israel, Oct 10 2016: (Start)

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

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

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

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

a(n) = A006519(A014445(n)). - Michel Marcus, Oct 10 2016

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

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

Maple

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

Mathematica

Table[2^(Length@ NestWhileList[#/2 &, Fibonacci[3 n], IntegerQ] - 2), {n, 120}] (* Michael De Vlieger, Oct 10 2016 *)
a[n_] := If[EvenQ[n], 2^(FactorInteger[n][[1]][[2]] + 2), 2]; Array[a, 90] (* Frank M Jackson, Jul 28 2018 *)

Prog

(PARI) a(n) = 2^valuation(fibonacci(3*n), 2); \\ Michel Marcus, Oct 10 2016

Crossrefs

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

Sequence in context: A098471 A341533 A341741 * A286455 A175183 A189217

Adjacent sequences: A074720 A074721 A074722 * A074724 A074725 A074726

Keyword

nonn

Author

Benoit Cloitre, Sep 04 2002

Status

approved