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A074723 Largest power of 2 dividing F(3n) where F(k) is the k-th Fibonacci number. 3
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

if m=1 or 2 (mod 3) 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 factorisation of 2m. Also a(n) = 2*A297402(n). - Frank M Jackson, Jul 28 2018

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, A006519, A014445, A297402.

Sequence in context: A222828 A222842 A098471 * A286455 A175183 A189217

Adjacent sequences:  A074720 A074721 A074722 * A074724 A074725 A074726

KEYWORD

nonn

AUTHOR

Benoit Cloitre, Sep 04 2002

STATUS

approved

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Last modified September 19 04:39 EDT 2019. Contains 327187 sequences. (Running on oeis4.)