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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 factorization 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 24 17:09 EDT 2020. Contains 337321 sequences. (Running on oeis4.)